Question

500. mL of a solution containing 1.5 M NH4Cl (aq) is mixed with 500.mL of a solution containing 0.50 M of NaOH (aq). What is the pH of the final solution (Kb(NH3)=1.8*10^-5)

Answer 1

Question

500 mL of a solution containing 1.5 M NH₄Cl(aq) is mixed with 500 mL of a solution containing 0.5 M of NaOH(aq). What is the pH of the final solution (K_b(NH_3)=1.8·10⁻⁵)?

Solution

NH₄Cl + NaOH → NaCl + NH₄OH

Initially:

n(NH₄Cl) init. = V(NH₄Cl)·C(NH₄Cl) = 0.5·1.5 = 0.75 mol

n(NaOH) init. = V(NaOH)·C(NaOH) = 0.5·0.5 = 0.25 mol

After mixing:

n(NaOH) = 0 mol – it has been completely reacted.

n(NH₄Cl) reacted = n(NaOH) init. = 0.25 mol

n(NH₄Cl) = n(NH₄Cl) init. - n(NH₄Cl) reacted = 0.75 – 0.25 = 0.50 mol

n(NH₄OH) = n(NH₄Cl) reacted = 0.25 mol

n(NaCl) = 0.25 mol, but it has no effect on the final solution.

The final solution volume V = V(NH₄Cl) + V(NaOH) = 0.5 + 0.5 = 1.0 L

The final solution is an ammonia buffer solution. The ammonia buffer solution pH is calculated by the formula

$$pH = 14 - pK_b - \lg \frac{C_{NH_4^+}}{C_{NH_3}}$$ $$C_{NH_4^+} = \frac{n(NH_4Cl)}{V} = \frac{0.5}{1.0} = 0.5 \text{ mol/L}$$ $$C_{NH_3} = \frac{n(NH_4OH)}{V} = \frac{0.25}{1.0} = 0.25 \text{ mol/L}$$ $$pK_b = - \lg K_b$$

Now we can calculate the pH of final solution

$$pH = 14 - pK_b - \lg \frac{C_{NH_4^+}}{C_{NH_3}} = 14 + \lg 1.8 \cdot 10^{-5} - \lg \frac{0.5}{0.25} = 8.95$$

Answer: pH = 8.95