$100 mL$ of a solution that is $0.040 M$ in $HCl$ and $0.060 M$ in $CH X_{3} COOH$ ( $p K_{a} = 4.74$ ) is titrated with $0.10 M$ $NaOH$ . What total volume of $NaOH$ (in $mL$ , nearest integer) is required to bring the pH to $5.00$ ? Take $1 0^{0.26} = 1.82$ .

Numerical answer type, answer in mL, tolerance 0.5.

Answer: 79 mL

Strong acid:

4.0 mmol

is neutralised first. Then pH

= p K_{a} + lo g ([A X^{-}] / [HA])

gives

[A X^{-}] / [HA] = 1 0^{0.26} = 1.82

. Of

6.0 mmol

acetic acid,

A X^{-} = 6.0 \times 1.82/2.82 = 3.87 mmol

. Total

NaOH = 4.0 + 3.87 = 7.87 mmol

, i.e. 78.7 mL, about 79 mL.