Question

Which of the following has the same pH as 1 L of a 0.1 M $C{H}_{3}COOH$ solution?
$K_a=1.8\times {10}^{-5}$ for $C{H}_{3}COOH$
Take $(1.34{)}^2=1.8$

• A. $3mMHCOOH(K_a=6\times {10}^{-4})$
• B. $0.1MC{H}_{3}COONa$
• C. $1.34mMHCl$
• D. $0.1MC{H}_{3}COOH$

Answer 1

Answer: (A), (C), (D)

The correct options are
A $3mMHCOOH(K_a=6\times {10}^{-4})$
C $1.34mMHCl$
D $0.1MC{H}_{3}COOH$

In each case we will have to calculate the concentration of hydronium ion
(a) $3mMHCOOH(K_a=6\times {10}^{-4})$
$\left[H^{+}\right]=c\alpha =\sqrt{K_a\times c}=\sqrt{6\times {10}^{-4}\times 3\times {10}^{-3}}$
$\left[H^{+}\right]=1.34\times {10}^{-3}$
(b)$0.1MC{H}_{3}COONa$
This will undergo salt hydrolysis
$K_h=\frac{\left[C{H}_{3}COOH\right]\left[O{H}^{-}\right]}{\left[C{H}_{3}CO{O}^{-}\right]}=\frac{K_w}{K_a}$
Now $\left[O{H}^{-}\right]=\sqrt{K_h\times c}$
$\left[H^{+}\right]=\sqrt{\frac{K_w\times K_a}{c}}$, which is not equal to $1.34\times {10}^{-3}$
(c)$1.34mMHCl$
$\left[H^{+}\right]=1.34\times {10}^{-3}$
(d)$0.1MC{H}_{3}COOH$
$\left[H^{+}\right]=c\alpha =\sqrt{K_a\times c}=\sqrt{1.8\times {10}^{-5}\times {10}^{-1}}$
$\left[H^{+}\right]=1.34\times {10}^{-3}$