Acid-Base Titrations

Acid-Base Titrations

Section 17.3

Introduction

• Definition:– In an acid-base titration, a solution containing a

known concentration of a base is slowly added to an acid.

– An indicator is used to signal the equivalence point of the titration.• This is the point at which stoichiometrically equivalent

amounts of acid and base have been mixed.– A pH meter can also be used to find the

equivalence point.

Introduction

• The typical titration apparatus includes:– a buret to hold the titrant– a beaker to hold the analyte– a pH meter to measure the pH

Introduction

• In this section, we will be looking at a series of titrations in detail to understand why acids and behave the way they do.– Strong acid-strong base titration– Weak acid-strong base titration– Polyprotic acid-strong base titration

Strong Acid-Strong Base Titrations

• The titration curve of a strong acid-strong base titration has the following shape.



We divide the curve into four regions:1. Initial pH

2. Initial pH to eq. point

3. Equivalence point

4. After eq. point



We divide the curve into four regions:1. Initial pH

The pH of the solution is determined by the concentration of the strong acid.



We divide the curve into four regions:

As base is added, pH increases slowly and then rapidly. The pH is determined by the concentration of the acid that is not neutralized.

2. Initial pH to eq. point




At the equivalence point, [OH−] = [H+]. The pH = 7.00.

3. Equivalence point




As more base is added, pH increases rapidly and then slowly. pH is determined by the concentration of the excess base.

4. After eq. point



Let’s see how this works in practice.

Sample Exercise 17.6 (page 731)

Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.a) 49.0 mLb) 51.0 mL


Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.a) 49.0 mL


Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.a) 49.0 mLThis is between the initial point and the equivalence point.


Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.a) 49.0 mLThis is between the initial point and the equivalence point. pH is determined by the amount of acid that has not been neutralized.


Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.a) 49.0 mLTherefore, we need to determine the number of mols of acid remaining, nacid, and the total volume, Vtotal, of the solution. (Remember, adding the NaOH increases the total volume.


Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.a) 49.0 mLnacid,i = Macid,i × Vacid,i


Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.a) 49.0 mLnacid,i = Macid,i × Vacid,i = (0.100 M)(0.0500 L)


Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.a) 49.0 mLnacid,i = Macid,i × Vacid,i = 5.00 × 10−3 mol


Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.a) 49.0 mLnacid,i = Macid,i × Vacid,i = 5.00 × 10−3 molnbase,added = Mbase,added × Vbase,added


Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.a) 49.0 mLnacid,i = Macid,i × Vacid,i = 5.00 × 10−3 molnbase,added = (0.100 M)(0.0490 L)


Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.a) 49.0 mLnacid,i = Macid,i × Vacid,i = 5.00 × 10−3 molnbase,added = 4.90 × 10−3 mol


Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.a) 49.0 mLnacid,i = Macid,i × Vacid,i = 5.00 × 10−3 molnbase,added = 4.90 × 10−3 molnacid,remaining = nacid,i − nbase,added


Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.a) 49.0 mLnacid,i = Macid,i × Vacid,i = 5.00 × 10−3 molnbase,added = 4.90 × 10−3 molnacid,remaining = (5.00 − 4.90) × 10−3 mol


Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.a) 49.0 mLnacid,i = Macid,i × Vacid,i = 5.00 × 10−3 molnbase,added = 4.90 × 10−3 molnacid,remaining = 0.10 × 10−3 mol


Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.a) 49.0 mLnacid,i = Macid,i × Vacid,i = 5.00 × 10−3 molnbase,added = 4.90 × 10−3 molnacid,remaining = 1.0 × 10−4 mol


Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.a) 49.0 mLnacid,remaining = 1.0 × 10−4 mol


Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.a) 49.0 mLnacid,remaining = 1.0 × 10−4 molVtotal = Vacid + Vbase


Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.a) 49.0 mLnacid,remaining = 1.0 × 10−4 molVtotal = Vacid + Vbase = 0.0500 L + 0.0490 L


Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.a) 49.0 mLnacid,remaining = 1.0 × 10−4 molVtotal = Vacid + Vbase = 0.0990 L


Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.a) 49.0 mLnacid,remaining = 1.0 × 10−4 molVtotal = 0.0990 L


Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.a) 49.0 mLnacid,remaining = 1.0 × 10−4 molVtotal = 0.0990 L[H+] = nacid,remaining/Vtotal


Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.a) 49.0 mLnacid,remaining = 1.0 × 10−4 molVtotal = 0.0990 L[H+] = (1.0 × 10−4 mol)/(0.0990 L)


Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.a) 49.0 mLnacid,remaining = 1.0 × 10−4 molVtotal = 0.0990 L[H+] = 1.0 × 10−3 M


Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.a) 49.0 mLnacid,remaining = 1.0 × 10−4 molVtotal = 0.0990 L[H+] = 1.0 × 10−3 MpH = −log[H+]


Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.a) 49.0 mLnacid,remaining = 1.0 × 10−4 molVtotal = 0.0990 L[H+] = 1.0 × 10−3 MpH = −log(1.0 × 10−3)


Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.a) 49.0 mLnacid,remaining = 1.0 × 10−4 molVtotal = 0.0990 L[H+] = 1.0 × 10−3 MpH = 3.00


Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.a) 49.0 mLpH = 3.00


Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.a) 49.0 mL


Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.a) 49.0 mLb) 51.0 mL


Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.

b) 51.0 mL


Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.b) 51.0 mL


Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.b) 51.0 mLThis is beyond the equivalence point.


Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.b) 51.0 mLThis is beyond the equivalence point. All of the strong acid has been used up.


Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.b) 51.0 mLThis is beyond the equivalence point. All of the strong acid has been used up. The pH is determined by the excess base that has been added.


Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.b) 51.0 mLWe already determined nacid,i = 5.00 × 10−3 mol


Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.b) 51.0 mLnacid,i = 5.00 × 10−3 mol


Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.b) 51.0 mLnacid,i = 5.00 × 10−3 molnbase,added = Mbase,added × Vbase,added


Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.b) 51.0 mLnacid,i = 5.00 × 10−3 molnbase,added = (0.100 M)(0.0510 L)


Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.b) 51.0 mLnacid,i = 5.00 × 10−3 molnbase,added = 5.10 × 10−3 mol


Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.b) 51.0 mLnacid,i = 5.00 × 10−3 molnbase,added = 5.10 × 10−3 molnbase,remaining = nbase,added − nacid,i


Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.b) 51.0 mLnacid,i = 5.00 × 10−3 molnbase,added = 5.10 × 10−3 molnbase,remaining = (5.10 − 5.00) × 10−3 mol


Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.b) 51.0 mLnacid,i = 5.00 × 10−3 molnbase,added = 5.10 × 10−3 molnbase,remaining = 0.10 × 10−3 mol


Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.b) 51.0 mLnbase,remaining = 1.0 × 10−4 mol


Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.b) 51.0 mLnbase,remaining = 1.0 × 10−4 molVtotal = Vacid + Vbase


Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.b) 51.0 mLnbase,remaining = 1.0 × 10−4 molVtotal = Vacid + Vbase = 0.0500 L + 0.0510 L


Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.b) 51.0 mLnbase,remaining = 1.0 × 10−4 molVtotal = Vacid + Vbase = 0.1010 L


Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.b) 51.0 mLnbase,remaining = 1.0 × 10−4 molVtotal = 0.1010 L


Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.b) 51.0 mLnbase,remaining = 1.0 × 10−4 molVtotal = 0.1010 L[OH−] = nbase,remaining/Vtotal


Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.b) 51.0 mLnbase,remaining = 1.0 × 10−4 molVtotal = 0.1010 L[OH−] = (1.0 × 10−4 mol)/(0.1010 L)


Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.b) 51.0 mLnbase,remaining = 1.0 × 10−4 molVtotal = 0.1010 L[OH−] = 9.9 × 10−4 M


Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.b) 51.0 mLnbase,remaining = 1.0 × 10−4 molVtotal = 0.1010 L[OH−] = 9.9 × 10−4 MpOH = −log[OH−]


Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.b) 51.0 mLnbase,remaining = 1.0 × 10−4 molVtotal = 0.1010 L[OH−] = 9.9 × 10−4 MpOH = −log(9.9 × 10−4)


Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.b) 51.0 mLnbase,remaining = 1.0 × 10−4 molVtotal = 0.1010 L[OH−] = 9.9 × 10−4 MpOH = 3.00


Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.b) 51.0 mLnbase,remaining = 1.0 × 10−4 molVtotal = 0.1010 L[OH−] = 9.9 × 10−4 MpOH = 3.00 pH = 14.00 − 3.00⇒


Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.b) 51.0 mLnbase,remaining = 1.0 × 10−4 molVtotal = 0.1010 L[OH−] = 9.9 × 10−4 MpOH = 3.00 pH = 14.00 − 3.00 = 11.00⇒


Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.b) 51.0 mLnbase,remaining = 1.0 × 10−4 molVtotal = 0.1010 L[OH−] = 9.9 × 10−4 MpH = 11.00


Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.b) 51.0 mLpH = 11.00

Weak Acid-Strong Base Titrations

• The curve for a weak acid-strong base titration is similar to that for a strong acid-weak base.



1. The initial pH is determined by the Ka of the acid.



2. To determine the pH from the initial point to the eq. point, we first neutralize the weak acid and then use the Henderson-Hasselbach equation.



3. At the eq. point, we have no HX, only X−. We need to use the Kb value to find pH.



4. Beyond the eq. point, we use the excess base to calculate pH.



Let’s see how this works in practice.


• Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH3COOH.



First, calculate the concentrations of materials before neutralization reaction.




CH3COOH + OH− CH➙ 3COO− + H2O




CH3COOH + OH− CH➙ 3COO− + H2Oninitial,acid = Macid × Vacid




CH3COOH + OH− CH➙ 3COO− + H2Oninitial,acid = Macid × Vacid = (0.100 M)(0.0500 L)




CH3COOH + OH− CH➙ 3COO− + H2Oninitial,acid = Macid × Vacid = 5.00 × 10−3 mol




CH3COOH + OH− CH➙ 3COO− + H2Oninitial,acid = 5.00 × 10−3 mol




CH3COOH + OH− CH➙ 3COO− + H2Oninitial,acid = 5.00 × 10−3 molnbase = Mbase × Vbase




CH3COOH + OH− CH➙ 3COO− + H2Oninitial,acid = 5.00 × 10−3 molnbase = Mbase × Vbase = (0.100 M)(0.0450 L)




CH3COOH + OH− CH➙ 3COO− + H2Oninitial,acid = 5.00 × 10−3 molnbase = Mbase × Vbase = 4.50 × 10−3 mol




CH3COOH + OH− CH➙ 3COO− + H2Oninitial,acid = 5.00 × 10−3 molnbase = 4.50 × 10−3 mol



Next, use the values to determine the concentrations after the neutralization.

CH3COOH + OH− CH➙ 3COO− + H2Oninitial,acid = 5.00 × 10−3 molnbase = 4.50 × 10−3 mol




CH3COOH + OH− CH➙ 3COO− + H2Oninitial 5.0 × 10−3 mol 4.5 × 10−3 mol 0.0 mol





nchange −4.5 × 10−3 mol −4.5 × 10−3 mol +4.5 × 10−3 mol





nchange −4.5 × 10−3 mol −4.5 × 10−3 mol +4.5 × 10−3 mol

nfinal 5.0 × 10−4 mol 0.0 mol 4.5 × 10−3 mol




CH3COOH + OH− CH➙ 3COO− + H2Onacid,final = 5.0 × 10−4 mol




CH3COOH + OH− CH➙ 3COO− + H2Onacid,final = 5.0 × 10−4 mol[acid] = nacid,final/Vtotal




CH3COOH + OH− CH➙ 3COO− + H2Onacid,final = 5.0 × 10−4 mol[acid] = (5.0 × 10−4 mol)/(0.0950 L)




CH3COOH + OH− CH➙ 3COO− + H2Onacid,final = 5.0 × 10−4 mol[acid] = 5.3 × 10−3 M




CH3COOH + OH− CH➙ 3COO− + H2O[acid] = 5.3 × 10−3 M[base] = nbase,final/Vtotal




CH3COOH + OH− CH➙ 3COO− + H2O[acid] = 5.3 × 10−3 M[base] = (4.50 × 10−3 mol)/(0.0950 L)




CH3COOH + OH− CH➙ 3COO− + H2O[acid] = 5.3 × 10−3 M[base] = 4.74 × 10−2 M




CH3COOH + OH− CH➙ 3COO− + H2O[acid] = 5.3 × 10−3 M [base] = 4.74 × 10−2 M




CH3COOH + OH− CH➙ 3COO− + H2O[acid] = 5.3 × 10−3 M [base] = 4.74 × 10−2 MpKa = −logKa




CH3COOH + OH− CH➙ 3COO− + H2O[acid] = 5.3 × 10−3 M [base] = 4.74 × 10−2 MpKa = −log(1.8 × 10−5)




CH3COOH + OH− CH➙ 3COO− + H2O[acid] = 5.3 × 10−3 M [base] = 4.74 × 10−2 MpKa = 4.74




CH3COOH + OH− CH➙ 3COO− + H2OpH = pKa + log([base]/[acid)




CH3COOH + OH− CH➙ 3COO− + H2OpH = 4.74 + log([4.74 × 10−2 M]/[5.3 × 10−3 M])




CH3COOH + OH− CH➙ 3COO− + H2OpH = 4.74 + log(9.00)




CH3COOH + OH− CH➙ 3COO− + H2OpH = 4.74 + 0.954




CH3COOH + OH− CH➙ 3COO− + H2OpH = 5.69


• Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH3COOH with 0.100 M NaOH.



At the equivalence point, all of the weak acid is converted to its conjugate weak base.



At the equivalence point, all of the weak acid is converted to its conjugate weak base. This means that we will be using the base hydrolysis expression to find [OH−], pOH, and pH.



The number of mols of acetate in solution at the equivalence point is equal to the number of mols of acetic acid at the start of the titration.



nbase,eq. pt. = ninitial,acid



nbase,eq. pt. = ninitial,acid = (0.100 M)(0.0500 L)



nbase,eq. pt. = ninitial,acid = 5.00 × 10−3 mol



nbase,eq. pt. = 5.00 × 10−3 mol



nbase,eq. pt. = 5.00 × 10−3 molThe concentration of the acetate is given by:



nbase,eq. pt. = 5.00 × 10−3 molThe concentration of the acetate is given by:[base] = (nbase,eq. pt.)/(Vtotal)



nbase,eq. pt. = 5.00 × 10−3 molThe concentration of the acetate is given by:[base] = (5.00 × 10−3 mol)/(0.100 L)



nbase,eq. pt. = 5.00 × 10−3 molThe concentration of the acetate is given by:[base] = 5.00 × 10−2 M



[base] = 5.00 × 10−2 M



[base] = 5.00 × 10−2 MNext, we do an i-c-e table on the base hydrolysis.



[base] = 5.00 × 10−2 MCH3COO− + H2O CH➙ 3COOH + OH−




i 5.00 × 10−2 0.0 0.0




i 5.00 × 10−2 0.0 0.0c −x +x +x




i 5.00 × 10−2 0.0 0.0c −x +x +xe 5.00 × 10−2 − x x x



[base] = 5.00 × 10−2 M, [acid] = [OH−] = xCH3COO− + H2O CH➙ 3COOH + OH−

i 5.00 × 10−2 0.0 0.0c −x +x +xe 5.00 × 10−2 − x x x



[base] = 5.00 × 10−2 M, [acid] = [OH−] = xKb = Kw/Ka = (1.0 × 10−14)/(1.8 × 10−5)



[base] = 5.00 × 10−2 M, [acid] = [OH−] = xKb = Kw/Ka = 5.6 × 10 −10



[base] = 5.00 × 10−2 M, [acid] = [OH−] = xKb = 5.6 × 10 −10



[base] = 5.00 × 10−2 M, [acid] = [OH−] = xKb = 5.6 × 10 −10 = ([acid][OH−])/[base]



[base] = 5.00 × 10−2 M, [acid] = [OH−] = xKb = 5.6 × 10 −10 = (x2)/(5.00 × 10−2)



[base] = 5.00 × 10−2 M, [acid] = [OH−] = xKb = 5.6 × 10 −10 = (x2)/(5.00 × 10−2)x2 = (5.6 × 10 −10)(5.00 × 10−2) = 2.8 × 10−11

x = (2.8 × 10−11)½




x = (2.8 × 10−11)½ = 5.3 × 10−6




x = (2.8 × 10−11)½ = 5.3 × 10−6 = [OH−]



[OH−] = 5.3 × 10−6 M



[OH−] = 5.3 × 10−6 MpOH = −log[OH−]



[OH−] = 5.3 × 10−6 MpOH = −log[OH−] = −log(5.3 × 10−6)



[OH−] = 5.3 × 10−6 MpOH = −log[OH−] = −log(5.3 × 10−6) = 5.28



[OH−] = 5.3 × 10−6 MpOH = 5.28



[OH−] = 5.3 × 10−6 MpOH = 5.28pH = 14.00 – pOH



[OH−] = 5.3 × 10−6 MpOH = 5.28pH = 14.00 – pOH = 14.00 − 5.28



[OH−] = 5.3 × 10−6 MpOH = 5.28pH = 14.00 – pOH = 14.00 − 5.28 = 8.72



pH = 8.72

• The pH of the titration at the equivalence point depends on the type of acids and bases being titrated.– Strong acid-Strong base: eq. pt. = 7.0– Weak acid-Strong base: eq. pt. > 7.0– Strong acid-Weak base: eq. pt. < 7.0

Titrations of Polyprotic Acids

• When weak acid contain more than one ionizable H atom, as in phosporous acid, H3PO3, reaction with OH− occurs in a series of steps.– H3PO3(aq) + H2O(l) H➙ 2PO3

−(aq) + H3O+(aq)– H2PO3

−(aq) + H2O(l) HPO➙ 32−(aq) + H3O+(aq)

– HPO32−(aq) + H2O(l) PO➙ 3

3−(aq) + H3O+(aq)

Titrations of Polyprotic Acids

• When the neutralization steps are sufficiently separated, the substance exhibits multiple equivalence points.

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Acid-Base Titrations

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