1.
Q3. A solution of sodium cyanate, NaCNO(aq), is prepared at a concentration of 0.20 M.
Calculate the equilibrium concentrations of OH (aq), HCNO (aq), CNO (aq), and H30* (aq) and
the pH of the solution at 25°C
Solution
NaCNO(aq) -----------------> Na^+ (aq) + CNO^- (aq)
0.2MÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â
               0.2M
CNO^- (aq) + H2O(l) -----------> HCNO(aq) + OH^- (aq)
IÂ Â Â Â Â Â Â Â
0.2Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â
       0                 0
CÂ Â Â Â Â Â Â Â -
x                                      Â
       +x                +x
EÂ Â Â Â Â Â Â Â 0.2-
x                                      Â
     +x                +x
Kb   = Kw/Ka
= 1*10^-14/3.5*10^-4Â Â Â = 2.86*10^-11
Kb     =   [HCNO][OH^-]/[CNO^-]
2.86*10^-11Â Â Â = x*x/0.2-x
2.86*10^-11*(0.2-x)Â Â = x^2
x   = 2.4*10^-6

2.
[CNO^-]  = x  = 2.4*10^-6M
[OH^-]  = x  = 2.4*10^-6M
[H3O^+]Â Â Â = Kw/[OH^-]
= 1*10^-14/2.4*10^-6Â Â = 4.16*10^-9M
PHÂ Â Â = -log[H3O^+]
= -log4.16*10^-9
= 8.38