Question

a 40mL weak base that has a concentration of 0.25M will be titrated with 0.62M HCl. Calculate the pH of the solution upon addition of 16.129mL HCl. The Kb of the base is 7.6E-6

Answer 1

Given:

M(HCl) = 0.62 M

V(HCl) = 16.129 mL

M(B) = 0.25 M

V(B) = 40 mL

mol(HCl) = M(HCl) * V(HCl)

mol(HCl) = 0.62 M * 16.129 mL = 10 mmol

mol(B) = M(B) * V(B)

mol(B) = 0.25 M * 40 mL = 10 mmol

We have:

mol(HCl) = 10 mmol

mol(B) = 10 mmol

10 mmol of both will react to form BH+ and H2O

BH+ here is strong acid

BH+ formed = 10 mmol

Volume of Solution = 16.129 + 40 = 56.129 mL

Ka of BH+ = Kw/Kb = 1.0E-14/7.6E-6 = 1.316*10^-9

concentration ofBH+,c = 10 mmol/56.129 mL = 0.1782 M

BH+ + H2O -----> B + H+

0.1782 0 0

0.1782-x x x

Ka = [H+][B]/[BH+]

Ka = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Ka = x*x/(c)

so, x = sqrt (Ka*c)

x = sqrt ((1.316*10^-9)*0.1782) = 1.531*10^-5

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

so, x = 1.531*10^-5 M

[H+] = x = 1.531*10^-5 M

use:

pH = -log [H+]

= -log (1.531*10^-5)

= 4.815

Answer: 4.81