Question

23. A 100.0 mL sample of 0.100 mol L-1 weak monoprotic acid is titrated with 0.0500 mol L-1 KOH. Determine the pH of the solution after the addition of 200.0 mL of KOH at 25 °C. The Ka of the weak monoprotic acid is 1.52 x 10-5. A) 5.330 B) 8.671 5.064 D) 11.091 E) 8.909

Answer 1

Given:

M(HA) = 0.1 M

V(HA) = 100 mL

M(KOH) = 0.05 M

V(KOH) = 200 mL

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

mol(HA) = 0.1 M * 100 mL = 10 mmol

mol(KOH) = M(KOH) * V(KOH)

mol(KOH) = 0.05 M * 200 mL = 10 mmol

We have:

mol(HA) = 10 mmol

mol(KOH) = 10 mmol

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

A- here is strong base

A- formed = 10 mmol

Volume of Solution = 100 + 200 = 300 mL

Kb of A- = Kw/Ka = 1*10^-14/1.52*10^-5 = 6.579*10^-10

concentration ofA-,c = 10 mmol/300 mL = 0.0333M

A- dissociates as

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

0.0333 0 0

0.0333-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 ((6.579*10^-10)*3.333*10^-2) = 4.683*10^-6

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

so, x = 4.683*10^-6 M

[OH-] = x = 4.683*10^-6 M

use:

pOH = -log [OH-]

= -log (4.683*10^-6)

= 5.3295

use:

PH = 14 - pOH

= 14 - 5.3295

= 8.671

Answer: B