Answer in General Chemistry for Abraham Alfeuss #144600
“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
