Question

A) A chemist titrates a 25.0 mL sample of 0.191 M benzoic acid (C6H5COOH) against a 0.100 M solution of NaOH. The overall reaction is shown by the equation below: $$C6​H5​COOH(aq)+NaOH(aq) C6​H5​COONa(aq)+H2​O(l) The Ka value for benzoic acid is 6.28 × 10–5. Calculate the pH at the start of the titration, before any NaOH has been added. Give your answer correctly to two places after the decimal.

B) What will the pH be when 12.0 mL of 0.100 M NaOH has been added to the weak acid solution? Give your answer to two places after the decimal.

Answer 1

A)

C6H5COOH dissociates as:

C6H5COOH -----> H+ + C6H5COO-

0.191 0 0

0.191-x x x

Ka = [H+][C6H5COO-]/[C6H5COOH]

Ka = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Ka = x*x/(c)

so, x = sqrt (Ka*c)

x = sqrt ((6.28*10^-5)*0.191) = 3.463*10^-3

since x is comparable c, our assumption is not correct

we need to solve this using Quadratic equation

Ka = x*x/(c-x)

6.28*10^-5 = x^2/(0.191-x)

1.199*10^-5 - 6.28*10^-5 *x = x^2

x^2 + 6.28*10^-5 *x-1.199*10^-5 = 0

This is quadratic equation (ax^2+bx+c=0)

a = 1

b = 6.28*10^-5

c = -1.199*10^-5

Roots can be found by

x = {-b + sqrt(b^2-4*a*c)}/2a

x = {-b - sqrt(b^2-4*a*c)}/2a

b^2-4*a*c = 4.798*10^-5

roots are :

x = 3.432*10^-3 and x = -3.495*10^-3

since x can't be negative, the possible value of x is

x = 3.432*10^-3

use:

pH = -log [H+]

= -log (3.432*10^-3)

= 2.4644

Answer: 2.46

B)

Given:

M(C6H5COOH) = 0.191 M

V(C6H5COOH) = 25 mL

M(NaOH) = 0.1 M

V(NaOH) = 12 mL

mol(C6H5COOH) = M(C6H5COOH) * V(C6H5COOH)

mol(C6H5COOH) = 0.191 M * 25 mL = 4.775 mmol

mol(NaOH) = M(NaOH) * V(NaOH)

mol(NaOH) = 0.1 M * 12 mL = 1.2 mmol

We have:

mol(C6H5COOH) = 4.775 mmol

mol(NaOH) = 1.2 mmol

1.2 mmol of both will react

excess C6H5COOH remaining = 3.575 mmol

Volume of Solution = 25 + 12 = 37 mL

[C6H5COOH] = 3.575 mmol/37 mL = 0.0966M

[C6H5COO-] = 1.2/37 = 0.0324M

They form acidic buffer

acid is C6H5COOH

conjugate base is C6H5COO-

Ka = 6.28*10^-5

pKa = - log (Ka)

= - log(6.28*10^-5)

= 4.202

use:

pH = pKa + log {[conjugate base]/[acid]}

= 4.202+ log {3.243*10^-2/9.662*10^-2}

= 3.728

Answer: 3.73