Acids, bases and titration — Question 44

At equivalence $n(HA)=0.100\times 0.0200=0.00200$ mol, so $c=0.00200/0.0250$. Half-equivalence has equal HA and A⁻, hence pH=pKa=5.00. At equivalence ${\mathrm{A}}^{-}+{\mathrm{H}}_2\mathrm{O}\rightleftharpoons \mathrm{H}\mathrm{A}+\mathrm{O}{\mathrm{H}}^{-}$.

• Calculates acid moles 0.00200 mol. [1 marks]
• Calculates concentration 0.0800 mol L⁻¹. [1 marks]
• Uses pH = pKa at half-equivalence. [1 marks]
• Calculates Ka = 1.0 × 10⁻⁵. [1 marks]
• Identifies the conjugate base as basic. [1 marks]
• Explains hydrolysis produces OH⁻. [1 marks]