Solution:

Given:
Ka for HCN = 5×10^-10 at 25 deg C
pH = 9

To find:
Volume of 5 M KCN solution required to be added to 10 ml of 1 M HCN solution

Explanation:

The pH of a solution is a measure of its acidity or alkalinity. It is a logarithmic scale ranging from 0 to 14, where pH 7 is considered neutral, pH less than 7 is acidic, and pH greater than 7 is alkaline. In this case, the pH is given as 9, which means the solution is alkaline.

To maintain a constant pH of 9, we need to consider the equilibrium between the weak acid HCN and its conjugate base CN-.

HCN ⇌ H+ + CN-

The equilibrium constant for this reaction is given by the expression Ka = [H+][CN-]/[HCN].

Given that the value of Ka for HCN is 5×10^-10, we can calculate the concentration of CN- using the equation:

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

Since the pH is 9, the concentration of H+ can be calculated using the equation:

[H+] = 10^(-pH)

Substituting the values, we have:

[H+] = 10^(-9) = 1×10^-9

Now, let's calculate the concentration of CN-:

[CN-] = (5×10^-10) * (1 M)/(1×10^-9 M) = 0.5 M

To find the volume of 5 M KCN solution required, we can use the equation:

[CN-] * V = [HCN] * V0

Where V is the volume of the 5 M KCN solution and V0 is the volume of the 1 M HCN solution.

Substituting the values, we have:

(0.5 M) * V = (1 M) * (10 ml)

V = (1 M * 10 ml)/(0.5 M) = 20 ml

Therefore, the volume of the 5 M KCN solution required to be added to 10 ml of 1 M HCN solution to maintain a constant pH of 9 is 20 ml.

In summary:

- The equilibrium constant Ka for HCN is 5×10^-10.
- The pH of the solution is 9, indicating an alkaline solution.
- The concentration of CN- is calculated using the equation [CN-] = Ka * [HCN]/[H+].
- The concentration of H+ is calculated using the equation [H+] = 10^(-pH).
- The concentration of CN- is found to be 0.5 M.
- The volume of the 5 M KCN solution required is calculated using the equation [CN-] * V = [HCN] * V0.
- The volume of the 5 M KCN solution required is determined to be 20 ml.