Question

Sodium cyanide is the salt of the weak acid HCN. Calculate the concentration of H3O+ ; OH- ; HCN and Na+ in solution prepared by dissolving 10.8 g of NaCN in enough water to make 5.00 x 102 mL of solution at 25 0C.

Answer 1

$M(NaCN)= 49 g/mol$ (ref.1)

The overall volume $V = 0.51 L$

$n(NaCN) = m(NaCN)/M(NaCN) =$

$= 10.8 g/49(g/mol) \approx 0.22041 mol$

$C(NaCN) = n(NaCN)/V =$

$= 0.22041mol / 0.51L \approx 0.4322mol/L$

$NaCN$ is a strong electrolyte, therefore assume it dissociates completely:

$NaCN \to Na^+ (aq.) + CN^-(aq.)$

Therefore, $C(NaCN) = C(Na^+(aq.)) =$

$= C(CN^-(aq.)) = 0.4322mol/L$

Cyanide ion undergoes hydrolysis:

$CN^-(aq.) + H_2O \leftrightarrows HCN + OH^-(aq.)$

The dissociation of water is ignored

$C_{reacted}(CN^-(aq.)) = [NCN] = [OH^-(aq.)]=$

$= x\ mol/L$

$[CN^-(aq.)]=$

$= C(CN^-(aq.)) - C_{reacted}(CN^-(aq.)) =$

$= 0.4322mol/L - x$

$K_h = [HCN][OH^-(aq.)]/[CN^-] =$

$= x^2/(0.4322 - x) = 2.5*10^{-5}$ (ref. 2)

$x^2 = 2.5*10^{-5} *(0.4322 - x)$

$x^2 = 1.0805*10^{-5} - 2.5*10^{-5}x$

$x^2 + 2.5*10^{-5}x - 1.0805*10^{-5} = 0$

Solving the equation gives $x \approx 0.00327462 \ mol/L = 3.27462 *10^{-3} \ mol/L$

$[HCN] = [OH^-(aq.)] = 3.27462 *10^{-3} \ mol/L$

at $t = 25 \ ^\omicron C$ the following holds:

$pH + pOH = 14$

$pH = 14 - pOH$

$pOH = - log[OH^-(aq.)] =$

$= - log(3.27462 *10^{-3}) \approx 2.48484$

$pH = 14 - 2.48484 = 11.51516$

$pH = - log[H_3O^+(aq.)]$

$[H_3O^+(aq.)] = 10^{-pH} =$

$=10^{-11.51516} mol/L \approx 3.0538*10^{-12} \ mol/L$

$[Na^+(aq.)] = C(Na^+(aq.)) = 0.4322\ mol/L$

Answer:

$[H_3O^+(aq.)] = 3.0538*10^{-12} \ mol/L;$

$[HCN] = [OH^-(aq.)] = 3.27462 *10^{-3} \ mol/L;$

$[Na^+(aq.)] = 0.4322\ mol/L.$

References:

1. https://pubchem.ncbi.nlm.nih.gov/compound/Sodium-cyanide#section=Computed-Properties

2. https://pubchem.ncbi.nlm.nih.gov/compound/Sodium-cyanide#section=Stability-Shelf-Life