Question

Calculate the pH at the equivalence point for the titration of 1.328 M HA (a weak acid with pKa = 6.95) with 1.328 M NaOH.

Answer 1

Let Volume of acid be 1 mL

use:

pKa = -log Ka

6.95 = -log Ka

Ka = 1.122*10^-7

find the volume of NaOH used to reach equivalence point

M(HA)*V(HA) =M(NaOH)*V(NaOH)

1.328 M *1.0 mL = 1.328M *V(NaOH)

V(NaOH) = 1 mL

Given:

M(HA) = 1.328 M

V(HA) = 1 mL

M(NaOH) = 1.328 M

V(NaOH) = 1 mL

mol(HA) = M(HA) * V(HA)

mol(HA) = 1.328 M * 1 mL = 1.328 mmol

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

mol(NaOH) = 1.328 M * 1 mL = 1.328 mmol

We have:

mol(HA) = 1.328 mmol

mol(NaOH) = 1.328 mmol

1.328 mmol of both will react to form A- and H2O

A- here is strong base

A- formed = 1.328 mmol

Volume of Solution = 1 + 1 = 2 mL

Kb of A- = Kw/Ka = 1*10^-14/1.122*10^-7 = 8.913*10^-8

concentration ofA-,c = 1.328 mmol/2 mL = 0.664M

A- dissociates as

A- + H2O -----> HA + OH-

0.664 0 0

0.664-x x x

Kb = [HA][OH-]/[A-]

Kb = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Kb = x*x/(c)

so, x = sqrt (Kb*c)

x = sqrt ((8.913*10^-8)*0.664) = 2.433*10^-4

since c is much greater than x, our assumption is correct

so, x = 2.433*10^-4 M

[OH-] = x = 2.433*10^-4 M

use:

pOH = -log [OH-]

= -log (2.433*10^-4)

= 3.6139

use:

PH = 14 - pOH

= 14 - 3.6139

= 10.3861

Answer: 10.39