How Do You Calculate NaCH3COO Concentration and Predict pH at Equivalence Point?

[cvc121]

Homework Statement

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 pK_b for NaCH₃COO or PK_a for CH₃COOH, predict the theoretical pH of the equivalence point.

Homework Equations

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:

n_titrant = n_analyte

n_HCl = n_NaCH3COO

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 molConcentration 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.

 

[Borek]

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).

 

[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?

 

[Borek]

Looks OK.

 

[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