In: Chemistry
Calculate the pH of a 0.19 M CH3NH3+ solution. CH3NH2+(aq) + H2O(l) = CH3NH3+(aq) + OH-(aq)
Species, Data CH3NH2, Kb = 3.70x10-4. CH3NH2+, Ka = 2.73x10-11.
CH3NH3+ --------------------------> CH3NH2 + H+
0.19 0 0
0.19-x x x
Ka = [CH3NH2][H+]/[CH3NH3+]
Ka = x^2 / 0.19- x
2.73 x 10^-11 = x^2 / 0.19- x
x^2 + 2.73 x 10^-11 x - 5.19 x 10^-12 = 0
x = 2.28 x 10^-6
[H+] = 2.28 x 10^-6 M
pH = -log [H+]
pH = -log (2.28 x 10^-6 )
pH = 5.64