Question

Calculate the pH when 30.0 mL of 0.200 M HBr is mixed with 30.0 mL of 0.400 M CH₃NH₂ (Kb = 4.4 × 10⁻⁴).

Answer 1

Given:

M(HBr) = 0.2 M

V(HBr) = 30 mL

M(CH3NH2) = 0.4 M

V(CH3NH2) = 30 mL

mol(HBr) = M(HBr) * V(HBr)

mol(HBr) = 0.2 M * 30 mL = 6 mmol

mol(CH3NH2) = M(CH3NH2) * V(CH3NH2)

mol(CH3NH2) = 0.4 M * 30 mL = 12 mmol

We have:

mol(HBr) = 6 mmol

mol(CH3NH2) = 12 mmol

6 mmol of both will react

excess CH3NH2 remaining = 6 mmol

Volume of Solution = 30 + 30 = 60 mL

[CH3NH2] = 6 mmol/60 mL = 0.1 M

[CH3NH3+] = 6 mmol/60 mL = 0.1 M

They form basic buffer

base is CH3NH2

conjugate acid is CH3NH3+

Kb = 4.4*10^-4

pKb = - log (Kb)

= - log(4.4*10^-4)

= 3.357

use:

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

= 3.357+ log {0.1/0.1}

= 3.357

use:

PH = 14 - pOH

= 14 - 3.3565

= 10.6435

Answer: 10.64