Concentration of CN- in 0.1 M HCN

Given:
HCN is a weak acid with a dissociation constant (Ka) of 4×10^-10.

To find the concentration of CN- in 0.1 M HCN, we need to consider the dissociation reaction of HCN in water:

HCN ⇌ H+ + CN-

- We can use the equilibrium expression to find the concentration of CN-.
- Let's assume the concentration of CN- at equilibrium is x M.
- Since HCN is a weak acid, we can assume that the initial concentration of HCN will be nearly equal to the concentration of CN- at equilibrium.
- Therefore, the initial concentration of CN- is also 0.1 M.

Using the dissociation constant (Ka), we can write the equilibrium expression:

Ka = [H+][CN-] / [HCN]

Substituting the given values:

4×10^-10 = [H+][x] / (0.1 - x)

- Since the value of x is very small compared to 0.1, we can assume that 0.1 - x ≈ 0.1.
- Simplifying the equation, we get:

4×10^-10 = [H+][x] / 0.1

- Rearranging the equation, we find:

[H+][x] = 4×10^-11

- Since HCN is a weak acid, we can assume that [H+] will be equal to the concentration of HCN that dissociates.
- Therefore, [H+] = 0.1 - x.

Substituting this value in the equation, we get:

(0.1 - x)[x] = 4×10^-11

- Expanding the equation, we have:

0.1x - x^2 = 4×10^-11

- Rearranging the equation, we get:

x^2 - 0.1x + 4×10^-11 = 0

- Solving this quadratic equation, we find two possible values for x: 2.5×10^-6 and 4.5×10^-6.

Therefore, the concentration of CN- in 0.1 M HCN can be either 2.5×10^-6 or 4.5×10^-6.

So, the correct options are:
a) 2.5×10^-6
b) 4.5×10^-6

Note: The exact value of x can be determined using the quadratic formula, but in this case, we can make an approximation due to the small value of x compared to 0.1.

We have H+= root Ka*C