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

`\boxed{11.13}`

Step-by-step 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]} = \frac{\text{moles}}{\text{litres}} = \frac{\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}} = \frac{\text{[BH}^(+)]\text{[OH}^(-)]}{\text{[B]}} = 1.8 * 10^(-5)\\\\(x^(2))/(0.100 - x) = 1.8 * 10^(-5)`

Check for negligibility:

\
`(0.100 )/(1.8 * 10^(-5)) = 5600 > 400\\\\ x \ll 0.100`

3. Solve for x

`(x^(2))/(0.100) = 1.8 * 10^(-5)\\\\x^(2) = 0.100 * 1.8 * 10^(-5)\\\\x^(2) = 1.80 * 10^(-6)\\\\x = \sqrt{1.80 * 10^(-6)}\\\\x = \text{[OH]}^(-) = 1.34 * 10^(-3) \text{ mol/L}`

4. Calculate the pH

`\text{pOH} = -\log \text{[OH}^(-)] = -\log(1.34 * 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}}`