Question

Calculate the pH when 65.0 mL of 0.269 M of a certain monoprotic weak acid, HA, is mixed with 65.0 mL of 0.269 M sodium hydroxide solution at 25 °C. For HA, the Ka is 6.1× 10–5.

Answer 1

Given:

M(HA) = 0.269 M

V(HA) = 65 mL

M(NaOH) = 0.269 M

V(NaOH) = 65 mL

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

mol(HA) = 0.269 M * 65 mL = 17.485 mmol

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

mol(NaOH) = 0.269 M * 65 mL = 17.485 mmol

We have:

mol(HA) = 17.485 mmol

mol(NaOH) = 17.485 mmol

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

A- here is strong base

A- formed = 17.485 mmol

Volume of Solution = 65 + 65 = 130 mL

Kb of A- = Kw/Ka = 1*10^-14/6.1*10^-5 = 1.639*10^-10

concentration ofA-,c = 17.485 mmol/130 mL = 0.1345M

A- dissociates as

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

0.1345 0 0

0.1345-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 ((1.639*10^-10)*0.1345) = 4.696*10^-6

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

so, x = 4.696*10^-6 M

[OH-] = x = 4.696*10^-6 M

use:

pOH = -log [OH-]

= -log (4.696*10^-6)

= 5.33

use:

PH = 14 - pOH

= 14 - 5.33

= 8.67

pH = 8.67