Question

Calculate the pH of a 0.21 M solution of C₅H₅NHCl (K_b for C₅H₅N = 1.7 x 10^-9). Record your pH value to 2 decimal places.

Calculate the pH of a 0.17 M solution of KNO₂ (K_a for HNO₂ = 4.0 x 10^-4). Record your pH value to 2 decimal places.

Answer 1

There are multiple questions here . i am allowed to answer only 1 at a time. I will answer 1st one for you. Please ask other as different question

C5H5NHCl is C5H5NH+ + Cl-

C5H5NH+ -----> C5H5N +    H+
0.21                             0                  0 (initial)
0.21-x                         x                  x   (at equilibrium)

C5H5NH+   is acid,
Ka = Kw/Kb
      = 10^-14 / (1.7*10^-9)
     = 5.88*10^-6

Ka = [C5H5N] [H+] / [C5H5NH+]
5.88*10^-6 = x*x / (0.21-x)
since Ka is very small, x will be small and it can be ignored as compared to 0.21
So, above expression becomes

5.88*10^-6 = x*x / (0.21)
x = 1.11*10^-3 M

This is concentration of H+
[H+] = 1.11*10^-3 M

pH = -log [H+]
     = -log (1.11*10^-3)
     = 2.95
Answer: 3.0