What is the pH of a 0.26 mol/L solution of methylamine ( ${CH}_{3} {NH}_{2}, K_{b} = 4.4 \times 10^{-4}$ $"CH"_3"NH"_2, K_text(b) = 4.4 × 10^"-4"$ )?

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What is the pH of a 0.26 mol/L solution of methylamine ( ${CH}_{3} {NH}_{2}, K_{b} = 4.4 \times 10^{-4}$ $"CH"_3"NH"_2, K_text(b) = 4.4 × 10^"-4"$ )?

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Answer 1 · 2017-04-21T00:38:52

pH = 12.03

Explanation:

We can use an ICE table to calculate the concentrations of the ions in solution.

The chemical equation is

${CH}_{3} {NH}_{2} + H_{2} O ⇌ {CH}_{3} {NH}_{3}^{+} + {OH}^{-}; K_{b} = 4.4 \times 10^{-4}$ $"CH"_3"NH"_2 + "H"_2"O" ⇌ "CH"_3"NH"_3^"+" + "OH"^"-"; K_text(b) = 4.4 × 10^"-4"$

Let's rewrite this as

$m m m m m m m m m {B + H}_{2} O ⇌ {BH}^{+} + {OH}^{-}$ $color(white)(mmmmmmmmm)"B + H"_2"O" ⇌ "BH"^"+" + "OH"^"-"$
${I/mol\cdotL}^{-1} : m m m 0.26 m m m m m m 0 m m m 0$ $"I/mol·L"^"-1":color(white)(mmm)0.26color(white)(mmmmmm)0color(white)(mmm)0$
${C/mol\cdotL}^{-1} : m m m - x m m m m m m + x m m + x$ $"C/mol·L"^"-1":color(white)(mmm)"-"xcolor(white)(mmmmmm)"+"xcolor(white)(mm)"+"x$
${E/mol\cdotL}^{-1} : m l 0.26 - x m m m m m l x m m l l x$ $"E/mol·L"^"-1":color(white)(ml)0.26-xcolor(white)(mmmmml)xcolor(white)(mmll)x$

$K_{b} = \frac{[{BH}^{+}] [{OH}^{-}]}{[B]} = \frac{x \times x}{0.26 - x} = \frac{x^{2}}{0.26 - x} = 4.4 \times 10^{-4}$ $K_text(b) = (["BH"^"+"]["OH"^"-"])/(["B"]) = (x × x)/(0.26-x) = x^2/(0.26-x) = 4.4 × 10^"-4"$

Check for negligibility:

$\frac{0.26}{4.4 \times 10^{-4}} = 5900 ≫ 400$ $0.26/(4.4 × 10^"-4") = 5900 ≫ 400$ .

∴ $x ≪ 0.26$ $x ≪ 0.26$

Then

$\frac{x^{2}}{0.26} = 4.4 \times 10^{-4}$ $x^2/0.26 = 4.4 × 10^"-4"$

$x^{2} = 0.26 \times 4.4 \times 10^{-4} = 1.14 \times 10^{-4}$ $x^2 = 0.26 × 4.4 × 10^"-4" = 1.14 × 10^"-4"$

$x = 1.07 \times 10^{-2}$ $x = 1.07 × 10^"-2"$

$[{OH}^{-}] = x l mol/L = 1.07 \times 10^{-2} l mol/L$ $["OH"^"-"] = x color(white)(l)"mol/L" = 1.07 × 10^"-2"color(white)(l)"mol/L"$

$pOH = -log [{OH}^{-}] = -log (1.07 \times 10^{-2}) = 1.97$ $"pOH" = "-log"["OH"^"-"] = "-log"(1.07 × 10^"-2") = 1.97$

$pH = 14.00 - pH = 14.00 - 1.97 = 12.03$ $"pH" = "14.00 - pH" = "14.00 - 1.97" = 12.03$

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What is the pH of a 0.26 mol/L solution of methylamine ( ${CH}_{3} {NH}_{2}, K_{b} = 4.4 \times 10^{-4}$ $"CH"_3"NH"_2, K_text(b) = 4.4 × 10^"-4"$ )?

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What is the pH of a 0.26 mol/L solution of methylamine (CH3NH2,Kb=4.4×10-4"CH"_3"NH"_2, K_text(b) = 4.4 × 10^"-4")?

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Explanation:

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What is the pH of a 0.26 mol/L solution of methylamine ( ${CH}_{3} {NH}_{2}, K_{b} = 4.4 \times 10^{-4}$ $"CH"_3"NH"_2, K_text(b) = 4.4 × 10^"-4"$ )?