Explanation

(a)
For 0.1 M acetic acid, or $C = 0.1\ mol\ L^{- 1}$
$K_{a} = 1.74 \times 10^{- 5}$
pH=?
According to Ostwald dilution formula,
$\left\lbrack H^{+} \right\rbrack = \sqrt{K_{a} \times C}$
$= \sqrt{1.7 \times 10^{- 5} \times 0.1}$
$= \sqrt{0.017 \times 10^{- 4}}$
$pH = - \log_{10}\left\lbrack H^{+} \right\rbrack$
$= - \log_{10}\sqrt{0.017 \times 10^{- 4}}$
pH=2.88