Question

calculate the pH of the solution resulting from the addition of 50 ml of .10 M NaOH to 50 ml of .10M HCN (k_a=4.9*10^-10) solution. The answer is 11.0 please show work and how to get to it

Answer 1

millimoles of HCN = 50x 0.1 =5

millimoles of NaOH = 50 x0.1 =5

HCN + NaOH ---------------------------------> NaCN + H2O

5             5                                                0                0 -------------------------initial

0            0                                                5                5   ------------------after reaction

in the solution only salt is remained .so we need to follow the concept salt hydrolysis

CN- concentration = millimoles /total volume = 5/(50+50) = 0.05 M

NaCN -------------------Na+ CN-

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

0.05                                              0            0

0.05-x                                           x            x

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

Kw/Ka= (x)(x)/(0.05-x)

10^-14/4.9 x10^-10 = (x)(x)/(0.05-x)

2.04 x 10^-5 = (x)(x)/(0.05-x)

x^2 +2.04 x 10^-5 x - 1.02 x 10^-6 =0

x= 9.9 x 10^-4

[OH-] =x= 9.9 x 10^-4

pOH = -log[OH-] =-log(9.9 x 10^-4)

pOH = 3.0

pH + pOH = 14

pH = 14-pOH =14-3.0

pH = 11.0