Question

Ammonia has a Kb of 1.8 × 10−5. Find [H3O+], [OH−], pH, and pOH for a 0.310 M ammonia solution.

Answer 1

Answer:

[H₃O⁺] = 4.3 × 10⁻¹² mol·L⁻¹; [OH⁻] = 2.4 × 10⁻³ mol·L⁻¹;

pH = 11.4; pOH = 2.6

Explanation:

The chemical equation is

$\rm NH _{3} + \text{H} _{2} O \, \rightleftharpoons \, NH _{4}^{+} + \,\text{OH} ^{-}$

For simplicity, let's re-write this as

$\rm B + H _{2} O \, \rightleftharpoons\, BH ^{+} + OH ^{-}$

1. Calculate [OH]⁻

(a) Set up an ICE table.  

   B + H₂O ⇌ BH⁺ + OH⁻

0.310               0        0

   -x                  +x      +x

0.310-x               x        x

$K_{\text{b}} = \dfrac{\text{[BH}^{+}]\text{[OH}^{-}]}{\text{[B]}} = 1.8 \times 10^{-5}\\\\\dfrac{x^{2}}{0.100 - x} = 1.8 \times 10^{-5}$

Check for negligibility:

$\dfrac{0.310}{1.8 \times 10^{-5}} = 17 000 > 400\\\\x \ll 0.310$

(b) Solve for [OH⁻]

$\dfrac{x^{2}}{0.310} = 1.8 \times 10^{-5}\\\\x^{2} = 0.310 \times 1.8 \times 10^{-5}\\x^{2} = 5.58 \times 10^{-6}\\x = \sqrt{5.58 \times 10^{-6}}\\x = \text{[OH]}^{-} = \mathbf{2.4 \times 10^{-3}} \textbf{ mol/L}$

2. Calculate the pOH

$\text{pOH} = -\log \text{[OH}^{-}] = -\log(2.4 \times 10^{-3}) = \mathbf{2.6}$

3. Calculate the pH

$\text{pH} = 14.00 - \text{pOH} = 14.00 - 2.6 = \mathbf{11.4}$

4 Calculate [H₃O⁺]

$\text{H _{3} O ^{+} } = 10^{-\text{pH}} = 10^{-11.4} = \mathbf{4.2 \times 10^{-12}} \textbf{ mol/L}$