$50.0 mL$ of $0.100 M$ acetic acid ( $K_{a} = 1.8 \times 1 0^{- 5}$ ) is titrated with $0.100 M$ NaOH using an indicator that changes colour at pH $7.00$ . Take the end point as pH $= 7.00$ . Calculate the titration error as a percentage of the equivalence volume (negative if the end point is reached before equivalence). Use the charge balance and give two decimals.

Numerical answer type, answer in %, tolerance 0.02.

Answer: -0.552486 %

Charge balance:

[Na X^{+}] + [H X^{+}] = [OH X^{-}] + [A X^{-}]

; at pH 7,

[H X^{+}] = [OH X^{-}]

, so

[Na X^{+}] = [A X^{-}] = c_{T} K_{a} / (K_{a} + [H X^{+}])

, where

c_{T}

is the total acetate species. Hence

V_{N a O H} = 50.0 \times K_{a} / (K_{a} + 1 0^{- 7}) = 50.0 \times 0.99448 = 49.72 mL

. Error

= (49.72 - 50.00) /50.0 = - 0.55%