$20.0 mL$ of $0.100 M$ acetic acid ( $p K_{a} = 4.74$ ) is titrated with $0.100 M$ NaOH. What volume of NaOH (in mL) has been added when the pH is $5.00$ ? Use Henderson's equation; give the answer to one decimal.

Numerical answer type, answer in mL, tolerance 0.1.

Answer: 12.907049 mL

lo g ([A X^{-}] / [HA]) = 5.00 - 4.74 = 0.26

, ratio

= 1.820

. Fraction neutralised

= r / (1 + r) = 0.6454

, so

V = 20.0 \times 0.6454 = 12.9 mL