Question

A 21.6 mL sample of 0.374 M ammonia, NH₃, is titrated with 0.331 M hydrochloric acid.

After adding 10.6 mL of hydrochloric acid, the pH is

.

Use the Tables link in the References for any equilibrium constants that are required.

Answer 1

Given:

M(HCl) = 0.331 M

V(HCl) = 10.6 mL

M(NH3) = 0.374 M

V(NH3) = 21.6 mL

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

mol(HCl) = 0.331 M * 10.6 mL = 3.5086 mmol

mol(NH3) = M(NH3) * V(NH3)

mol(NH3) = 0.374 M * 21.6 mL = 8.0784 mmol

We have:

mol(HCl) = 3.5086 mmol

mol(NH3) = 8.0784 mmol

3.5086 mmol of both will react

excess NH3 remaining = 4.5698 mmol

Volume of Solution = 10.6 + 21.6 = 32.2 mL

[NH3] = 4.5698 mmol/32.2 mL = 0.1419 M

[NH4+] = 3.5086 mmol/32.2 mL = 0.109 M

They form basic buffer

base is NH3

conjugate acid is NH4+

Kb = 1.8*10^-5

pKb = - log (Kb)

= - log(1.8*10^-5)

= 4.745

use:

pOH = pKb + log {[conjugate acid]/[base]}

= 4.745+ log {0.109/0.1419}

= 4.63

use:

PH = 14 - pOH

= 14 - 4.63

= 9.37

Answer: 9.37