Question

What is the pH of the solution when 0.2 mol of $HCl$ is added to one litre of a solution containing 1M acetic acid and acetate ions. Assume that the total volume is one litre. $K_a\left(C{H}_{3}COOH\right)=1.8\times {10}^{-5}$
$log\left(1.8\right)=0.255,log\left(0.66\right)=-0.181$

• A. 9.04
• B. 4.56
• C. 7.2
• D. 6.5

Answer 1

The correct option is B 4.56

Given, $n_{HCl}=0.2mol,\left[C{H}_{3}COOH\right]=\left[C{H}_{3}CO{O}^{-}\right]=1M$
$K_a\left(C{H}_{3}COOH\right)=1.8\times {10}^{-5}$
$\left[HCl\right]=\frac{0.2}{1}=0.2M$
On adding $HCl$, the free hydrogen ion will combine with $C{H}_{3}CO{O}^{-}$ ions to form $C{H}_{3}COOH$.
Thus, the concentration of acetic acid increases while that of $C{H}_{3}CO{O}^{-}$ decreases.
$\left[C{H}_{3}COOH\right]=1+0.2=1.2mol{L}^{-1}$
$\left[Salt\right]=1-0.2=0.8mol{L}^{-1}$
Applying Henderson's equation:
$pH=pKa+log\frac{\left[Salt\right]}{\left[Acid\right]}$
$pH=-log{K}_a+log\frac{\left[Salt\right]}{\left[Acid\right]}$
$pH=-log(1.8\times {10}^{-5})+log\frac{0.8}{1.2}$
$pH=5-log\left(1.8\right)+log\left(0.66\right)$
$pH=5-0.255-0.181$
$pH=4.56$