Question

What is the pH of a solution of 0.400 M CH₃NH₂ containing 0.250 M CH₃NH₃I? (Kb of CH₃NH₂ is 4.4 × 10⁻⁴)

Answer 1

Answer:

$\boxed{\text{10.84}}$

Explanation:

A solution of a weak base and its conjugate acid is a buffer.

The equation for the equilibrium is

$\rm CH _3 NH _2 + H _2 O \, \rightleftharpoons \, CH _3 NH _2 + H _{3} O ^{+} \\\text{For ease of typing, let's rewrite this equation as}\\\rm B + H _2 O \longrightarrow \, BH ^{+} + OH ^{-} ; K_{\text{b}} = 4.4 \times 10^{-4}$

The Henderson-Hasselbalch equation for a basic buffer is

$\text{pOH} = \text{p}K_{\text{b}} + \log\dfrac{[\text{BH}^{+}]}{\text{[B]}}$

Data:

   [B] = 0.400 mol·L⁻¹

[BH⁺] = 0.250 mol·L⁻¹

    Kb = 4.4 × 10⁻⁴

Calculations:

(a) Calculate pKb

pKb = -log(4.4× 10⁻⁴)  = 3.36

(b) Calculate the pH

$\text{pOH} = 3.36 + \log \dfrac{0.250}{0.400} = 3.36 + \log 0.625 = 3.36 - 0.204 = 3.16\\\\\text{pH} =14.00 -3.16 = \mathbf{10.84}\\\\\text{The pH of the solution is }\boxed{\textbf{10.84}}$