Hydrocyanic acid has a Ka of 4.0 X 10-10. What is the percent of ionization of a 1.0 molar solution?
my answer is 0.004% is this correct?
a) 0.004%
b) 0.0020%
c) 0.0004%
d) 0.02%
How did you get 0.004%? That isn't my answer.
Hydrocyanic acid has a Ka of 4.0 X 10-10. What is the percent of ionization of a 1.0 molar solution?
my answer is 0.004% is this correct?
(4.0 x 10-10)/1.0 x 100 this wrong?
a) 0.004%
b) 0.0020%
c) 0.0004%
d) 0.02%
To find the percent ionization of a weak acid, you can use the equation:
% Ionization = (concentration of ionized acid / initial concentration of acid) x 100
In this case, the weak acid is hydrocyanic acid (HCN) with a Ka of 4.0 x 10^-10, and the initial concentration is 1.0 M.
First, we need to calculate the concentration of the ionized HCN. Since HCN ionizes into H+ and CN- ions, the concentration of H+ is equal to the concentration of ionized HCN, while the concentration of CN- is negligible compared to the initial concentration of HCN.
Therefore, we can assume that the concentration of ionized HCN is equal to [H+] after equilibrium is established.
To find [H+], we can use the expression for Ka:
Ka = [H+][CN-] / [HCN]
Since the concentration of CN- is much smaller than the initial concentration of HCN, we can approximate [HCN] ≈ [H+].
Therefore, Ka ≈ [H+]^2 / [HCN]
Solving for [H+], we get:
[H+] ≈ √(Ka * [HCN])
[H+] ≈ √((4.0 x 10^-10) * (1.0 M))
[H+] ≈ √(4.0 x 10^-10)
[H+] ≈ 2.0 x 10^-5 M
Now, we can substitute the values into the percent ionization equation:
% Ionization = (2.0 x 10^-5 M / 1.0 M) x 100
% Ionization ≈ 2.0 x 10^-3%
Therefore, the correct answer is b) 0.0020%.