Question

Compare the pH at equivalence point for the titration of: a) 10.0 mL of 0.0250 M acetic acid (pka = 4.756) with the standard 0.0125 M NaOH solution and b) 10.0 mL of 0.0250 M HCl with the standard 0.0125 M NaOH solution

Answer 1

1)

use:

pKa = -log Ka

4.756 = -log Ka

Ka = 1.754*10^-5

find the volume of NaOH used to reach equivalence point

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

0.025 M *10.0 mL = 0.0125M *V(NaOH)

V(NaOH) = 20 mL

Given:

M(CH3COOH) = 0.025 M

V(CH3COOH) = 10 mL

M(NaOH) = 0.0125 M

V(NaOH) = 20 mL

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

mol(CH3COOH) = 0.025 M * 10 mL = 0.25 mmol

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

mol(NaOH) = 0.0125 M * 20 mL = 0.25 mmol

We have:

mol(CH3COOH) = 0.25 mmol

mol(NaOH) = 0.25 mmol

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

CH3COO- here is strong base

CH3COO- formed = 0.25 mmol

Volume of Solution = 10 + 20 = 30 mL

Kb of CH3COO- = Kw/Ka = 1*10^-14/1.754*10^-5 = 5.702*10^-10

concentration ofCH3COO-,c = 0.25 mmol/30 mL = 0.0083M

CH3COO- dissociates as

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

0.0083 0 0

0.0083-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 ((5.702*10^-10)*8.333*10^-3) = 2.18*10^-6

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

so, x = 2.18*10^-6 M

[OH-] = x = 2.18*10^-6 M

use:

pOH = -log [OH-]

= -log (2.18*10^-6)

= 5.6616

use:

PH = 14 - pOH

= 14 - 5.6616

= 8.3384

Answer: 8.34

b)

This is titration of strong acid and strong base

At equivalence point, equal mol of strong acid and strong base would react to form neutral and water

So, pH at equivalence point would be 7.00

Answer: 7.00