Chemistry - Ionic Equilibrium Question with Solution | TestHub

(A) Charge balance gives $[{\mathrm{H}}^{+}]$ = $[{\mathrm{HA}}^{-}]$ + $2[{\mathrm{A}}^{2-}]$, so the factor on $[{\mathrm{A}}^{2-}]$ is wrong: false.

(B) $K_{a1}$ >> $K_{a2}$, so $[{\mathrm{H}}^{+}]$ = √($K_{a1}$ C) = √($10^{-6}$ × 0.01) = $10^{-4}$: pH = 4. True.

(C) 100 mL × 0.01 M = 1 mmol ${\mathrm{H}}_2\mathrm{A}$. Adding 1 mmol NaHA gives ${\mathrm{H}}_2\mathrm{A}$ : ${\mathrm{HA}}^{-}$ = 1 : 1, so pH = $\mathrm{p}{K}_{a1}$ = 6. The pH rises from 4 to 6: 2 units, true.

(D) ${\mathrm{H}}_2\mathrm{A}$ is dibasic, so complete neutralisation needs 2 mmol NaOH = 200 mL of 0.01 M: false.

Answer: (B), (C).