Question

The weak base ammonia, NH3, and the strong acid hydrochloric acid react to form the salt ammonium chloride, NH4Cl. Given that the value of Kb for ammonia is 1.8×10−5, what is the pH of a 0.289 M solution of ammonium chloride at 25∘C

Answer 1

Answer:

$\boxed{2.64}$

Explanation:

The equation for the equilibrium is

NH₄⁺ + H₂O ⇌ NH₃ + H₃O⁺

1. Set up an ICE table

$\begin{array}{ccccccc}\text{NH _{4}^{+} } & + & \text{H _{2} O}& \, \rightleftharpoons \, & \text{NH _{3} } & + & \text{H _{3} O ^{+} }\\0.289 & & & &0 & & 0 \\-x & & & &+x & & +x \\0.289-x & & & &x & & x \\\end{array}$

2. Solve for x

$K_{\text{c}} = \dfrac{\text{[NH _{4}^{+} ][OH ^{-} ]}}{\text{[NH _3 ]}} = 1.8 \times 10^{-5}\\\\\dfrac{x^{2}}{0.289-x} = 1.8 \times 10^{-5}\\\\\text{Check for negligibility of }x\\\\\dfrac{0.289 }{1.8 \times 10^{-5}} = 16 000 > 400\\\\\therefore x \ll 0.289\\\\\begin{array}{rcl}\\\dfrac{x^{2}}{0.289}& = & 1.8 \times 10^{-5}\\\\x^{2}& = & 0.289 \times1.8 \times 10^{-5}\\x^{2}& = & 5.202 \times 10^{-6}\\x & = &2.281 \times 10^{-3}\\\end{array}$

3. Calculate the pH

$\text{[H _{3} O ^{+} ]}= x \text{ mol \cdot L ^{-1} } = 2.281 \times 10^{-3} \text{ mol \cdot L ^{-1} }\\\text{pH} = -\log{\rm[H_{3}O^{+}]} = -\log{2.281 \times 10^{-3}} = \boxed{\mathbf{2.64}}\\\text{The pH of the solution is } \boxed{\textbf{2.64}}$