Question

What is the $pH$ of acetic acid at equilibrium, given that acetic acid concentration is $0.1\, M$ 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

$CH _{3} COOH \rightleftharpoons CH _{3} COO ^{-}+ H ^{+}$

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

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

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

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

$\left(\because\right.$ Concentrating $o \left[ CH _{3} COOH =0.1\, M \right)$

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

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

$pH =-\log [0.03]$

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

$pH =-0.47+2$

$pH =1.53$