K_a for HCN is 5×10^(-10) at 25°C . For maintaining a constant pH of 9, the volume in ml of 5 M KCN solution required to be added to 10 ml of 2 M HCN…

Question

K_a for HCN is 5×10^(-10) at 25°C . For maintaining a constant pH of 9, the volume in ml of 5 M KCN solution required to be added to 10 ml of 2 M HCN solution is

MARKS Content Team · Accepted Answer

Correct answer: 2.00. Suppose, volume of KCN added = x ml Total volume of solution after mixing = (10 + x) ml Molarity of HCN in the final solution =M_1V_1/V_2=(2×10)/((10+x))=20/((10+x))M Molarity of HCN in the final solution =M_1V_1/V_2=(5×x)/((10+x))=5x/((10+x))M pH =-log⁡k_a+log⁡([KCN])/([HCN]) 9 = −log (5×10^(−10) ) + log 5x/(10 + x) 20/(10 + x) On solving, x = 2 mL

$K_{a}$ for HCN is $5 \times 1 0^{- 10}$ at $25 ° C$ . For maintaining a constant pH of 9, the volume in ml of 5 M KCN solution required to be added to 10 ml of 2 M HCN solution is

Solution

More Ionic Equilibrium Questions

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

Solution

More Ionic Equilibrium Questions

$K_{a}$ for HCN is $5 \times 1 0^{- 10}$ at $25 ° C$ . For maintaining a constant pH of 9, the volume in ml of 5 M KCN solution required to be added to 10 ml of 2 M HCN solution is