Question

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

In 0.1 molar solutions of acetic acid and hydrocyanic acid, which of the following is true?

A.
[H+] equals [OH-] in each solution.
B.
[H+] exceeds [OH-] in each solution.
C.
[H+] of the acetic acid solution is less than that of the hydrocyanic acid solution.
D.
[H+] of the acetic acid solution is greater than that of the hydrocyanic acid solution.

Answer 1

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

Step-by-step 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*10^(-5)= \frac {x^2}{(0.1M-x)}`

(-x is neglected) so we get

`1.8*10^(-5)*0.1=x^2\\\\x^2=1.8*10^(-6)`

`x=√(x^2)=1.34*10^(-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*10^(-10)= \frac {x^2}{(0.1M-x)}`

(-x is neglected) so we get

`4.0*10^(-10)*0.1=x^2\\\\x^2=4.0*10^(-11)`

`x=√(x^2)=6.32*10^(-6) M=H^3 O^(+)` is the
`[H^+ ]` concnetration of Hydrocyanic acid

Thus we see here, [H+] of the acetic acid solution
`1.34 * 10^(-3) M` is greater than that of the hydrocyanic acid solution
`6.32 *10^(-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