Acids, bases and titration — Question 128

Answer
$[B{]}_{initial}=0.0800\mathrm{m}\mathrm{o}\mathrm{l}{\mathrm{L}}^{-1}$; $p{K}_b=4.80$ and $K_b=1.58\times {10}^{-5}$. The conjugate acid reacts with water to form hydronium ions.

Working
At equivalence, $n(B)=0.100(0.0200)=0.00200$ mol, so $c=0.00200/0.0250$. At half-equivalence $p{K}_b=pOH=14.00-9.20=4.80$; $K_b={10}^{-4.80}$. The salt contains the weak conjugate acid: $\mathrm{B}{\mathrm{H}}^{+}(\mathrm{a}\mathrm{q})+{\mathrm{H}}_2\mathrm{O}(\mathrm{l})\rightleftharpoons \mathrm{B}(\mathrm{a}\mathrm{q})+{\mathrm{H}}_3{\mathrm{O}}^{+}(\mathrm{a}\mathrm{q})$. Hydronium formation makes the equivalence solution acidic.

Marking criteria

• Calculates initial base amount = 0.00200 mol. [1 mark]
• Calculates base concentration = 0.0800 mol L⁻¹. [1 mark]
• Uses pKb = pOH = 4.80 at half-equivalence. [1 mark]
• Calculates Kb = 1.58 × 10⁻⁵. [1 mark]
• Identifies BH⁺ as the weak conjugate acid. [1 mark]
• Gives its reaction with water producing hydronium and links this to pH < 7. [1 mark]