Question

What is [CN-] in a solution that contain 0.02M HCN (Ka=1.8 ×10^-5) and 0.01M HCl ?
1). 8×10^-6. 2). 1.8×10^-5
3) 9×10^-5. 4). 3.6× 10^-5

Answer 1

HCN which is an weak acid disassociates as,

HCN ---- H⁺ + CN⁻

Let the degree of disassociation be alpha.

Then concentration of H⁺ = 0.02α

Conc. of CN ion = 0.02α

HCl is also there which disassociates completely.

Therefore, H⁺ = 0.01 M

Total H⁺ = 0.01 + 0.02α

≈ 0.01         [Since, α for weak acid is very small.]

Conc. of HCN left = 0.02 - 0.02α

= 0.02(1 - α) ≈ 0.02

Ka = [H⁺][CN⁻]/[HCN]

1.8 × 10⁻⁵ = 0.01 × 0.02α/0.02

α = 1.8 × 10⁻³

Therefore, Conc. of CN⁻ = 0.02 × 1.8 × 10⁻³

= 3.6 × 10⁻⁵

Hence, Option (4.) is correct.

Hope it helps.