Calculate the % hydrolysis in a 10-2 M solution of NaCN (Ka = 6.2×10-10).

To calculate the % hydrolysis in a 10-2 M solution of NaCN, we need to use the equilibrium constant (Ka) for the hydrolysis reaction. The Ka value given is 6.2×10-10.

Step 1: Write the balanced equation for the hydrolysis reaction of NaCN: NaCN + H2O ⇌ NaOH + HCN

Step 2: Set up the expression for the equilibrium constant (Ka): Ka = [NaOH][HCN] / [NaCN][H2O]

Step 3: Since we are given the initial concentration of NaCN as 10-2 M, we can assume that the concentration of NaOH and HCN at equilibrium is negligible compared to the initial concentration of NaCN. Therefore, we can simplify the expression for Ka as follows: Ka ≈ [NaOH][HCN] / [NaCN]

Step 4: Let x be the extent of hydrolysis, which represents the concentration of NaOH and HCN formed. Since NaCN is a strong electrolyte, it dissociates completely, so the concentration of NaCN at equilibrium will be (10-2 - x) M.

Step 5: Substitute the values into the simplified expression for Ka: 6.2×10-10 = x^2 / (10-2 - x)

Step 6: Solve the equation for x. This can be done by assuming that x is small compared to 10-2, so we can neglect x in the denominator: 6.2×10-10 ≈ x^2 / (10-2)

Step 7: Rearrange the equation and solve for x: x^2 ≈ 6.2×10-10 * (10-2) x^2 ≈ 6.2×10-8 x ≈ √(6.2×10-8) x ≈ 7.87×10-5 M

Step 8: Calculate the % hydrolysis by dividing the concentration of NaOH and HCN formed (x) by the initial concentration of NaCN (10-2) and multiplying by 100: % hydrolysis = (x / 10-2) * 100 % hydrolysis ≈ (7.87×10-5 / 10-2) * 100 % hydrolysis ≈ 0.787%

Therefore, the % hydrolysis in a 10-2 M solution of NaCN is approximately 0.787%.