Titrations and Balancing by Half Reactions

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Reactions and Calculations with Acids and Bases Neutralization Reactions - when stoichiometrically equivalent amounts of acid and base react to form salt and water. (The net ionic equation of a neutralization is H+ + OH- → H2O)

Molarity =

mol or

mmols or

L mL 1000 mL

Examples of Neutralization Reactions:

If 100. mL of 0.100 M HCl and 100. mL of 0.100 M NaOH react, then what is the molarity of the salt produced?

Rice Table (use mols!)

The salt is NaCl, the moles of salt formed is 0.0100 mols, and the volume of the solution is 200. mL (be sure to add your volumes for the results of a reaction).

Molarity =

0.0100 mol = 0.0500 M NaClL 0.200 L

Molarity =

10.0 mmol = 0.0500 M NaClmL 200. mL

If 100. mL of 1.00 M HCl and 100. mL of 0.800 M NaOH react, then what is the molarity of the solutes? (remember water is a solvent)

Rice Table, then Molarity

If 100. mL of 1.00 M HCl and 100. mL of 1.00 M Ca(OH)2 react, then what is the molarity of the solutes?

Your turn: If 100. mL of 1.00 M H2SO4 and 200. mL of 1.00 M KOH react, then what is the salt and what is the concentration of the salt?

What Volume of 0.00300 M HCl would neutralize 30.0 mL of 0.00100 M Ca(OH)2?

2HCl + Ca(OH)2 CaCl2 + 2HOH

0.00100 mol Ca(OH)2

30.0 mL 2 mol HCl 1000 mL= 20.0

mL1000 mL 1 mol

Ca(OH)2

0.00300 mol HCl

Your turn: What volume of 0.0150 M acetic acid would neutralize 18.7 mL of 0.0105 M Ba(OH)2?

Titration – adding a known solution (titrant) to an unknown solution, to determine the unknown’s concentration.

Standard Solution – titrating one solution with another to get the exact concentration of a titrant (that will later be used for titrating).

Often our common acids and bases can decompose or become contaminated while they sit in the storeroom, so even though you might have measured out enough solute to make a particular concentration of solution, it may not end up at the molarity you expected. Once standardized, the exact concentration is known and that is used to calculate your titration results, not the amount you were originally expecting to use.

Typical standardizing solutions are sodium bicarbonate for acids and potassium hydrogen phthalate (KHP) for bases.

Equivalence point – When [H+] = [OH-] End Point – point in a titration where the indicate color changes infers you have reached the equivalence point. Pick indicators wisely!

Examples of Titration Reactions:

What is the molarity of HCl if 36.7 mL of HCl was titrated with 43.2 mL of 0.236 M NaOH?

HCl + NaOH → NaCl + H2O

43.2 mL NaOH

0.236 mols NaOH 1 mol HCl

= 0.0102 mols HCl1000 mL 1 mol

43.2 mL NaOH

0.236 mols NaOH 1 mol HCl

= 0.0102 mols HCl1000 mL 1 mol

For HCl, 0.0102 mol HCl = 0.278 M

HCl0.0367 L

Your turn: What is the concentration of H2SO4 if 43.2 mL of 0.236 M NaOH neutralized 36.7 mL of H2SO4?

Normality – the number of equivalents per liter

Good for: qualitative work – like determining the freshness of milk

Not good for: quantitative work – like the titrations of complete unknowns Examples:

HCl = 1 eq. H2SO4 = 2 eq.NaOH = 1 eq. Ca(OH)2 = 2 eq.

Redox reactions – not only does the mass balance (like before) but the charges (oxidation numbers) must also. To balance redox reactions, the half-reaction method is commonly used.

To Balance by Half Reactions:1.Write unbalanced net ionic equations, as

completely as possible2.Split into two reactions, the oxidation and

reduction reactions3.Balance everything except for O and H4.Appropriately add O then H, if needed (see list

below)5.Add electrons as needed6.Balance the number of electrons gained and

lost between the two reactions, multiplying as needed

7.Add the two equations together, simplifying as much as possible

In acid solution:For O, add H2O where O is neededthen for H, add H+ where needed In basic solution:For O, add 2xOH- and H2O on the other side

then for H, add H2O where needed and OH- to the other side

Examples of Redox Reactions:

I2 and S2O32- react to form I- and S4O6

I2 + S2O3

2- →I- + S4O62-

Fe2+ and MnO4- react to form Fe3+ and Mn2+ in a

sulfuric acid solution

Fe2+ + MnO4- → Fe3+ + Mn2+

In a basic solution, ClO- will oxidize CrO2- to CrO4

2- and be reduced to Cl-.

CrO2- + ClO- → CrO4

2- + Cl-

Your turn: Balance using HR: Al + NO3

- + OH- + H2O → Al(OH)4- + NH3

Examples of Redox Titration Reactions:

What volume of 0.0200 M KMnO4 is required to oxidize 40.0 mL of 0.100 M FeSO4?

5Fe2+ + MnO4- + 8H+ → 5Fe3+ + Mn2- + 4H2O

What volume of 0.0200 M KMnO4 is required to oxidize 40.0 mL of 0.100 M FeSO4?

5Fe2+ + MnO4- + 8H+ → 5Fe3+ + Mn2- + 4H2O

40.0 mL FeSO4

0.100 mol FeSO4

1 mol Fe2+

1 mol MnO4

-1 mol KMnO4

1000 mL = 40.0 mL KMnO41000 mL

1 mol FeSO4

5 mols Fe2+ 1 mol MnO4

-0.0200 mol

What is the molarity of Na2SO3 if 20.00 mL Na2SO3 are titrated with 36.30 mL of 0.05130 M K2Cr2O7 in H2SO4?

8H+ + Cr2O72- + 3SO3

2- → 2Cr3+ + 3SO42- + 4H2O

8H+ + Cr2O72- + 3SO3

2- → 2Cr3+ + 3SO42- + 4H2O

36.30 mL K2Cr2O7

0.05130 mols K2Cr2O7

1 mol Cr2O7

2- 3 mol SO32- 1 mol

Na2SO3 = 0.005586 mol Na2SO31000 mL

K2Cr2O7

1 mol K2Cr2O7

1 mol Cr2O7

2-1 mol SO3

8H+ + Cr2O72- + 3SO3

2- → 2Cr3+ + 3SO42- + 4H2O

36.30 mL K2Cr2O7

0.05130 mols K2Cr2O7

1 mol Cr2O7

2- 3 mol SO32- 1 mol

Na2SO3 = 0.005586 mol Na2SO31000 mL

K2Cr2O7

1 mol K2Cr2O7

1 mol Cr2O7

2-1 mol SO3

For Na2SO30.005586 mol Na2SO3 = 0.2793 M Na2SO30.02000 L

Your turn: 40.00 mL of 0.1442 M Na2S2O3 reacts with 26.36 mL of I2. What is the molarity of the I2 solution? 2 Na2S2O3 + I2 → Na2S4O6 + 2NaI

Titrations and Balancing by Half Reactions

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