CHAPTER FOUR TYPES OF CHEMICAL REACTIONS AND … · TYPES OF CHEMICAL REACTIONS AND SOLUTION...

61

CHAPTER FOUR

TYPES OF CHEMICAL REACTIONS AND SOLUTIONSTOICHIOMETRY

Questions

9. "Slightly soluble" refers to substances that dissolve only to a small extent. A slightly soluble salt maystill dissociate completely to ions and, hence, be a strong electrolyte. An example of such a substance

2is Mg(OH) . It is a strong electrolyte, but not very soluble. A weak electrolyte is a substance thatdoesn't dissociate completely to produce ions. A weak electrolyte may be very soluble in water, or itmay not be very soluble. Acetic acid is an example of a weak electrolyte that is very soluble in water.

10. Measure the electrical conductivity of a solution and compare it to the conductivity of a solution ofequal concentration of a strong electrolyte.

Exercises

Aqueous Solutions: Strong and Weak Electrolytes

211. a. NaBr(s) ÷ Na (aq) + Br (aq) b. MgCl (s) ÷ Mg (aq) + 2 Cl (aq)+ - 2+ -

Your drawing should show equal Your drawing should show twice the numbernumbers of Na and Br ions. of Cl ions as Mg ions.+ - - 2+

62 CHAPTER 4 SOLUTION STOICHIOMETRY62

3 3 3 4 2 4 4 4c. Al(NO ) (s) ÷ Al (aq) + 3 NO (aq) d. (NH ) SO (s) ÷ 2 NH (aq) + SO (aq)3+ - + 2-

For e-i, your drawings should show equal numbers of the cations and anions present as each salt is a1:1 salt. The ions present are listed in the following dissolution reactions.

4 4e. NaOH(s) ÷ Na (aq) + OH (aq) f. FeSO (s) ÷ Fe (aq) + SO (aq)+ - 2+ 2-

4 4 4 4g. KMnO (s) ÷ K (aq) + MnO (aq) h. HClO (aq) ÷ H (aq) + ClO (aq)+ - + -

4 2 3 2 4 2 3 2i. NH C H O (s) ÷ NH (aq) + C H O (aq)+ -

3 2 3 312. a. Ba(NO ) (aq) ÷ Ba (aq) + 2 NO (aq); Picture iv represents the Ba and NO ions present in2+ - 2+ -

3 2Ba(NO ) (aq).

b. NaCl(aq) ÷ Na (aq) + Cl (aq); Picture ii represents NaCl(aq).+ -

2 3 3 2 3c. K CO (aq) ÷ 2 K (aq) + CO (aq); Picture iii represents K CO (aq).+ 2-

4 4 4d. MgSO (aq) ÷ Mg (aq) + SO (aq); Picture i represents MgSO (aq).2+ 2-

213. CaCl (s) ÷ Ca (aq) + 2 Cl (aq)2+ -

4 4 4 3 4 314. MgSO (s) ÷ Mg (aq) + SO (aq); NH NO (s) ÷ NH (aq) + NO (aq)2+ 2- + -

Solution Concentration: Molarity

3 315. a. 5.623 g NaHCO × = 6.693 × 10 mol NaHCO-2

3M = = 0.2677 M NaHCO

2 2 7 2 2 7b. 0.1846 g K Cr O × = 6.275 × 10 mol K Cr O-4

2 2 7M = = 1.255 × 10 M K Cr O-3

CHAPTER 4 SOLUTION STOICHIOMETRY 63

c. 0.1025 g Cu × = 1.613 × 10 mol Cu = 1.613 × 10 mol Cu-3 -3 2+

M = = 8.065 × 10 M Cu-3 2+

2 5 2 516. 75.0 mL × = 1.3 mol C H OH; Molarity = = 5.2 M C H OH

217. a. CaCl (s) ÷ Ca (aq) + 2 Cl (aq); = 0.15 M; = 2(0.15) = 0.30 M2+ -

3 3 3b. Al(NO ) (s) ÷ Al (aq) + 3 NO (aq); = 0.26 M; = 3(0.26) = 0.78 M3+ -

2 2 7 2 7c. K Cr O (s) ÷ 2 K (aq) + Cr O (aq); = 2(0.25) = 0.50 M; = 0.25 M+ 2-

2 4 3 4d. Al (SO ) (s) ÷ 2 Al (aq) + 3 SO (aq)3+ 2-

= 4.0 × 10 M-3

= 6.0 × 10 M-3

18. a. = = 1.00 M

3 2 3Ca(NO ) (s) ÷ Ca (aq) + 2 NO (aq); = 1.00 M; = 2(1.00) = 2.00 M2+ -

b. = = 2.0 M

2 4 4Na SO (s) ÷ 2 Na (aq) + SO (aq); = 2(2.0) = 4.0 M; = 2.0 M+ 2-

4 4c. 5.00 g NH Cl × = 0.0935 mol NH Cl

= M

4 4NH Cl(s) ÷ NH (aq) + Cl (aq); = = 0.187 M+ -

3 4 3 4d. 1.00 g K PO × = 4.71 × 10 mol K PO-3


= = 0.0188 M

3 4 4K PO (s) ÷ 3 K (aq) + PO (aq); = 3(0.0188) = 0.0564 M; = 0.0188 M+ 3-

319. mol solute = volume (L) × molarity ; AlCl (s) ÷ Al (aq) + 3 Cl (aq) 3+ -

mol Cl = 0.1000 L × = 9.0 × 10 mol Cl- -2 -

2MgCl (s) ÷ Mg (aq) + 2 Cl (aq)2+ -

mol Cl = 0.0500 L × = 6.0 × 10 mol Cl- -2 -

NaCl(s) ÷ Na (aq) + Cl (aq)+ -

mol Cl = 0.2000 L × = 8.0 × 10 mol Cl- -2 -

3100.0 mL of 0.30 M AlCl contains the most moles of Cl ions.-

20. NaOH(s) ÷ Na (aq) + OH (aq), 2 total mol of ions (1 mol Na and 1 mol Cl ) per mol NaOH.+ - + -

0.1000 L × = 2.0 × 10 mol ions-2

2 2BaCl (s) ÷ Ba (aq) + 2 Cl (aq), 3 total mol of ions per mol BaCl .2+ -

0.0500 L × = 3.0 × 10 mol ions-2

3 4 4 3 4Na PO (s) ÷ 3 Na (aq) + PO (aq), 4 total mol of ions per mol Na PO .+ 3-

0.0750 L × = 4.50 × 10 mol ions-2

3 475.0 mL of 0.150 M Na PO contains the largest number of ions.

321. Molar mass of NaHCO = 22.99 + 1.008 + 12.01 + 3(16.00) = 84.01 g/mol


3Volume = 0.350 g NaHCO × = 0.0417 L = 41.7 mL

3 341.7 mL of 0.100 M NaHCO contains 0.350 g NaHCO .

22. Molar mass of NaOH = 22.99 + 16.00 + 1.008 = 40.00 g/mol

Mass NaOH = 0.2500 L × = 4.00 g NaOH

23. a. 2.00 L × = 20.0 g NaOH

Place 20.0 g NaOH in a 2 L volumetric flask; add water to dissolve the NaOH, and fill to the markwith water, mixing several times along the way.

b. 2.00 L × = 0.500 L

Add 500. mL of 1.00 M NaOH stock solution to a 2 L volumetric flask; fill to the mark withwater, mixing several times along the way.

2 4c. 2.00 L × = 38.8 g K CrO

2 4Similar to the solution made in part a, instead using 38.8 g K CrO .

d. 2.00 L × = 0.114 L

2 4 Similar to the solution made in part b, instead using 114 mL of the 1.75 M K CrO stock solution.

2 424. a. 1.00 L solution × = 0.50 mol H SO

2 4 2 40.50 mol H SO × = 2.8 × 10 L conc. H SO or 28 mL-2

2 4Dilute 28 mL of concentrated H SO to a total volume of 1.00 L with water.

b. We will need 0.50 mol HCl.

0.50 mol HCl × = 4.2 × 10 L = 42 mL -2

Dilute 42 mL of concentrated HCl to a final volume of 1.00 L.

2c. We need 0.50 mol NiCl .


2 2 20.50 mol NiCl × = 118.8 g NiCl C6H O . 120 g

2 2Dissolve 120 g NiCl C6H O in water, and add water until the total volume of the solution is1.00 L.

3d. 1.00 L × = 0.50 mol HNO

30.50 mol HNO × = 0.031 L = 31 mL

Dissolve 31 mL of concentrated reagent in water. Dilute to a total volume of 1.00 L.

2 3e. We need 0.50 mol Na CO .

2 3 2 30.50 mol Na CO × = 53 g Na CO

2 3 Dissolve 53 g Na CO in water, dilute to 1.00 L.

4 2 4 4 2 425. 10.8 g (NH ) SO × = 8.17 × 10 mol (NH ) SO-2

4 2 4Molarity = = 0.817 M (NH ) SO

4 2 4Moles of (NH ) SO in final solution

10.00 × 10 L × = 8.17 × 10 mol-3 -3

4 2 4Molarity of final solution = = 0.136 M (NH ) SO

4 2 4 4 4(NH ) SO (s) ÷ 2 NH (aq) + SO (aq); = 2(0.136) = 0.272 M; = 0.136 M+ 2-

2 3 2 326. mol Na CO = 0.0700 L × = 0.21 mol Na CO

2 3 3Na CO (s) ÷ 2 Na (aq) + CO (aq); mol Na = 2(0.21) = 0.42 mol+ 2- +

3 3mol NaHCO = 0.0300 L × = 0.030 mol NaHCO

3 3NaHCO (s) ÷ Na (aq) + HCO (aq); mol Na = 0.030 mol+ - +

= = = 4.5 M Na+

27. Stock solution =


100.0 × 10 L stock × = 2.00 × 10 g steroid-6 -6

This is diluted to a final volume of 100.0 mL.

= 5.94 × 10 M steroid-8

28. Stock solution:

1.584 g Mn × = 2.883 × 10 mol Mn ; = 2.883 × 10 M2+ -2 2+ -2

Solution A contains:

50.00 mL × = 1.442 × 10 mol Mn-3 2+

Molarity = = 1.442 × 10 M-3

Solution B contains:

10.0 mL × = 1.442 × 10 mol Mn-5 2+

Molarity = = 5.768 × 10 M-5

Solution C contains:

10.00 × 10 L × = 5.768 × 10 mol Mn-3 -7 2+

Molarity = = 1.154 × 10 M-6

Precipitation Reactions

29. In these reactions, soluble ionic compounds are mixed together. To predict the precipitate, switch theanions and cations in the two reactant compounds to predict possible products; then use the solubilityrules in Table 4.1 to predict if any of these possible products are insoluble (are the precipitate).

4 4a. Possible products = BaSO and NaCl; precipitate = BaSO (s)

2 3 2b. Possible products = PbCl and KNO ; precipitate = PbCl (s)

3 4 3 3 4c. Possible products = Ag PO and NaNO ; precipitate = Ag PO (s)

3 3 3d. Possible products = NaNO and Fe(OH) ; precipitate = Fe(OH) (s)

2 2 430. a. Possible products = FeCl and K SO ; Both salts are soluble so no precipitate forms.

3 3 2 3b. Possible products = Al(OH) and Ba(NO ) ; precipitate = Al(OH) (s)

4 4c. Possible products = CaSO and NaCl; precipitate =CaSO (s)

3d. Possible products = KNO and NiS; precipitate = NiS(s)


31. For the following answers, the balanced molecular equation is first, followed by the complete ionicequation, then the net ionic equation.

2 2 4 4a. BaCl (aq) + Na SO (aq) ÷ BaSO (s) + 2 NaCl(aq)

4 4Ba (aq) + 2 Cl (aq) + 2 Na (aq) + SO (aq) ÷ BaSO (s) + 2 Na (aq) + 2 Cl (aq)2+ - + 2- + -

4 4Ba (aq) + SO (aq) ÷ BaSO (s)2+ 2-

3 2 2 3b. Pb(NO ) (aq) + 2 KCl(aq) ÷ PbCl (s) + 2 KNO (aq)

3 2 3Pb (aq) + 2 NO (aq) + 2 K (aq) + 2 Cl (aq) ÷ PbCl (s) + 2 K (aq) + 2 NO (aq)2+ - + - + -

2Pb (aq) + 2 Cl (aq) ÷ PbCl (s)2+ -

3 3 4 3 4 3c. 3 AgNO (aq) + Na PO (aq) ÷ Ag PO (s) + 3 NaNO (aq)

3 4 3 4 33 Ag (aq) + 3 NO (aq) + 3 Na (aq) + PO (aq) ÷ Ag PO (s) + 3 Na (aq) + 3 NO (aq)+ - + 3- + -

4 3 43 Ag (aq) + PO (aq) ÷ Ag PO (s)+ 3-

3 3 3 3d. 3 NaOH(aq) + Fe(NO ) (aq) ÷ Fe(OH) (s) + 3 NaNO (aq)

3 3 33 Na (aq) + 3 OH (aq) + Fe (aq) + 3 NO (aq) ÷ Fe(OH) (s) + 3 Na (aq) + 3 NO (aq)+ - 3+ - + -

3Fe (aq) + 3 OH (aq) ÷ Fe(OH) (s)3+ -

32. a. No reaction occurs since all possible products are soluble salts.

3 3 2 3 3 2b. 2 Al(NO ) (aq) + 3 Ba(OH) (aq) ÷ 2 Al(OH) (s) + 3 Ba(NO ) (aq)

3 3 32 Al (aq) + 6 NO (aq) + 3 Ba (aq) + 6 OH (aq) ÷2 Al(OH) (s) + 3 Ba (aq) + 6 NO (aq)3+ - 2+ - 2+ -

3Al (aq) + 3 OH (aq) ÷ Al(OH) (s)3+ -

2 2 4 4c. CaCl (aq) + Na SO (aq) ÷ CaSO (s) + 2 NaCl(aq)

4 4Ca (aq) + 2 Cl (aq) + 2 Na (aq) + SO (aq) ÷ CaSO (s) + 2 Na (aq) + 2 Cl (aq)2+ - + 2- + -

4 4Ca (aq) + SO (aq) ÷ CaSO (s)2+ 2-

2 3 2 3d. K S(aq) + Ni(NO ) (aq) ÷ 2 KNO (aq) + NiS(s)

3 32 K (aq) + S (aq) + Ni (aq) + 2 NO (aq) ÷2 K (aq) + 2 NO (aq) + NiS(s)+ 2- 2+ - + -

Ni (aq) + S (aq) ÷ NiS(s)2+ 2-


4 233. a. When CuSO (aq) is added to Na S(aq), the precipitate that forms is CuS(s). Therefore, Na (the+

4grey spheres) and SO (the blueish-green spheres) are the spectator ions. 2-

4 2 2 4CuSO (aq) + Na S(aq) ÷ CuS(s) + Na SO (aq); Cu (aq) + S (aq) ÷ CuS(s)2+ 2-

2 2b. When CoCl (aq) is added to NaOH(aq), the precipitate that forms is Co(OH) (s). Therefore, Na+

(the grey spheres) and Cl (the green spheres) are the spectator ions.-

2 2 2CoCl (aq) + 2 NaOH(aq) ÷ Co(OH) (s) + 2 NaCl(aq); Co (aq) + 2 OH (aq) ÷ Co(OH) (s)2+ -

3c. When AgNO (aq) is added to KI(aq), the precipitate that forms is AgI(s). Therefore, K (the red+

3spheres) and NO (the blue spheres) are the spectator ions.-

3 3AgNO (aq) + KI(aq) ÷ AgI(s) + KNO (aq); Ag (aq) + I (aq) ÷ AgI(s)+ -

334. There are many acceptable choices for spectator ions. We will generally choose Na and NO as the+ -

spectator ions because sodium salts and nitrate salts are usually soluble in water.

3 3 3 3a. Fe(NO ) (aq) + 3 NaOH(aq) ÷ Fe(OH) (s) + 3 NaNO (aq)

2 3 2 2 2 3b. Hg (NO ) (aq) + 2 NaCl(aq) ÷ Hg Cl (s) + 2 NaNO (aq)

3 2 2 4 4 3c. Pb(NO ) (aq) + Na SO (aq) ÷ PbSO (s) + 2 NaNO (aq)

2 2 4 4d. BaCl (aq) + Na CrO (aq) ÷ BaCrO (s) + 2 NaCl(aq)

4 2 4 3 2 4 3 435. a. (NH ) SO (aq) + Ba(NO ) (aq) ÷ 2 NH NO (aq) + BaSO (s)

4 4Ba (aq) + SO (aq) ÷ BaSO (s)2+ 2-

3 2 2 3b. Pb(NO ) (aq) + 2 NaCl(aq) ÷ PbCl (s) + 2 NaNO (aq)

2Pb (aq) + 2 Cl (aq) ÷ PbCl (s)2+ -

c. Potassium phosphate and sodium nitrate are both soluble in water. No reaction occurs.

d. No reaction occurs since all possible products are soluble.

2 2e. CuCl (aq) + 2 NaOH(aq) ÷ Cu(OH) (s) + 2 NaCl(aq)

2Cu (aq) + 2 OH (aq) ÷ Cu(OH) (s)2+ -

3 336. a. CrCl (aq) + 3 NaOH(aq) ÷ Cr(OH) (s) + 3 NaCl(aq)

3Cr (aq) + 3 OH (aq) ÷ Cr(OH) (s)3+ -

3 4 2 3 2 3 4 3b. 2 AgNO (aq) + (NH ) CO (aq) ÷ Ag CO (s) + 2 NH NO (aq)

3 2 32 Ag (aq) + CO (aq) ÷ Ag CO (s)+ 2-


4 2 3 2 3 2 2 4c. CuSO (aq) + Hg (NO ) (aq) ÷ Cu(NO ) (aq) + Hg SO (s)

2 4 2 4Hg (aq) + SO (aq) ÷ Hg SO (s)2+ 2-

2 3d. No reaction occurs since all possible products (SrI and KNO ) are soluble.

37. Three possibilities are:

2 4 4Addition of K SO solution to give a white ppt. of PbSO . Addition of NaCl solution to give a

2 2 4 4white ppt. of PbCl . Addition of K CrO solution to give a bright yellow ppt. of PbCrO .

2 4 238. Since no precipitates formed upon addition of NaCl or Na SO , we can conclude that Hg and Ba2+ 2+

2 2 4are not present in the sample since Hg Cl and BaSO are insoluble salts. However, Mn may be2+

2 4present since Mn does not form a precipitate with either NaCl or Na SO . Since a precipitate formed2+

2with NaOH, the solution must contain Mn because it forms a precipitate with OH [Mn(OH) (s)].2+ -

3 2 4 2 4 339. 2 AgNO (aq) + Na CrO (aq) ÷ Ag CrO (s) + 2 NaNO (aq)

2 40.0750 L × = 0.607 g Na CrO

3 4 3 2 3 4 2 340. 2 Na PO (aq) + 3 Pb(NO ) (aq) ÷ Pb (PO ) (s) + 6 NaNO (aq)

0.1500 L × = 0.250 L

3 4 = 250. mL Na PO

3 3 3 341. A1(NO ) (aq) + 3 KOH(aq) ÷ Al(OH) (s) + 3 KNO (aq)

3 30.0500 L × = 0.0100 mol Al(NO )

0.2000 L × = 0.0200 mol KOH

3 3 From the balanced equation, 3 mol of KOH are required to react with 1 mol of Al(NO ) (3:1 mol

3 3ratio). The actual KOH to Al(NO ) mol ratio present is 0.0200/0.0100 = 2 (2:1). Since the actualmol ratio present is less than the required mol ratio, KOH is the limiting reagent.

30.0200 mol KOH × = 0.520 g Al(OH)

2 2 4 3 4 342. The balanced equation is: 3 BaCl (aq) + Fe (SO ) (aq) ÷ 3 BaSO (s) + 2 FeCl (aq)

2 2100.0 mL BaCl × = 1.00 × 10 mol BaCl-2


2 4 3 2 4 3100.0 mL Fe (SO ) × = 1.00 × 10 mol Fe (SO )-2

2 2 4 3The required mol BaCl to mol Fe (SO ) ratio from the balanced reaction is 3:1. The actual mol ratio

2is 0.0100/0.0100 = 1 (1:1). This is well below the required mol ratio, so BaCl is the limiting reagent.

2 40.0100 mol BaCl × = 2.33 g BaSO

3 2 3 243. 2 AgNO (aq) + CaCl (aq) ÷ 2 AgCl(s) + Ca(NO ) (aq)

3 3mol AgNO = 0.1000 L × = 0.020 mol AgNO

2 2mol CaCl = 0.1000 L × = 0.015 mol CaCl

3 2The required mol AgNO to mol CaCl ratio is 2:1 (from the balanced equation). The actual mol ratio

3present is 0.020/0.015 = 1.3 (1.3:1). Therefore, AgNO is the limiting reagent.

3mass AgCl = 0.020 mol AgNO × = 2.9 g AgCl

The net ionic equation is: Ag (aq) + Cl (aq) ÷ AgCl(s). The ions remaining in solution are the+ -

3unreacted Cl ions and the spectator ions, NO and Ca (all Ag is used up in forming AgCl). The- - 2+ +

mol of each ion present initially (before reaction) can be easily determined from the mol of each

3 3 2reactant. 0.020 mol AgNO dissolves to form 0.020 mol Ag and 0.020 mol NO . 0.015 mol CaCl+ -

dissolves to form 0.015 mol Ca and 2(0.015) = 0.030 mol Cl .2+ -

mol unreacted Cl = 0.030 mol Cl initially - 0.020 mol Cl reacted = 0.010 mol Cl unreacted- - - -

= 0.050 M Cl-

The molarity of the spectator ions are:

3 = 0.10 M NO ; = 0.075 M Ca- 2+

3 2 2 344. a. Cu(NO ) (aq) + 2 KOH(aq) ÷ Cu(OH) (s) + 2 KNO (aq)

3 2Solution A contains 2.00 L × 2.00 mol/L = 4.00 mol Cu(NO ) and solution B contains 2.00 L × 3.00 mol/L = 6.00 mol KOH. Lets assume in our picture that we have 4 formula units

3 2 3of Cu(NO ) (4 Cu ions and 8 NO ions) and 6 formula units of KOH (6 K ions and 6 OH2+ - + -

ions). With 4 Cu ions and 6 OH ions present, then OH is limiting. One Cu ion remains as 32+ - - 2+


2Cu(OH) (s) formula units form as precipitate. The following drawing summarizes the ions that

3remain in solution and the relative amount of precipitate that forms. Note that K and NO ions+ -

1 2are spectator ions. In the drawing, V is the volume of solution A or B and V is the volume of

2 1the combined solutions with V = 2 V . The drawing exaggerates the amount of precipitate thatwould actually form.

b. The spectator ion concentrations will be one-half of the original spectator ion concentrations in

the individual beakers because the volume was doubled. Or using moles, =

= 1.50 M and = 2.00 M. The concentration of OH ions will be zero-

since OH is the limiting reagent. From the drawing, the number of Cu ions will decrease by a- 2+

factor of four as the precipitate forms. Since the volume of solution doubled, the concentration

of Cu ions will decrease by a factor of eight after the two beakers are mixed: 2+

= 0.250 M

Alternately, one could certainly use moles to solve for :

mol Cu reacted = 2.00 L × = 3.00 mol Cu reacted2+ 2+

mol Cu present initially = 2.00 L × = 4.00 mol Cu present initially2+ 2+

excess Cu present after reaction = 4.00 mol - 3.00 mol = 1.00 mol Cu excess2+ 2+

= 0.250 M

2mass of precipitate = 6.00 mol KOH × = 293 g Cu(OH)

Acid-Base Reactions

45. All the bases in this problem are ionic compounds containing OH . The acids are either strong or weak-

electrolytes. The best way to determine if an acid is a strong or weak electrolyte is to memorize all thestrong electrolytes (strong acids). Any other acid you encounter that is not a strong acid will be a weak


electrolyte (a weak acid) and the formula should be left unaltered in the complete ionic and net ionic

3 4 2 4equations. The strong acids to recognize are HCl, HBr, HI, HNO , HClO and H SO . For theanswers below, the order of the equations are molecular, complete ionic and net ionic.

4 2 2 4 2a. 2 HClO (aq) + Mg(OH )(s) ÷ 2 H O(l) + Mg(ClO ) (aq)

4 2 2 42 H (aq) + 2 ClO (aq) + Mg(OH) (s) ÷ 2 H O(l) + Mg (aq) + 2 ClO (aq)+ - 2+ -

2 22 H (aq) + Mg(OH) (s) ÷ 2 H O(l) + Mg (aq)+ 2+

2b. HCN(aq) + NaOH(aq) ÷ H O(l) + NaCN(aq)

2HCN(aq) + Na (aq) + OH (aq) ÷ H O(l) + Na (aq) + CN (aq)+ - + -

2HCN(aq) + OH (aq) ÷ H O(l) + CN (aq)- -

2c. HCl(aq) + NaOH(aq) ÷ H O(l) + NaCl(aq)

2H (aq) + Cl (aq) + Na (aq) + OH (aq) ÷ H O(l) + Na (aq) + Cl (aq)+ - + - + -

2H (aq) + OH (aq) ÷ H O(l)+ -

3 3 2 3 346. a. 3 HNO (aq) + Al(OH) (s) ÷ 3 H O(l) + Al(NO ) (aq)

3 3 2 33 H (aq) + 3 NO (aq) + Al(OH) (s) ÷ 3 H O(l) + Al (aq) + 3 NO (aq)+ - 3+ -

3 23 H (aq) + Al(OH) (s) ÷ 3 H O(l) + Al (aq)+ 3+

2 3 2 2 2 3 2b. HC H O (aq) + KOH(aq) ÷ H O(l) + KC H O (aq)

2 3 2 2 2 3 2HC H O (aq) + K (aq) + OH (aq) ÷ H O(l) + K (aq) + C H O (aq)+ - + -

2 3 2 2 2 3 2HC H O (aq) + OH (aq) ÷ H O(l) + C H O (aq)- -

2 2 2c. Ca(OH) (aq) + 2 HCl(aq) ÷ 2 H O(l) + CaCl (aq)

2Ca (aq) + 2 OH (aq) + 2 H (aq) + 2 Cl (aq) ÷ 2 H O(l) + Ca (aq) + 2 Cl (aq)2+ - + - 2+ -

2 22 H (aq) + 2 OH (aq) ÷ 2 H O(l) or H (aq) + OH (aq) ÷ H O(l)+ - + -

47. All the acids in this problem are strong electrolytes. The acids to recognize as strong electrolytes are

3 4 2 4HCl, HBr, HI, HNO , HClO and H SO .

3 2 3a. KOH(aq) + HNO (aq) ÷ H O(l) + KNO (aq)

3 2 3K (aq) + OH (aq) + H (aq) + NO (aq) ÷ H O(l) + K (aq) + NO (aq)+ - + - + -

2OH (aq) + H (aq) ÷ H O(l)- +


2 2 2b. Ba(OH) (aq) + 2 HCl(aq) ÷ 2 H O(l) + BaCl (aq)

2Ba (aq) + 2 OH (aq) + 2 H (aq) + 2 Cl (aq) ÷ 2 H O(l) + Ba (aq) + 2 Cl (aq)2+ - + - 2+ -

2 22 OH (aq) + 2 H (aq) ÷ 2 H O(l) or OH (aq) + H (aq) ÷ H O(l)- + - +

4 3 2 4 3c. 3 HClO (aq) + Fe(OH) (s) ÷ 3 H O(l) + Fe(ClO ) (aq)

4 3 2 43 H (aq) + 3 ClO (aq) + Fe(OH) (s) ÷ 3 H O(l) + Fe (aq) + 3 ClO (aq)+ - 3+ -

3 23 H (aq) + Fe(OH) (s) ÷ 3 H O(l) + Fe (aq)+ 3+

248. a. AgOH(s) + HBr(aq) ÷ AgBr(s) + H O(l)

2AgOH(s) + H (aq) + Br (aq) ÷ AgBr(s) + H O(l)+ -

2AgOH(s) + H (aq) + Br (aq) ÷ AgBr(s) + H O(l)+ -

2 2 2b. Sr(OH) (aq) + 2 HI(aq) ÷ 2 H O(l) + SrI (aq)

2Sr (aq) + 2 OH (aq) + 2 H (aq) + 2 I (aq) ÷ 2 H O(l) + Sr (aq) + 2 I (aq)2+ - + - 2+ -

2 22 OH (aq) + 2 H (aq) ÷ 2 H O(l) or OH (aq) + H (aq) ÷ H O(l)- + - +

3 3 2 3 3c. Cr(OH) (s) + 3 HNO (aq) ÷ 3 H O(l) + Cr(NO ) (aq)

3 3 2 3Cr(OH) (s) + 3 H (aq) + 3 NO (aq) ÷ 3 H O(l) + Cr (aq) + 3 NO (aq)+ - 3+ -

3 2Cr(OH) (s) + 3 H (aq) ÷ 3 H O(l) + Cr (aq)+ 3+

49. If we begin with 50.00 mL of 0.200 M NaOH, then:

50.00 × 10 L × = 1.00 × 10 mol NaOH is to be neutralized.-3 -2

2a. NaOH(aq) + HCl(aq) ÷ NaCl(aq) + H O(l)

1.00 × 10 mol NaOH × = 0.100 L or 100. mL-2

3 2 3b. HNO (aq) + NaOH(aq) ÷ H O(l) + NaNO (aq)

1.00 × 10 mol NaOH × = 6.67 × 10 L or 66.7 mL-2 -2

2 3 2 2 2 3 2c. HC H O (aq) + NaOH(aq) ÷ H O(l) + NaC H O (aq)


1.00 × 10 mol NaOH × = 5.00 × 10 L or 50.0 mL-2 -2

50. We begin with 25.00 mL of 0.200 M HCl or 25.00 × 10 L × 0.200 mol/L = 5.00 × 10 mol HCl.-3 -3

2a. HCl(aq) + NaOH(aq) ÷ H O(l) + NaCl(aq)

5.00 × 10 mol HCl × = 5.00 × 10 L or 50.0 mL-3 -2

2 2 2b. 2 HCl(aq) + Ba(OH) (aq) ÷ 2 H O(l) + BaCl (aq)

5.00 × 10 mol HCl × = 5.00 × 10 L = 50.0 mL-3 -2

2c. HCl(aq) + KOH(aq) ÷ H O(l) + KCl(aq)

5.00 × 10 mol HCl × = 2.00 × 10 L or 20.0 mL-3 -2

3 3 251. HNO (aq) + NaOH(aq) ÷ NaNO (aq) + H O(l)

15.0 g NaOH × = 0.375 mol NaOH

30.1500 L × = 0.0375 mol HNO

3 3We have added more moles of NaOH than mol of HNO present. Since NaOH and HNO react in a1:1 mol ratio, NaOH is in excess and the solution will be basic. The ions present after reaction will

3be the excess OH ions and the spectator ions, Na and NO . The moles of ions present initially are:- + -

mol NaOH = mol Na = mol OH = 0.375 mol+ -

3 3mol HNO = mol H = mol NO = 0.0375 mol+ -

2The net ionic reaction occurring is: H (aq) + OH (aq) ÷ H O(l)+ -

The mol of excess OH remaining after reaction will be the initial mol of OH minus the amount of OH- - -

neutralized by reaction with H :+

mol excess OH = 0.375 mol - 0.0375 mol = 0.338 mol OH excess- -

The concentration of ions present is:

= 2.25 M OH-


3 = 0.250 M NO ; = 2.50 M Na- +

2 2 2 252. Ba(OH) (aq) + 2 HCl(aq) ÷ BaCl (aq) + 2 H O(l); H (aq) + OH (aq) ÷ H O(l)+ -

75.0 × 10 L × = 1.88 × 10 mol HCl = 1.88 × 10 mol H + 1.88 × 10 mol Cl-3 -2 -2 + -2 -

2225.0 × 10 L × = 1.24 × 10 mol Ba(OH) = 1.24 × 10 mol Ba-3 -2 -2 2+

+ 2.48 × 10 mol OH-2 -

The net ionic equation requires a 1:1 mol ratio between OH and H . The actual mol OH to mol H- + - +

ratio is greater than 1:1 so OH is in excess.-

Since 1.88 × 10 mol OH will be neutralized by the H , we have (2.48 - 1.88) × 10 = -2 - + -2

0.60 × 10 mol OH remaining in excess.-2 -

= 2.0 × 10 M OH-2 -

253. HCl(aq) + NaOH(aq) ÷ H O(l) + NaCl(aq)

24.16 × 10 L NaOH × = 2.56 × 10 mol HCl-3 -3

Molarity of HCl = = 0.102 M HCl

3 2 2 3 254. 2 HNO (aq) + Ca(OH) (aq) ÷ 2 H O(l) + Ca(NO ) (aq)

335.00 × 10 L HNO × -3

2 = 0.0438 L = 43.8 mL Ca(OH)

255. Since KHP is a monoprotic acid, the reaction is: NaOH(aq) + KHP(aq) ÷ H O(l) + NaKP(aq)

Mass KHP = 0.02046 L NaOH × = 0.4178 g KHP

256. NaOH(aq) + KHP(aq) ÷ NaKP(aq) + H O(l)

0.1082 g KHP × = 5.298 × 10 mol NaOH-4


There are 5.298 × 10 mol of sodium hydroxide in 34.67 mL of solution. Therefore, the concentration-4

of sodium hydroxide is:

= 1.528 × 10 M NaOH-2

Oxidation-Reduction Reactions

57. Apply rules in Table 4.2.

4 4a. KMnO is composed of K and MnO ions. Assign oxygen a value of -2, which gives manganese+ -

4a +7 oxidation state since the sum of oxidation states for all atoms in MnO must equal the -1-

4charge on MnO . K, +1; O, -2; Mn, +7.-

b. Assign O a -2 oxidation state, which gives nickel a +4 oxidation state. Ni, +4; O, -2.

4 6 6 6c. K Fe(CN) is composed of K cations and Fe(CN) anions. Fe(CN) is composed of iron and+ 4- 4-

CN anions. For an overall anion charge of -4, iron must have a +2 oxidation state.-

4 2 4 4 4d. (NH ) HPO is made of NH cations and HPO anions. Assign +1 as the oxidation state of H+ 2-

4and -2 as the oxidation state of O. In NH , x + 4(+1) = +1, x = -3 = oxidation state of N. In+

4HPO , +1 + y + 4(-2) = -2, y = +5 = oxidation state of P.2-

e. O, -2; P, +3 f. O, -2; Fe, + 8/3

g. O, -2; F, -1; Xe, +6 h. F, -1; S, +4

i. O, -2; C, +2 j. Na, +1; O, -2; C, +3

258. a. UO : O, -2; For U, x + 2(-2) = +2, x = +62+

2 3b. As O : O, -2; For As, 2(x) + 3(-2) = 0, x = +3

3c. NaBiO : Na, +1; O, -2; For Bi, +1 + x + 3(-2) = 0, x = +5

4d. As : As, 0

2e. HAsO : assign H = +1 and O = -2; For As, +1 + x + 2(-2) = 0; x = +3

2 2 7 2 7f. Mg P O : Composed of Mg ions and P O ions. Oxidation states are:2+ 4-

Mg, +2; O, -2; P, +5

2 2 3 2 3g. Na S O : Composed of Na ions and S O ions. Na, +1; O, -2; S, +2+ 2-

2 2h. Hg Cl : Hg, +1; Cl, -1

3 2 3i. Ca(NO ) : Composed of Ca ions and NO ions. Ca, +2; O, -2; N, +52+ -


59. a. HBr: H, +1; Br, -1

b. HOBr: H, +1; O, -2; For Br, +1 + 1(-2) + x = 0, x = +1

2c. Br : Br, 0

4d. HBrO : H, +1; O, -2; For Br, +1 + 4(-2) + x = 0, x = +7

3e. BrF : F, -1; For Br, x + 3(-1) = 0, x = +3

60. a. -3 b. -3 c. 2(x) + 4(+1) = 0, x = -2d. +2 e. +1 f. +4g. +3 h. +5 i. 0

61. To determine if the reaction is an oxidation-reduction reaction, assign oxidation numbers. If theoxidation numbers change for some elements, the reaction is a redox reaction. If the oxidationnumbers do not change, the reaction is not a redox reaction. In redox reactions, the species oxidized(called the reducing agent) shows an increase in oxidation numbers and the species reduced (called theoxidizing agent) shows a decrease in oxidation numbers.

Redox? Oxidizing Reducing Substance Substance Agent Agent Oxidized Reduced

2 4 4 2a. Yes O CH CH (C) O (O)b. Yes HCl Zn Zn HCl (H)c. No - - - -

3 3d. Yes O NO NO (N) O (O)

2 2 2 2 2 2 2 2e. Yes H O H O H O (O) H O (O)f. Yes CuCl CuCl CuCl (Cu) CuCl (Cu)

In c, no oxidation numbers change from reactants to products.

62. Redox? Oxidizing Reducing Substance Substance Agent Agent Oxidized Reduced

a. Yes Ag Cu Cu Ag+ +

b. No - - - -c. No - - - -

4 4d. Yes SiCl Mg Mg SiCl (Si)e. No - - - -

In b, c, and e, no oxidation numbers change.


63. Use the method of half-reactions described in Section 4.10 of the text to balance these redox reactions.The first step always is to separate the reaction into the two half-reactions, then balance each half-reaction separately.

2a. Zn ÷ Zn + 2 e 2e + 2 HCl ÷ H + 2 Cl2+ - - -

2Adding the two balanced half-reactions, Zn(s) + 2 HCl(aq) ÷ H (g) + Zn (aq) + 2 Cl (aq)2+ -

3b. 3 I ÷ I + 2e ClO ÷ Cl- - - - -

22e + 2H + ClO ÷ Cl + H O- + - -

Adding the two balanced half-reactions so electrons cancel:

3 23 I (aq) + 2 H (aq) + ClO (aq) ÷ I (aq) + Cl (aq) + H O(l)- + - - -

2 3 3 4 3 2c. As O ÷ H AsO NO ÷ NO + 2 H O-

2 3 3 4 3 2As O ÷ 2 H AsO 4 H + NO ÷ NO + 2 H O+ -

3 2Left 3 - O; Right 8 - O (3 e + 4 H + NO ÷ NO + 2 H O) × 4- + -

Right hand side has 5 extra O.

2Balance the oxygen atoms first using H O, then balance H using H , and finally balance charge+

using electrons.

2 2 3 3 4(5 H O + As O ÷ 2 H AsO + 4 H + 4 e ) × 3+ -

Common factor is a transfer of 12 e . Add half-reactions so electrons cancel.-

3 212 e + 16 H + 4 NO ÷ 4 NO + 8 H O- + -

2 2 3 3 4 15 H O + 3 As O ÷ 6 H AsO + 12 H + 12 e+ -

2 2 3 3 3 4 7 H O(l) + 4 H (aq) + 3 As O (s) + 4 NO (aq) ÷ 4 NO(g) + 6 H AsO (aq)+ -

2 4 2d. (2 Br ÷ Br + 2 e ) × 5 MnO ÷ Mn + 4 H O- - - 2+

4 2 (5 e + 8 H + MnO ÷ Mn + 4 H O) × 2- + - 2+

Common factor is a transfer of 10 e .-

2 10 Br ÷ 5 Br + 10 e- -

4 2 10 e + 16 H + 2 MnO ÷ 2 Mn + 8 H O- + - 2+

4 2 216 H (aq) + 2 MnO (aq) + 10 Br (aq) ÷ 5 Br (l) + 2 Mn (aq) + 8 H O(l)+ - - 2+

3 2 2 7e. CH OH ÷ CH O Cr O ÷ Cr2- 3+

3 2 2 7 2(CH OH ÷ CH O + 2 H + 2 e ) × 3 14 H + Cr O ÷ 2 Cr + 7 H O+ - + 2- 3+

2 7 26 e + 14 H + Cr O ÷ 2 Cr + 7 H O- + 2- 3+



3 2 3 CH OH ÷ 3 CH O + 6 H + 6 e+ -

2 7 2 6 e + 14 H + Cr O ÷ 2 Cr + 7 H O- + 2- 3+

3 2 7 2 2 8 H (aq) + 3 CH OH(aq) + Cr O (aq) ÷ 2 Cr (aq) + 3 CH O(aq) + 7 H O(l)+ 2- 3+

3 264. a. (Cu ÷ Cu + 2 e ) × 3 NO ÷ NO + 2 H O2+ - -

3 2(3 e + 4 H + NO ÷ NO + 2 H O) × 2- + -

Adding the two balanced half-reactions so electrons cancel:

3 Cu ÷ 3 Cu + 6 e 2+ -

3 2 6 e + 8 H + 2 NO ÷ 2 NO + 4 H O- + -

3 23 Cu(s) + 8 H (aq) + 2 NO (aq) ÷ 3 Cu (aq) + 2 NO(g) + 4 H O(l)+ - 2+

2 2 7 2b. (2 Cl ÷ Cl + 2 e ) × 3 Cr O ÷ 2 Cr + 7 H O- - 2- 3+

2 7 26 e + 14 H + Cr O ÷ 2 Cr + 7 H O- + 2- 3+

Add the two half-reactions with six electrons transferred:

2 6 Cl ÷ 3 Cl + 6 e- -

2 7 2 6 e + 14 H + Cr O ÷ 2 Cr + 7 H O- + 2- 3+

2 7 2 214 H (aq) + Cr O (aq) + 6 Cl (aq) ÷ 3 Cl (g) + 2 Cr (aq) + 7 H O(l)+ 2- - 3+

4 2 4c. Pb ÷ PbSO PbO ÷ PbSO

2 4 4 2 2 4 4 2Pb + H SO ÷ PbSO + 2 H PbO + H SO ÷ PbSO + 2 H O+

2 4 4 2 2 4 4 2Pb + H SO ÷ PbSO + 2 H + 2 e 2 e + 2 H + PbO + H SO ÷ PbSO + 2 H O+ - - +

Add the two half-reactions with two electrons transferred:

2 2 4 4 2 2 e + 2 H + PbO + H SO ÷ PbSO + 2 H O- +

2 4 4 Pb + H SO ÷ PbSO + 2 H + 2 e+ -

2 4 2 4 2Pb(s) + 2 H SO (aq) + PbO (s) ÷ 2 PbSO (s) + 2 H O(l)

This is the reaction that occurs in an automobile lead-storage battery.

4d. Mn ÷ MnO2+ -

2 4(4 H O + Mn ÷ MnO + 8 H + 5 e ) × 22+ - + -


3 NaBiO ÷ Bi + Na3+ +

3 26 H + NaBiO ÷ Bi + Na + 3 H O+ 3+ +

3 2 (2 e + 6 H + NaBiO ÷ Bi + Na + 3 H O) × 5- + 3+ +

2 4 8 H O + 2 Mn ÷ 2 MnO + 16 H + 10 e2+ - + -

3 2 10 e + 30 H + 5 NaBiO ÷ 5 Bi + 5 Na + 15 H O- + 3+ +

2 3 4 28 H O + 30 H + 2 Mn + 5 NaBiO ÷ 2 MnO + 5 Bi + 5 Na + 15 H O + 16 H+ 2+ - 3+ + +

Simplifying :

3 4 214 H (aq) + 2 Mn (aq) + 5 NaBiO (s) ÷ 2 MnO (aq) + 5 Bi (aq) + 5 Na (aq) + 7 H O(l)+ 2+ - 3+ +

3 4 3e. H AsO ÷ AsH (Zn ÷ Zn + 2 e ) × 42+ -

3 4 3 2 H AsO ÷ AsH + 4 H O

3 4 3 28 e + 8 H + H AsO ÷ AsH + 4 H O- +

3 4 3 2 8 e + 8 H + H AsO ÷ AsH + 4 H O- +

4 Zn ÷ 4 Zn + 8 e2+ -

3 4 3 28 H (aq) + H AsO (aq) + 4 Zn(s) ÷ 4 Zn (aq) + AsH (g) + 4 H O(l)+ 2+

65. Use the same method as with acidic solutions. After the final balanced equation, convert H to OH+ -

2as described in section 4.10 of the text. The extra step involves converting H into H O by adding+

equal moles of OH to each side of the reaction. This converts the reaction to a basic solution while-

keeping it balanced.

4 4 2a. Al ÷ Al(OH) MnO ÷ MnO- -

2 4 4 2 2 4 H O + Al ÷ Al(OH) + 4 H 3 e + 4 H + MnO ÷ MnO + 2 H O- + - + -

2 4 4 H O + Al ÷ Al(OH) + 4 H + 3 e- + -

2 4 4 H O + Al ÷ Al(OH) + 4 H + 3 e- + -

4 2 2 3 e + 4 H + MnO ÷ MnO + 2 H O- + -

2 4 4 22 H O(l) + Al(s) + MnO (aq) ÷ Al(OH) (aq) + MnO (s)- -

Since H doesn’t appear in the final balanced reaction, we are done.+

2 2b. Cl ÷ Cl Cl ÷ OCl- -

2 2 22 e + Cl ÷ 2 Cl 2 H O + Cl ÷ 2 OCl + 4 H + 2 e- - - + -

2 2 e + Cl ÷ 2 Cl- -

2 2 2 H O + Cl ÷ 2 OCl + 4 H + 2 e- + -

2 2 2 H O + 2 Cl ÷ 2 Cl + 2 OCl + 4 H- - +

Now convert to a basic solution. Add 4 OH to both sides of the equation. The 4 OH will react- -

2with the 4 H on the product side to give 4 H O. After this step, cancel identical species on both+


2 2 2sides (2 H O). Applying these steps gives: 4 OH + 2 Cl ÷ 2 Cl + 2 OCl + 2 H O, which can- - -

be further simplified to:

2 22 OH (aq) + Cl (g) ÷ Cl (aq) + OCl (aq) + H O(l)- - -

2 3 2c. NO ÷ NH Al ÷ AlO- -

2 3 2 2 26 e + 7 H + NO ÷ NH + 2 H O (2 H O + Al ÷ AlO + 4 H + 3 e ) × 2- + - - + -


2 3 2 6e + 7 H + NO ÷ NH + 2 H O- + -

2 2 4 H O + 2 Al ÷ 2 AlO + 8 H + 6 e- + -

2 2 3 2OH + 2 H O + NO + 2 Al ÷ NH + 2 AlO + H + OH- - - + -

2 2 3 2Reducing gives: OH (aq) + H O(l) + NO (aq) + 2 Al(s) ÷ NH (g) + 2 AlO (aq)- - -

3 4 366. a. Cr ÷ Cr(OH) CrO ÷ Cr(OH)2-

2 3 4 3 23 H O + Cr ÷ Cr(OH) + 3 H + 3 e 3 e + 5 H + CrO ÷ Cr(OH) + H O+ - - + 2-

2 33 H O + Cr ÷ Cr(OH) + 3 H + 3 e+ -

4 3 23 e + 5 H + CrO ÷ Cr(OH) + H O- + 2-

2 4 32 OH + 2 H + 2 H O + Cr + CrO ÷ 2 Cr(OH) + 2 OH- + 2- -

Two OH were added above to each side to convert to a basic solution. The two OH react with- -

2the 2 H on the reactant side to produce 2 H O. The overall balanced equation is:+

2 4 34 H O(l) + Cr(s) + CrO (aq) ÷ 2 Cr(OH) (s) + 2 OH (aq)2- -

4b. S ÷ S MnO ÷ MnS2- -

4(S ÷ S + 2 e ) × 5 MnO + S ÷ MnS2- - - 2-

4 2 ( 5 e + 8 H + MnO + S ÷ MnS + 4 H O) × 2- + - 2-

Common factor is a transfer of 10 e . -

5 S ÷ 5 S + 10 e2- -

4 2 10 e + 16 H + 2 MnO + 2 S ÷ 2 MnS + 8 H O- + - 2-

4 216 OH + 16 H + 7 S + 2 MnO ÷ 5 S + 2 MnS + 8 H O + 16 OH- + 2- - -

2 4 216 H O + 7 S + 2 MnO ÷ 5 S + 2 MnS + 8 H O + 16 OH2- - -

2 4Reducing gives: 8 H O(l) + 7 S (aq) + 2 MnO (aq) ÷ 5 S(s) + 2 MnS(s) + 16 OH (aq)2- - -


c. CN ÷ CNO- -

2 (H O + CN ÷ CNO + 2 H + 2 e ) × 3- - + -

4 2MnO ÷ MnO-

4 2 2(3 e + 4 H + MnO ÷ MnO + 2 H O) × 2- + -

Common factor is a transfer of 6 electrons.

2 3 H O + 3 CN ÷ 3 CNO + 6 H + 6 e- - + -

4 2 2 6 e + 8 H + 2 MnO ÷ 2 MnO + 4 H O- + -

4 2 22 OH + 2 H + 3 CN + 2 MnO ÷ 3 CNO + 2 MnO + H O + 2 OH- + - - - -

Reducing gives:

2 4 2H O(l) + 3 CN (aq) + 2 MnO (aq) ÷ 3 CNO (aq) + 2 MnO (s) + 2 OH (aq)- - - -

2 4 2 2 4 2 2 267. NaCl + H SO + MnO ÷ Na SO + MnCl + Cl + H O

We could balance this reaction by the half-reaction method or by inspection. Let’s try inspection. Tobalance Cl , we need 4 NaCl:-

2 4 2 2 4 2 2 24 NaCl + H SO + MnO ÷ Na SO + MnCl + Cl + H O

4Balance the Na and SO ions next:+ 2-

2 4 2 2 4 2 2 24 NaCl + 2 H SO + MnO ÷ 2 Na SO + MnCl + Cl + H O

2On the left side: 4-H and 10-O; On the right side: 8-O not counting H O

2We need 2 H O on the right side to balance H and O:

2 4 2 2 4 2 2 24 NaCl(aq) + 2 H SO (aq) + MnO (s) ÷ 2 Na SO (aq) + MnCl (aq) + Cl (g) + 2 H O(l)

3 468. Au + HNO + HCl ÷ AuCl + NO-

3 4Only deal with ions that are reacting (omit H ): Au + NO + Cl ÷ AuCl + NO+ - - -

The balanced half-reactions are:

4 3 2Au + 4 Cl ÷ AuCl + 3 e 3 e + 4 H + NO ÷ NO + 2 H O- - - - + -

Adding the two balanced half-reactions:

3 4 2Au(s) + 4 Cl (aq) + 4 H (aq) + NO (aq) ÷ AuCl (aq) + NO(g) + 2 H O(l)- + - -


Additional Exercises

69. 0.100 g Ca × = 4.99 × 10 mol OH-3 -

Molarity = = 1.11 × 10 M OH-2 -

2 2 270. mol CaCl present = 0.230 L CaCl × = 6.33 × 10 mol CaCl-2

2The volume of CaCl solution after evaporation is:

2 26.33 × 10 mol CaCl × = 5.75 × 10 L = 57.5 mL CaCl-2 -2

2 2Volume H O evaporated = 230. mL - 57.5 mL = 173 mL H O evaporated71. There are other possible correct choices for the following answers. We have listed only three possible

reactants in each case.

3 3 2 2 3 2a. AgNO , Pb(NO ) , and Hg (NO ) would form precipitates with the Cl ion.-

2Ag (aq) + Cl (aq) ÷ AgCl(s); Pb (aq) + 2 Cl (aq) ÷ PbCl (s); + - 2+ -

2 2 2Hg (aq) + 2 Cl (aq) ÷ Hg Cl (s)2+ -

2 4 2 3 3 4b. Na SO , Na CO , and Na PO would form precipitates with the Ca ion.2+

4 4 3 3Ca (aq) + SO (aq) ÷ CaSO (s); Ca + CO (aq) ÷ CaCO (s)2+ 2- 2+ 2-

4 3 4 23 Ca (aq) + 2 PO (aq)÷ Ca (PO ) (s)2+ 3-

2 2 3c. NaOH, Na S, and Na CO would form precipitates with the Fe ion.3+

3 2 3Fe (aq) + 3 OH (aq) ÷ Fe(OH) (s); 2 Fe (aq) + 3 S (aq) ÷ Fe S (s);3+ - 3+ 2-

3 2 3 32 Fe (aq) + 3 CO (aq) ÷ Fe (CO ) (s)3+ 2-

2 3 2 3 2 4d. BaCl , Pb(NO ) , and Ca(NO ) would form precipitates with the SO ion.2-

4 4 4 4Ba (aq) + SO (aq) ÷ BaSO (s); Pb (aq) + SO (aq) ÷ PbSO (s);2+ 2- 2+ 2-

4 4Ca (aq) + SO (aq) ÷ CaSO (s)2+ 2-

2 4 2e. Na SO , NaCl, and NaI would form precipitates with the Hg ion.2+

2 4 2 4 2 2 2Hg (aq) + SO (aq) ÷ Hg SO (s); Hg (aq) + 2 Cl (aq) ÷ Hg Cl (s);2+ 2- 2+ -

2 2 2Hg (aq) + 2 I (aq) ÷ Hg I (s)2+ -

2 4 3 4f. NaBr, Na CrO , and Na PO would form precipitates with the Ag ion.+

4 2 4Ag (aq) + Br (aq) ÷ AgBr(s); 2 Ag (aq) + CrO (aq) ÷ Ag CrO (s); + - + 2-

4 3 43 Ag (aq) + PO (aq) ÷ Ag PO (s)+ 3-

2 3 3 272. a. MgCl (aq) + 2 AgNO (aq) ÷ 2 AgCl(s) + Mg(NO ) (aq)


20.641 g AgCl × = 0.213 g MgCl

2 × 100 = 14.2% MgCl

2b. 0.213 g MgCl ×

3 = 8.95 mL AgNO

73. 1.00 L × = 18.8 g AgBr

2 4 2 474. M SO (aq) + CaCl (aq) ÷ CaSO (s) + 2 MCl(aq)

4 2 41.36 g CaSO × = 9.99 × 10 mol M SO-3

2 4From the problem, 1.42 g M SO was reacted so:

2 4 2 41.42 g M SO = 9.99 × 10 mol M SO , molar mass = = 142 g/mol-3

142 amu = 2(atomic mass M) + 32.07 + 4(16.00), atomic mass M = 23 amu

From periodic table, M = Na(sodium).

4 475. All the sulfur in BaSO came from the saccharin. The conversion from BaSO to saccharin utilizesthe molar masses of each.

40.5032 g BaSO × = 0.3949 g saccharin

Avg. mass % = × 100 = 67.00% saccharin by mass

376. a. Fe (aq) + 3 OH (aq) ÷ Fe(OH) (s)3+ -

3Fe(OH) : 55.85 + 3(16.00) + 3(1.008) = 106.87 g/mol

30.107 g Fe(OH) × = 0.0559 g Fe

3 3b. Fe(NO ) : 55.85 + 3(14.01) + 9(16.00) = 241.86 g/mol


3 30.0559 g Fe × = 0.242 g Fe(NO )

3 3c. Mass % Fe(NO ) = × 100 = 53.1%

3 3 3 377. Cr(NO ) (aq) + 3 NaOH(aq) ÷ Cr(OH) (s) + 3 NaNO (aq)

3mol NaOH used = 2.06 g Cr(OH) × = 6.00 × 10 mol NaOH-2

to form precipitate

2NaOH(aq) + HCl(aq) ÷ NaCl(aq) + H O(l)

mol NaOH used = 0.1000 L × = 4.00 × 10 mol NaOH-2

to react with HCl

NaOHM = = 2.00 M NaOH

78. a. Perchloric acid reacted with potassium hydroxide is a possibility.

4 2 4HClO (aq) + KOH(aq) ÷ H O(l) + KClO (aq)

b. Nitric acid reacted with cesium hydroxide is a possibility.

3 2 3HNO (aq) + CsOH(aq) ÷ H O(l) + CsNO (aq)

c. Hydroiodic acid reacted with calcium hydroxide is a possibility.

2 2 22 HI(aq) + Ca(OH) (aq) ÷ 2 H O(l) + CaI (aq)

2 3 2 2 2 3 279. HC H O (aq) + NaOH(aq) ÷ H O(l) + NaC H O (aq)

a. 16.58 × 10 L soln × = 8.393 × 10 mol acetic acid-3 -3

Concentration of acetic acid = = 0.8393 M

b. If we have 1.000 L of solution: total mass = 1000. mL × = 1006 g

2 3 2Mass of HC H O = 0.8393 mol × = 50.40 g

Mass % acetic acid = × 100 = 5.010%

2 280. Mg(s) + 2 HCl(aq) ÷ MgCl (aq) + H (g)

3.00 g Mg × = 0.0494 L = 49.4 mL HCl


281. Let HA = unknown acid; HA(aq) + NaOH(aq) ÷ NaA(aq) + H O(l)

mol HA present = 0.0250 L × = 0.0125 mol HA

, x = molar mass of HA = 176 g/mol

Empirical formula weight . 3(12) + 4(1) + 3(16) = 88 g/mol

3 4 3 2 6 8 6Since 176/88 = 2.0, the molecular formula is (C H O ) = C H O .

3 2 3 282. a. Al(s) + 3 HCl(aq) ÷ AlCl (aq) + 3/2 H (g) or 2 Al(s) + 6 HCl(aq) ÷ 2 AlCl (aq) + 3 H (g)

Hydrogen is reduced (goes from +1 oxidation state to 0 oxidation state) and aluminum Al isoxidized (0 ÷ +3).

b. Balancing S is most complicated since sulfur is in both products. Balance C and H first thenworry about S.

4 2 2CH (g) + 4 S(s) ÷ CS (l) + 2 H S(g)

Sulfur is reduced (0 ÷ -2) and carbon is oxidized (-4 ÷ +4).

c. Balance C and H first, then balance O.

3 8 2 2 2C H (g) + 5 O (g) ÷ 3 CO (g) + 4 H O(l)

Oxygen is reduced (0 ÷ -2) and carbon is oxidized (-8/3 ÷ +4).

d. Although this reaction is mass balanced, it is not charge balanced. We need 2 mol of silver oneach side to balance the charge.

Cu(s) + 2 Ag (aq) ÷ 2 Ag(s) + Cu (aq)+ 2+

Silver is reduced (+1 ÷ 0) and copper is oxidized (0 ÷ +2).

3 283. Mn + HNO ÷ Mn + NO2+

3 2 Mn ÷ Mn + 2 e HNO ÷ NO2+ -

3 2 2 HNO ÷ NO + H O

3 2 2 (e + H + HNO ÷ NO + H O) × 2- +

Mn ÷ Mn + 2 e2+ -

3 2 2 2 e + 2 H + 2 HNO ÷ 2 NO + 2 H O- +

3 2 22 H (aq) + Mn(s) + 2 HNO (aq) ÷ Mn (aq) + 2 NO (g) + 2 H O(l)+ 2+

4 4 3Mn + IO ÷ MnO + IO2+ - - -


2 4 4 3 2(4 H O + Mn ÷ MnO + 8 H + 5 e ) × 2 (2 e + 2 H + IO ÷ IO + H O) × 52+ - + - - + - -

2 4 8 H O + 2 Mn ÷ 2 MnO + 16 H + 10 e2+ - + -

4 3 2 10 e + 10 H + 5 IO ÷ 5 IO + 5 H O- + - -

2 4 4 33 H O(l) + 2 Mn (aq) + 5 IO (aq) ÷ 2 MnO (aq) + 5 IO (aq) + 6 H (aq)2+ - - - +

Challenge Problems

84. a. 5.0 ppb Hg in water =

= 2.5 × 10 M Hg-8

3b. = 8.4 × 10 M CHCl-9

c. 10.0 ppm As =

= 1.33 × 10 M As-4

d. = 2.8 × 10 M DDT-7

85. a. 0.308 g AgCl × = 0.0761 g Cl; %Cl = × 100 = 29.7% Cl

2 3Cobalt(III) oxide, Co O : 2(58.93) + 3(16.00) = 165.86 g/mol

2 30.145 g Co O × = 0.103 g Co; %Co = × 100 = 24.8% Co

The remainder, 100.0 - (29.7 + 24.8) = 45.5%, is water.

Assuming 100.0 g of compound:

245.5 g H O × = 5.09 g H; %H = × 100 = 5.09% H

245.5 g H O × = 40.4 g O; %O = × 100 = 40.4% O

The mass percent composition is 24.8% Co, 29.7% Cl, 5.09% H and 40.4% O.

b. Out of 100.0 g of compound, there are:

24.8 g Co × = 0.421 mol Co; 29.7 g Cl × = 0.838 mol Cl


5.09 g H × = 5.05 mol H; 40.4 g O × = 2.53 mol O

2 2Dividing all results by 0.421, we get CoCl C6H O.

2 2 3 3 2 2c. CoCl C6H O(aq) + 2 AgNO (aq) ÷ 2 AgCl(s) + Co(NO ) (aq) + 6 H O(l)

2 2 2 2CoCl C6H O(aq) + 2 NaOH(aq) ÷ Co(OH) (s) + 2 NaCl(aq) + 6 H O(l)

2 2 3Co(OH) ÷ Co O This is an oxidation-reduction reaction. Thus, we also need to include

2an oxidizing agent. The obvious choice is O .

2 2 2 3 24 Co(OH) (s) + O (g) ÷ 2 Co O (s) + 4 H O(l)

12 10-n n86. a. C H Cl + n Ag ÷ n AgCl; molar mass (AgCl) = 143.4 g/mol+

molar mass (PCB) = 12(12.01) + (10-n) (1.008) + n(35.45) = 154.20 + 34.44 n

Since n mol AgCl are produced for every 1 mol PCB reacted, then n(143.4) grams of AgCl willbe produced for every (154.20 + 34.44 n) grams of PCB reacted.

AgCl PCB or mass (154.20 + 34.44 n) = mass (143.4 n)

b. 0.4791 (154.20 + 34.44 n) = 0.1947 (143.4 n), 73.88 + 16.50 n = 27.92 n

73.88 = 11.42 n, n = 6.469

3 2 4 2 4 387. a. 2 AgNO (aq) + K CrO (aq) ÷ Ag CrO (s) + 2 KNO (aq)

Molar mass: 169.9 g/mol 194.20 g/mol 331.8 g/mol

2 4The molar mass of Ag CrO is 331.8 g/mol, so one mol of precipitate was formed.

3 2 4 3We have equal masses of AgNO and K CrO . Since the molar mass of AgNO is less than that

2 4 3of K CrO , then we have more mol of AgNO present. However, we will not have twice the mol

3 2 4of AgNO present as compared to K CrO as required by the balanced reaction; this is because

3 2 4the molar mass of AgNO is no where near one-half the molar mass of K CrO . Therefore,

3AgNO is limiting.

3 2 4 3mass AgNO = 1.000 mol Ag CrO × = 339.8 g AgNO

2 4Since equal masses of reactants are present, then 339.8 g K CrO were present initially.


= = 7.000 M K+

4 2 4 4b. mol CrO = 339.8 g K CrO × = 1.750 mol CrO2- 2-

present initially

4 2 4 4mol CrO in = 1.000 mol Ag CrO × = 1.000 mol CrO2- 2-

precipitate

= = 0.750 M

88. Molar masses: KCl, 39.10 + 35.45 = 74.55 g/mol; KBr, 39.10 + 79.90 = 119.00 g/mol

AgCl, 107.9 + 35.45 = 143.4 g/mol; AgBr, 107.9 + 79.90 = 187.8 g/mol

Let x = number of moles of KCl in mixture and y = number of moles of KBr in mixture.Since Ag + Cl ÷ AgCl and Ag + Br ÷ AgBr, then x = moles AgCl and y = moles AgBr. + - + -

Setting up two equations:

0.1024 g = 74.55 x + 119.0 y and 0.1889 g = 143.4 x + 187.8 y

Multiply the first equation by , and subtract from the second.

0.1889 = 143.4 x + 187.8 y-0.1616 = -117.7 x - 187.8 y 0.0273 = 25.7 x, x = 1.06 × 10 mol KCl-3

1.06 × 10 mol KCl × = 0.0790 g KCl-3

% KCl = × 100 = 77.1%, % KBr = 100.0 - 77.1 = 22.9%

4 489. 0.298 g BaSO × = 0.123 g SO ; % sulfate = = 60.0%2-

2 4 2 4Assume we have 100.0 g of the mixture of Na SO and K SO . There are:

4 460.0 g SO × = 0.625 mol SO2- 2-

4There must be 2 × 0.625 = 1.25 mol of +1 cations to balance the -2 charge of SO .2-


Let x = number of moles of K and y = number of moles of Na ; then x + y = 1.25.+ +

The total mass of Na and K must be 40.0 g in the assumed 100.0 g of mixture. Setting up an+ +

equation:

x mol K × + y mol Na × = 40.0 g+ +

So, we have two equations with two unknowns: x + y = 1.25 and 39.10 x + 22.99 y = 40.0

Since x = 1.25 - y, then 39.10(1.25 - y) + 22.99 y = 40.0

48.9 - 39.10 y + 22.99 y = 40.0, -16.11 y = -8.9

y = 0.55 mol Na and x = 1.25 - 0.55 = 0.70 mol K+ +

Therefore:

2 4 2 4 2 40.70 mol K × = 0.35 mol K SO ; 0.35 mol K SO × = 61 g K SO+

2 4 2 4Since we assumed 100.0 g, the mixture is 61% K SO and 39% Na SO .

3 4 2 3 490. a. H PO (aq) + 3 NaOH(aq) ÷ 3 H O(l) + Na PO (aq)

2 4 3 2 2 4 3b. 3 H SO (aq) + 2 Al(OH) (s) ÷ 6 H O(l) + Al (SO ) (aq)

2 2 2c. H Se(aq) + Ba(OH) (aq) ÷ 2 H O(l) + BaSe(s)

2 2 4 2 2 2 4d. H C O (aq) + 2 NaOH(aq) ÷ 2 H O(l) + Na C O (aq)

3 4 2 2 3 4 291. 2 H PO (aq) + 3 Ba(OH) (aq) ÷ 6 H O(l) + Ba (PO ) (s)

0.01420 L × = 0.0576 L

2 = 57.6 mL Ba(OH)

2 492. 35.08 mL NaOH × = 3.72 × 10 mol H SO-2

2 4Molarity = = 3.72 M H SO

2 293. a. MgO(s) + 2 HCl(aq) ÷ MgCl (aq) + H O(l)

2 2 2Mg(OH) (s) + 2 HCl(aq) ÷ MgCl (aq) + 2 H O(l)

3 3 2Al(OH) (s) + 3 HCl(aq) ÷ AlCl (aq) + 3 H O(l)


b. Let's calculate the number of moles of HCl neutralized per gram of substance. We can get thesedirectly from the balanced equations and the molar masses of the substances.

Therefore, one gram of magnesium oxide would neutralize the most 0.10 M HCl.

94. We get the empirical formula from the elemental analysis. Out of 100.00 g carminic acid there are:

53.66 g C × = 4.468 mol C; 4.09 g H × = 4.06 mol H

42.25 g O × = 2.641 mol O

Dividing the moles by the smallest number gives:

These numbers don’t give obvious mol ratios. Let’s determine the mol C to mol H ratio.

= 1.10 =

So let's try = 0.406 as a common factor: = 11.0; = 10.0; = 6.50

22 20 13Therefore, C H O is the empirical formula.

We can get molar mass from the titration data.

18.02 × 10 L soln × = 7.32 × 10 mol carminic acid-3 -4

Molar mass =

22 20 13The empirical formula mass of C H O . 22(12) + 20(1) + 13(16) = 492 g.

22 20 13Therefore, the molecular formula of carminic acid is also C H O .

6 8 7 6 8 7 6 8 795. mol C H O = 0.250 g C H O × = 1.30 × 10 mol C H O-3


xLet H A represent citric acid where x is the number of acidic hydrogens. The balanced neutralizationreaction is:

x 2H A(aq) + x OH (aq) ÷ x H O(l) + A (aq)- x-

mol OH reacted = 0.0372 L × = 3.91 × 10 mol OH- -3 -

x = = 3.01

3Therefore, the general acid formula for citric acid is H A, meaning that citric acid has three acidichydrogens per citric acid molecule (citric acid is a triprotic acid).

96. a. HCl(aq) dissociates to H (aq) + Cl (aq). For simplicity let's use H and Cl separately.+ - + -

2 4 H ÷ H Fe ÷ HFeCl+

2 4(2 H + 2 e ÷ H ) × 3 (H + 4 Cl + Fe ÷ HFeCl + 3 e ) × 2+ - + - -

2 6 H + 6 e ÷ 3 H+ -

4 2 H + 8 Cl + 2 Fe ÷ 2 HFeCl + 6 e+ - -

4 2 8 H + 8 Cl + 2 Fe ÷ 2 HFeCl + 3 H+ -

4 2or 8 HCl(aq) + 2 Fe(s) ÷ 2 HFeCl (aq) + 3 H (g)

3 3 3b. IO ÷ I I ÷ I- - - -

3 3 3 3 IO ÷ I (3 I ÷ I + 2 e ) × 8- - - - -

3 3 2 3 IO ÷ I + 9 H O- -

3 3 216 e + 18 H + 3 IO ÷ I + 9 H O- + - -

3 3 2 16 e + 18 H + 3 IO ÷ I + 9 H O- + - -

3 24 I ÷ 8 I + 16 e- - -

3 3 218 H + 24 I + 3 IO ÷ 9 I + 9 H O+ - - -

3 3 2Reducing: 6 H (aq) + 8 I (aq) + IO (aq) ÷ 3 I (aq) + 3 H O(l)+ - - -

c. (Ce + e ÷ Ce ) × 974+ - 3+

6 3 2 4 Cr(NCS) ÷ Cr + NO + CO + SO4- 3+ - 2-

2 6 3 2 454 H O + Cr(NCS) ÷ Cr + 6 NO + 6 CO + 6 SO + 108 H4- 3+ - 2- +

Charge on left -4. Charge on right = +3 + 6(-1) + 6(-2) + 108(+1) = +93. Add 97 e to the right,-

then add the two balanced half-reactions with a common factor of 97 e transferred.-


2 6 3 2 454 H O + Cr(NCS) ÷ Cr + 6 NO + 6 CO + 6 SO + 108 H + 97 e4- 3+ - 2- + -

97 e + 97 Ce ÷ 97 Ce- 4+ 3+

2 6 3 2 97 Ce (aq) + 54 H O(l) + Cr(NCS) (aq) ÷ 97 Ce (aq) + Cr (aq) + 6 NO (aq) + 6 CO (g)4+ 4- 3+ 3+ -

4 + 6 SO (aq) + 108 H (aq)2- +

This is very complicated. A check of the net charge is a good check to see if the equation isbalanced. Left: charge = 97(+4) - 4 = +384. Right: charge = 97(+3) + 3 + 6(-1) + 6(-2) + 108(+1) = +384.

3 4 4 2d. CrI ÷ CrO + IO Cl ÷ Cl2- - -

2 3 4 4 2(16 H O + CrI ÷ CrO + 3 IO + 32 H + 27 e ) × 2 (2 e + Cl ÷ 2 Cl ) × 272- - + - - -


254 e + 27 Cl ÷ 54 Cl- -

2 3 4 4 32 H O + 2 CrI ÷ 2 CrO + 6 IO + 64 H + 54 e2- - + -

2 3 2 4 4 32 H O + 2 CrI + 27 Cl ÷ 54 Cl + 2 CrO + 6 IO + 64 H- 2- - +

2Add 64 OH to both sides and convert 64 H into 64 H O.- +

2 3 2 4 4 264 OH + 32 H O + 2 CrI + 27 Cl ÷ 54 Cl + 2 CrO + 6 IO + 64 H O- - 2- -

Reducing gives:

3 2 4 4 264 OH (aq) + 2 CrI (s) + 27 Cl (g) ÷ 54 Cl (aq) + 2 CrO (aq) + 6 IO (aq) + 32 H O(l)- - 2- -

3e. Ce ÷ Ce(OH)4+

2 3(e + 3 H O + Ce ÷ Ce(OH) + 3 H ) × 61- 4+ +

6 3 3 3 Fe(CN) ÷ Fe(OH) + CO + NO4- 2- -

6 3 3 3 Fe(CN) ÷ Fe(OH) + 6 CO + 6 NO4- 2- -

2There are 39 extra O atoms on right. Add 39 H O to left, then add 75 H to right to balance H .+ +

2 6 3 3 3 39 H O + Fe(CN) ÷ Fe(OH) + 6 CO + 6 NO + 75 H4- 2- - +

net charge = -4 net charge = +57

Add 61 e to the right; then add the two balanced half-reactions with a common factor of 61 e- -

transferred.


2 6 3 3 339 H O + Fe(CN) ÷ Fe(OH) + 6 CO + 6 NO + 75 H + 61 e4- 2- - + -

2 3 61 e + 183 H O + 61 Ce ÷ 61 Ce(OH) + 183 H- 4+ +

2 6 3 3 3 3 222 H O + Fe(CN) + 61 Ce ÷ 61 Ce(OH) + Fe(OH) + 6 CO + 6 NO + 258 H4- 4+ 2- - +

Adding 258 OH to each side then reducing gives:-

6 3 3258 OH (aq) + Fe(CN) (aq) + 61 Ce (aq) ÷ 61 Ce(OH) (s) + Fe(OH) (s)- 4- 4+

3 3 2+ 6 CO (aq) + 6 NO (aq) + 36 H O(l)2- -

2 3 2 2 2f. Fe(OH) ÷ Fe(OH) H O ÷ H O

2 2 3 2 2 2(H O + Fe(OH) ÷ Fe(OH) + H + e ) × 2 2 e + 2 H + H O ÷ 2 H O+ - - +

2 2 3 2 H O + 2 Fe(OH) ÷ 2 Fe(OH) + 2 H + 2 e+ -

2 2 2 2 e + 2 H + H O ÷ 2 H O- +

2 2 2 2 3 22 H O + 2 H + 2 Fe(OH) + H O ÷ 2 Fe(OH) + 2 H O + 2 H+ +

2 2 2 3Reducing gives: 2 Fe(OH) (s) + H O (aq) ÷ 2 Fe(OH) (s)

97. The amount of KHP used = 0.4016 g × = 1.967 × 10 mol KHP-3

Since one mole of NaOH reacts completely with one mole of KHP, the NaOH solution contains 1.967× 10 mol NaOH.-3

Molarity of NaOH =

Maximum molarity =

Minimum molarity =

We can express this as 0.07849 ± 0.00016 M. An alternative is to express the molarity as 0.0785 ±0.0002 M. The second way shows the actual number of significant figures in the molarity. Theadvantage of the first method is that it shows that we made all of our individual measurements to foursignificant figures.


Date post:	01-Nov-2019
Category:	Documents
Upload:	others
View:	4 times
Download:	0 times

CHAPTER FOUR TYPES OF CHEMICAL REACTIONS AND … · TYPES OF CHEMICAL REACTIONS AND SOLUTION...

Documents