Ka for HCN is 5 × 10-10 at 25°C. For maintaining a constant pH = 9, the volume of 5 M KCN solution required to be added to 10 mL of 2 M HCN solution is

Question

Ka for HCN is 5 × 10-10 at 25°C. For maintaining a constant pH = 9, the volume of 5 M KCN solution required to be added to 10 mL of 2 M HCN solution is (A) 4 mL (B) 2.5 mL (C) 2 mL (D) 6.4 mL

Tardigrade Admin · Accepted Answer

pH = pKa+ log ([Salt]/[Acid]) =pKa+ log ([KCN]/[HCN]) ...(i) Let the volume of KCN solution required be V mL ∴ [KCN]=(5× V/V+10) and [HCN] = (10×2/V +10) Now from eqn. (i), pH=- log (5×10-10)+log [(5× V/V+10) /(10×2/V+10)] 9 = -log(5×10-10)+log (V/4) On solving, V = 1.99 ≈ 2 mL

Q. Ka​ for HCN is 5×10−10 at 25∘C. For maintaining a constant pH=9, the volume of 5MKCN solution required to be added to 10mL of 2MHCN solution is

4 mL

2.5 mL

2 mL

6.4 mL

pH=pKa​+log[Acid][Salt]​ =pKa​+log[HCN][KCN]​...(i) Let the volume of KCN solution required be V mL ∴[KCN]=V+105×V​and[HCN]=V+1010×2​ Now from eqn. (i), pH=−log(5×10−10)+log[V+105×V​/V+1010×2​] 9=−log(5×10−10)+log4V​ On solving, V=1.99≈2mL

Q. $K_{a}$ for $H CN$ is $5 \times 1 0^{- 10}$ at $2 5^{\circ} C$ . For maintaining a constant $p H = 9$ , the volume of $5 M K CN$ solution required to be added to $10 m L$ of $2 M H CN$ solution is