Question

The ionization constant for acetic acid is 1.8 × 10-5, and that for hydrocyanic acid is 4 × 10-10.

Answer 1

Answer:

D.  [H+] of the acetic acid solution is greater than that of the hydrocyanic acid solution.

Explanation:

$CH_3 COOH(aq)+ H_2 O(l) CH_3 COO^- (aq)+H_3 O^+ (aq)$

Initial                0.1M                                 0                                 0

Change                -x                                 +x                               +x

Equilibrium    0.1M-x                               +x                               +x

$Ka =\frac {((x)(x))}{(0.1M-x)}$

$1.8\times10^{-5}= \frac {x^2}{(0.1M-x)}$

(-x is neglected) so we get

$1.8\times10^{-5}\times0.1=x^2\\\\x^2=1.8\times10^{-6}$

$x=\sqrt{x^2}=1.34\times10^{-3} M=H^3 O^{+}$ is the $[H^+ ]$ concnetration of acetic acid

$HCN(aq)+ H_2 O(l) CN^- (aq)+H_3 O^+ (aq)$

Initial                     0.1M                                 0                                 0

Change                     -x                                 +x                               +x

Equilibrium         0.1M-x                               +x                               +x

$Ka =\frac {((x)(x))}{(0.1M-x)}$

$4.0\times10^{-10}= \frac {x^2}{(0.1M-x)}$

(-x is neglected) so we get

$4.0\times10^{-10}\times0.1=x^2\\\\x^2=4.0\times10^{-11}$

$x=\sqrt{x^2}=6.32\times10^{-6} M=H^3 O^{+}$ is the $[H^+ ]$ concnetration of Hydrocyanic acid

Thus we see here, [H+] of the acetic acid solution $1.34 \times 10^{-3} M$ is greater than that of the hydrocyanic acid solution $6.32 \times10^{-6} M$

Please note∶

To find $[OH^- ]$ we use formula $[OH^- ]= \frac {kw}{([H^+])}$  

The value of kw is 1.0×10^(-14)

The $[OH^- ]$ is lesser in each solution