# Homework Help: Acid-Base Titration Problems

1. Apr 8, 2017

### cvc121

1. The problem statement, all variables and given/known data
a) Determine the concentration of NaCH3COO. The volume of HCl used to reach the equivalence point is 25.05 mL, the concentration of HCl is 0.10 M, and the volume of NaCH3COO solution used is 10.0 mL.

b) Using the initial concentrations of NaCH3COO and HCl, and a reliable literature value for the pKb for NaCH3COO or PKa for CH3COOH, predict the theoretical pH of the equivalence point.

2. Relevant equations

3. The attempt at a solution
Here is my reasoning and calculations:

a)

The equivalence point is the point in the titration when the number of moles of a standard solution (titrant) is equal to the number of moles of a solution of unknown concentration (analyte).

Therefore, at the equivalence point:

ntitrant = nanalyte

nHCl = nNaCH3COO

moles of HCl = (concentration)(volume)

moles of HCl = (0.10 M)(0.02505 L)

moles of HCl = 2.5 x 10-3 mol

moles of HCl = moles of NaCH3COO = 2.5 x 10-3 mol

Concentration of NaCH3COO = moles / total volume

Concentration of NaCH3COO = 2.5 x 10-3 mol / (0.01 L + 0.02505 L)

Concentration of NaCH3COO = 0.07 M

b)

A reliable literature value for the pKa for acetic acid is 4.756.

pKa = -log Ka

Ka = 10-pKa

Ka = 10-4.756

Ka = 1.754 x 10-5

At the equivalence point of the titration of a weak base with a strong acid, the general reaction H3O+ (aq) + B (aq) ---> BH+ (aq) + H2O (l) has gone approximately to completion. Therefore, approximately all sodium acetate (NaCH3COO) will be converted to its conjugate acid, acetic acid (CH3COOH), in a 1:1 molar ratio.

Ka = [CH3COO-][H3O+] / [CH3COOH]
1.754 x 10-5 = x2 / (0.07 M – x)

Ka is very small. Therefore, the x value in the denominator is negligible and can be omitted.

1.754 x 10-5 = x2 / 0.07 M

x2 = (1.754 x 10-5 )(0.07 M)

x2 = 1.228 x 10-6 M

x = 1.108 x 10-3 M

[H3O+] = x = 1.108 x 10-3 M

Validity check:
(1.108 x 10-3 M / 0.07 M )(100%) = 1.58%

Theoretical pH = -log[H3O+]
Theoretical pH = -log[1.108 x 10-3]
Theoretical pH = 2.96

Could anyone verify my reasoning and calculations? Thanks. All help is very much appreciated.

2. Apr 8, 2017

### Staff: Mentor

There is no CH3COONa at the endpoint so your calculation of concentration (using total volume at the endpoint) doesn't make much sense. Question most likely asks for the initial CH3COONa concentration.

b looks reasonalby accurate.

Question is a bit idiotic to be honest. Good luck detecting the end point in this titration (not your fault though).

3. Apr 8, 2017

### cvc121

Thank you for your response. Assuming that the question is asking for the initial concentration of CH3COONa, does the question provide enough information to solve for this? Would I simply use the volume of CH3COONa used instead of the total volume?

Concentration of NaCH3COO = 2.5 x 10-3 mol / 0.01 L = 0.25 M?

4. Apr 8, 2017

Looks OK.

5. Apr 9, 2017

### epenguin

I concur with BoreK that the question does not make a lot of sense. Are you sure you have copied it and any background out exactly? One titrates an acid with a base or vice versa;, one does not titrate a salt which is what sodium acetate is.

Your calculation in #1 is just giving you the pH of 0.07 M acetic acid.

Maybe that was not the question but until we know exactly what it was it is not really worth further thought