Question

sodium cyanide is the salt of the weak acid HCN. Calculate the concentration of H3), OH, HCN, and NA in a solution prepared by dissolving 10.8 g of NaCN in enough water to make 5.00x10^2 ml of solution at 25 degrees celsius

Answer 1

mol NaCN = mass/MW = 10.8/49.0072 = 0.220375mol

V = 5*10^2 = 500 ml = 0.5 L

M = 0.220375/(0.5) = 0.44075M

then

NaCN <-> Na+ and CN-

[NA+] = 0.44075 M

CN- + H2O <--> HCN + OH-

[CN-] = 0.44075 - x

[HCN] = [OH-] = x

then

Kb =[HCN ][OH]/[CN-]

Ka = Kw/Kb = (10^-14)/(6.2*10^-10) = 1.612*10^-5

solve

Kb =[HCN ][OH]/[CN-]

1.612*10^-5 = x*x/(0.44075 -x)

x = 0.002657

[HCN] = [OH-] = 0.002657 M

[CN-] = 0.44075 - x = 0.44075-0.002657 = 0.438093 M

[H3O+] = (10^-14)/(OH-) = (10 ^-14)/0.002657 = 3.76364*10^-12 M