Calculate the pH of each of the following solutions. (a) 0.500 M HONH2 (Kb = 1.1

Question

Calculate the pH of each of the following solutions.
(a) 0.500 M HONH2 (Kb = 1.1 10-8)---I got that the pH of this was 9.87. I don't know how to do b and d.

(b) 0.500 M HONH3Cl

(c) pure H2O---I got the pH to be 7.

(d) a mixture containing 0.500 M HONH2 and 0.500 M HONH3Cl

Explanation / Answer

b) HONH3Cl => HONH3+ + Cl- HONH3+ NH2OH + H+ we need the Ka = 9.1 x 10^-7 (search for the value from some standard table, or it should be provided). so, (x^2 / (0.500-x)) = 9.1 x 10^-7 x = [H+]= 6.7 x 10^-4 M so, the pH = -log[H+] = 3.17 d) pOH = 14-pKa + log(0.5/0.5) = 14-pKa = 6.04. (Henderson-Hasselbach) pH = 7.96.

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