Question

For a solution equimolar in HCN and NaCN, which statement is false? a. [H+] is larger than it would be if only the HCN were in solution. b. Addition of NaOH will increase [CN–] and decrease [HCN]. c. Addition of more NaCN will shift the acid-dissociation equilibrium of HCN to the left. d. [H+] is equal to Ka. e. Addition of more HCN will shift the acid-dissociation equilibrium of HCN to the right.

Answer 1

Answer:

option  (d) is false.

Explanation:

Acid dissociation equilibrium of HCN is represented as-

 $HCN\rightleftharpoons H^{+}+CN^{-}$

Acid dissociation constant, $K_{a}$, is represented as-

$K_{a}=\frac{[H^{+}][CN^{-}]}{[HCN]}$

where species inside third bracket represents equilibrium concentrations of respective species

So, evidently, presence of excess $CN^{-}$ (or NaCN) in solution will combine with $H^{+}$ to produce HCN. Hence $H^{+}$ will be larger that it would be if only the HCN solution were present.

According to Le-chatlier principle, addition of HCN will shift equilibrium towards right and addition of NaCN will shift equilibrium towards left to keep constant $K_{a}$ value at a particular temperature.

NaOH gives acid-base reaction with HCN to produce NaCN and water. So, addition of NaOH will increase concentration of $CN^{-}$ and decrease concentration of HCN