A solution is 1.00 x 10^-3 M in HCl and 0.100 M in CH3COOH (K_a = 1.8 x 10^-5). Using the exact quadratic for the ionisation of acetic acid, the pH is

JEE Main levelSingle correct optionDifficulty 2 of 3ia097

A solution is $1.00 \times 1 0^{- 3} M$ in $HCl$ and $0.100 M$ in $CH X_{3} COOH$ ( $K_{a} = 1.8 \times 1 0^{- 5}$ ). Using the exact quadratic for the ionisation of acetic acid, the pH is

Let

x

[H X^{+}]

from the acid:

x (0.001 + x) / (0.100 - x) = 1.8 \times 1 0^{- 5}

, so

x = 9.26 \times 1 0^{- 4} M

. Total

[H X^{+}] = 1.93 \times 1 0^{- 3} M

and pH = 2.72. Ignoring the common-ion suppression would give an overestimate of

[H X^{+}]