Question

250.0 mL of 0.10 M HA (Ka = 1.0 x 10-4 ) is mixed with 100.0 mL 0.25 M KOH. What is the pH?

Answer 1

Given:

M(HA) = 0.1 M

V(HA) = 250 mL

M(KOH) = 0.25 M

V(KOH) = 100 mL

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

mol(HA) = 0.1 M * 250 mL = 25 mmol

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

mol(KOH) = 0.25 M * 100 mL = 25 mmol

We have:

mol(HA) = 25 mmol

mol(KOH) = 25 mmol

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

A- here is strong base

A- formed = 25 mmol

Volume of Solution = 250 + 100 = 350 mL

Kb of A- = Kw/Ka = 1*10^-14/1*10^-4 = 1*10^-10

concentration ofA-,c = 25 mmol/350 mL = 0.0714M

A- dissociates as

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

0.0714 0 0

0.0714-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*10^-10)*7.143*10^-2) = 2.673*10^-6

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

so, x = 2.673*10^-6 M

[OH-] = x = 2.673*10^-6 M

use:

pOH = -log [OH-]

= -log (2.673*10^-6)

= 5.5731

use:

PH = 14 - pOH

= 14 - 5.5731

= 8.4269

Answer: 8.43