determine the pH of each of the following solutions. a)0.25M KCHO2 b)0.24M CH3NH

Question

determine the pH of each of the following solutions. a)0.25M KCHO2 b)0.24M CH3NH3I c)0.16M KI can you please explain how to find the ph

Explanation / Answer

1) HCOOH(aq) --> HCOO-(aq) + H+(aq) Ka = 10-pKa = 10-3.744 = 1.8 * 10^-4 Kb = Kw/Ka = 10^-14/1.8 * 10^-4 = 5.55 * 10^-11 HCOO-(aq) + H2O(l) --> HCOOH(aq) + OH-(aq) let x = [OH-] then x = [HCOOH] and 0.25 - x = [HCOO-] x^2/(0.25-x) = 5.55 * 10^-11 x^2 - 5.55 * 10-11x - 1.38 * 10^-11 = 0 x = [OH-] = 3.71 * 10^-6 M ==> pOH = -log[OH-] = -log(3.71 * 10^-6) = 5.43 pH = 14 - pOH = 14 - 5.43 = 8.57 ------- 2) CH3NH3+(aq) --> H+(aq) + CH3NH2(aq) let x = [H+], then x = [CH3NH2] and (0.24 - x) = [CH3NH3+] CH3NH2(aq) + H2O(l) --> CH3NH3+(aq) + OH-(aq) Kb = 10-pKb = 10-3.36 = 4.37 * 10^-4 Kb = Kw/Ka ==> Ka = Kw/Kb = [H+][OH-][CH3NH2]/[CH3NH3+][OH-] = [H+][CH3NH2]/[CH3NH3+] = 10^-14/4.37 * 10^-4 = 2.29 * 10^-11 CH3NH3+(aq) --> CH3NH2(aq) + H+(aq) let = x= [H+] = [CH2NH2] and 0.24 - x = [CH3NH3+] x^2/(0.24 - x) = 2.29 * 10^-11 x^2 + 2.29 * 10^-11 - 5.49 * 10^-12 = 0 x = [H+] = 2.34 * 10^-6 M ==> pH = -log[H+] = -log(2.34 * 10^-6) = 5.63 3) do the same as 1 except use HI as the acid Everything you need to do the calculations is in the question. So, for [IO3-]o,you are taking 5.45 mL of 0.10 M KIO3- and diluting it and then mixing it with the other solution. The final total volume is 99.57 mL. So, M1V1 = M2V2 (0.10 M) (5.45 mL) = M2 (99.57 mL) M2 = 5.47 X 10^-3 M = [IO3-]o You can do basically the same thing for [I-]o The problem says that buffer solutions don't change pH appreciably on dilution, so the final pH will be 5.0 Since pH = -log [H+], [H+] = 10^-pH So, [H+] = 10^-5 = 1 X 10^-5 M

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