Question

What is the pH at the equivalence point when 95.0 mL of a 0.275 M solution of acetic acid (CH3COOH) is titrated with 0.100 M NaOH to its end point?

Answer 1

find the volume of NaOH used to reach equivalence point

M(CH3COOH)*V(CH3COOH) =M(NaOH)*V(NaOH)

0.275 M *95.0 mL = 0.1M *V(NaOH)

V(NaOH) = 261.25 mL

Given:

M(CH3COOH) = 0.275 M

V(CH3COOH) = 95 mL

M(NaOH) = 0.1 M

V(NaOH) = 261.25 mL

mol(CH3COOH) = M(CH3COOH) * V(CH3COOH)

mol(CH3COOH) = 0.275 M * 95 mL = 26.125 mmol

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

mol(NaOH) = 0.1 M * 261.25 mL = 26.125 mmol

We have:

mol(CH3COOH) = 26.125 mmol

mol(NaOH) = 26.125 mmol

26.125 mmol of both will react to form CH3COO- and H2O

CH3COO- here is strong base

CH3COO- formed = 26.125 mmol

Volume of Solution = 95 + 261.25 = 356.25 mL

Kb of CH3COO- = Kw/Ka = 1*10^-14/1.8*10^-5 = 5.556*10^-10

concentration ofCH3COO-,c = 26.125 mmol/356.25 mL = 0.0733M

CH3COO- dissociates as

CH3COO- + H2O -----> CH3COOH + OH-

0.0733 0 0

0.0733-x x x

Kb = [CH3COOH][OH-]/[CH3COO-]

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 ((5.556*10^-10)*7.333*10^-2) = 6.383*10^-6

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

so, x = 6.383*10^-6 M

[OH-] = x = 6.383*10^-6 M

use:

pOH = -log [OH-]

= -log (6.383*10^-6)

= 5.195

use:

PH = 14 - pOH

= 14 - 5.195

= 8.805

Answer: 8.805