Question

Calculate the pH of a .10M NH4CL solution. The kb value for NH3 is 1.8×10^-5

Answer 1

pH = 4.63.

Explanation

NH₄Cl is a soluble salt. It dissociates completely into its ions when dissolved in water.

NH₄Cl → NH₄⁺ + Cl⁻.

All NH₄Cl in the 0.10 M solution dissociates. There's one mole of NH₄⁺ ions in each mole formula unit of NH₄Cl. In the end, the concentration of NH₄⁺ will also be 0.10 M.

The NH₄Cl salt comes from a weak base (NH₃) and a strong acid (HCl). It  would undergo hydrolysis after dissolving. Its solution will be slightly acidic.

The Kb value of NH₃ (or NH₄OH as in water) is quite small. As a result, NH₄OH tend to act as a weak base. The conjugate acid of NH₄OH is NH₄⁺. NH₄⁺ will undergo hydrolysis to produce NH₄OH and H⁺. In other words, NH₄⁺ will combine with water to produce the weak base NH₄OH and make the solution slightly acidic.

NH₄⁺ + H₂O → NH₄OH + H⁺

The initial concentration of NH₄⁺ ions is the same as that of NH₄Cl. How acidic is this solution? The acidity of this solution depends on the H⁺ concentration, which depends on the Ka value of NH₄⁺. Only the Kb of NH₃ was given. Luckily, the two species are conjugates acid and base of each other. The Kb of NH₃ relates to the Ka of NH₄⁺ by Kw, the ionization constant of water. The value of Kw under standard conditions is around 1.0 × 10⁻¹⁴.

$\text{K}_a({\text{NH}_4}^{+}) = \text{K}_w / \text{K}_b(\text{NH}_3) \\\phantom{\text{K}_a({\text{NH}_4}^{+})} = 1.0 \times 10^{-14} / (1.8 \times 10^{-5})$

Therefore, Ka(NH₄⁺) = 5.56 × 10⁻⁹.

Now make a RICE table for this reaction. Let x be the increase in the concentration of $\text{H}^{+}$ with the unit M (the same as $\text{mol} \cdot \text{L}^{-1}$).

$\begin{array}{cclclclc}\text{R} & {\text{NH}_4}^{+} & + & \text{H}_2\text{O} & \to & \text{NH}_4\text{OH} & + & \text{H}^{+}\\ \text{I} & 0.10 \\ \text{C} & - x & & & & +x & & + x\\ \text{E} & 0.10 -x & & & & x & & x \end{array}$

Ka of NH₄⁺ is so small that the reaction barely takes place. Note the expression for Ka.

$\text{K}_a = [\text{NH}_4\text{OH}] \cdot [\text{H}^{+}] / [{\text{NH}_4}^{+}] \\\phantom{\text{K}_a} = x^{2} / (0.10 - x)$

$x^{2} / (0.10 - x) = 5.56 \times 10^{-9}$

x will be much smaller than 0.10. (0.10 - x) will be nearly the same as 0.10. However, the x² in the numerator is a standalone term. The same approximation shall not work for x². To simplify the calculation:

$x^{2} / (0.10 - x ) \approx x^{2} / 0.10$

$x^{2} / 0.10 \approx \text{K}_a({\text{NH}_4}^{+}) = 5.56 \times 10^{-9}$

$x \approx \sqrt{5.56 \times 10^{-9} \times 0.10} = 2.36 \times 10^{-5}$.

As a result

$[\text{H}^{+}] = x = 2.36 \times 10^{-5} \; \text{mol}\cdot \text{L}^{-1}$,

$\text{pH} = -\log_{10}{[\text{H}^{+}] = 4.63$.