Question

Calculate $H^{+}$ ion concentration from acetic acid $\left(C{H}_{3}COOH\right)$ if 0.2 mole of $HCl$ is added to 1 L of the solution
For acetic acid $K_a=1.8\times {10}^{-5}$

• A. $2\times {10}^{-5}M$
• B. $5.4\times {10}^{-7}M$
• C. $3.6\times {10}^{-6}M$
• D. $1.8\times {10}^{-7}M$

Answer 1

The correct option is D $1.8\times {10}^{-7}M$

Let x be the concentration (in molarity) of acetic acid reacted i.e. x moles of acetic acid is present in $1L$ of solution.
Since $HCl$ is a strong acid, so it fully dissociates.
$\therefore [H^{+}{]}_{HCl}=0.2M$

$C{H}_{3}COOH\rightleftharpoons C{H}_{3}CO{O}^{-}+H^{+}Initial:0.0200Equilibrium:0.02-xxx+0.2\left[C{H}_{3}COOH\right]=\frac{\left(x\right)(x+0.2)}{(0.02-x)}SinceC{H}_{3}COOH$ is a weak acid so, 'x' will be very small and can be neglected
so, $0.02-x\approx 0.02andx+0.2\approx 0.2K_a=\frac{x\times 0.2}{0.02}x=1.8\times {10}^{-7}[H^{+}{]}_{ion}=1.8\times {10}^{-7}M$