Question

What is the $pH$ of acetic acid at equilibrium, given that acetic acid concentration is $0.1M$ and it is $30\%$ dissociated at equilibrium? $(\log 3=0.47)$

• A. 2.00
• B. 1.53
• C. 3.53
• D. 3.00

Answer 1

Answer: (B)

The required relation is

$C{H}_{3}COOH\rightleftharpoons C{H}_{3}CO{O}^{-}+H^{+}$

$\because$ Dissociation occurs $=30\%$, means 100 moles of

$C{H}_{3}COOH$, gives $=30$ moles of $H^{+}$ -ions.

Thus, $0.1$ mole of $C{H}_{3}COOH$ gives

$=\frac{30\times 0.1}{100}=0.03\text{ mole of }{H}^{+}$ ions.

$(\because$ Concentrating $o\left[C{H}_{3}COOH=0.1M\right)$

Also, $\because pH=-\log$ [Concentration]

$pH=-\log \left[H^{+}\right]$

$pH=-\log [0.03]$

$pH=-\log \left[\left(3\times 10^{-2}\right)\right]$

$pH=-0.47+2$

$pH=1.53$