Ka for HCN is 5 × 10⁻¹⁰ at 25⁰C. 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:

pH=pKa+log[Salt][Acid]
pH=pKa+log[KCN][HCN].. (i)
pKa=−logKa=−log(5×10⁻¹⁰)=9.30
Let the volume of 5 M KCN solution required to be V mL.
KCN is added to 10 mL of 2 M HCN solution.
∴[KCN]=5M×VV+10mL and [HCN]=10mL×2MV+10mL
Now from eqn (i) ,9=9.30+log⎛⎝⎜⎜⎜5×VV+1010×2V+10⎞⎠⎟⎟⎟
9=9.30+logV4
logV4=−0.30
V4=0.5
V=2mL
Hence, the correct option is 2 mL