Question

You start with 0.050 moles of ammonia in 500. mL of water. The equilibrium constant Keq is 1.8 × 10–5. What is the pH of this solution at equilibrium? Please show work!

Answer 1

Answer:

$\boxed{11.13}$

Explanation:

The chemical equation is

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

For simplicity, let's rewrite this as

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

1. Initial concentration of NH₃

$\text{[B]} = \dfrac{\text{moles}}{\text{litres}} = \dfrac{\text{0.050 mol}}{\text{0.500 L}} = \text{0.100 mol/L}$

2. Calculate [OH]⁻

We can use an ICE table to do the calculation.

                      B + H₂O ⇌ BH⁺ + OH⁻

I/mol·L⁻¹:     0.100              0         0

C/mol·L⁻¹:       -x                +x       +x

E/mol·L⁻¹:  0.100 - 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{0.100 }{1.8 \times 10^{-5}} = 5600 > 400\\\\ x \ll 0.100$

3. Solve for x

$\dfrac{x^{2}}{0.100} = 1.8 \times 10^{-5}\\\\x^{2} = 0.100 \times 1.8 \times 10^{-5}\\\\x^{2} = 1.80 \times 10^{-6}\\\\x = \sqrt{1.80 \times 10^{-6}}\\\\x = \text{[OH]}^{-} = 1.34 \times 10^{-3} \text{ mol/L}$

4. Calculate the pH

$\text{pOH} = -\log \text{[OH}^{-}] = -\log(1.34 \times 10^{-3}) = 2.87\\\\\text{pH} = 14.00 - \text{pOH} = 14.00 - 2.87 = \mathbf{11.13}\\\\\text{The pH of the solution at equilibrium is } \boxed{\mathbf{11.13}}$