K a for HCN is 5 x 10 - 10 at 25 o C. For maintaining a constant pH of 9, the volume of 5 M KCN solution required to be added to 10 mL of 2 M HCN solution is:

Question

K a for HCN is 5 x 10 - 10 at 25 o C. For maintaining a constant pH of 9, the volume of 5 M KCN solution required to be added to 10 mL of 2 M HCN solution is:, 4 mL, 7.95 mL, 9.3 mL, 2 mL

MARKS App · Accepted Answer

On mixing the given solutions, acidic buffer will be formed. Thus applying Henderson equation to find the pH- pH = pK a + log 10 [ CN - ] [ HCN ] pH = - log 5 × 1 0 - 1 0 + log 5 × V / V + 1 0 1 0 × 2 / V + 1 0 9 = - log 5 × 1 0 - 1 0 + log V 4 9 = 1 0 - 0 . 7 + log V 4 - 0 . 3 = log V 4 ∴ log V 4 = - 0 . 3 = log 1 2 V 4 = 1 2 ∴ V = 2 mL.

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