Question

Calculate the pH when 15.0 mL of 0.20 M benzoic acid (HC7H5O2, Ka = 6.5 ×10–5) is titrated with 20.0 mL of 0.15 M sodium hydroxide (NaOH).

Answer 1

we have:

Molarity of C6H5COOH = 0.2 M

Volume of C6H5COOH = 15 mL

Molarity of NaOH = 0.15 M

Volume of NaOH = 20 mL

mol of C6H5COOH = Molarity of C6H5COOH * Volume of C6H5COOH

mol of C6H5COOH = 0.2 M * 15 mL = 3 mmol

mol of NaOH = Molarity of NaOH * Volume of NaOH

mol of NaOH = 0.15 M * 20 mL = 3 mmol

We have:

mol of C6H5COOH = 3 mmol

mol of NaOH = 3 mmol

3 mmol of both will react to form C6H5COO- and H2O

C6H5COO- here is strong base

C6H5COO- formed = 3 mmol

Volume of Solution = 15 + 20 = 35 mL

Kb of C6H5COO- = Kw/Ka = 1*10^-14/6.5*10^-5 = 1.538*10^-10

concentration ofC6H5COO-,c = 3 mmol/35 mL = 0.0857M

C6H5COO- dissociates as

C6H5COO- + H2O -----> C6H5COOH + OH-

0.0857 0 0

0.0857-x x x

Kb = [C6H5COOH][OH-]/[C6H5COO-]

Kb = x*x/(c-x)

Assuming small x approximation, that is lets assume that x can be ignored as compared to c

So, above expression becomes

Kb = x*x/(c)

so, x = sqrt (Kb*c)

x = sqrt ((1.538*10^-10)*8.571*10^-2) = 3.631*10^-6

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

so, x = 3.631*10^-6 M

[OH-] = x = 3.631*10^-6 M

we have below equation to be used:

pOH = -log [OH-]

= -log (3.631*10^-6)

= 5.4399

we have below equation to be used:

PH = 14 - pOH

= 14 - 5.4399

= 8.5601

Answer: 8.56