Calculate the pH of a 0.19 M CH3NH3+ solution. CH3NH2+(aq) + H2O(l) = CH3NH3+(aq) + OH-(aq)...

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.

Solutions

Expert Solution

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

queen_honey_blossom answered 1 year ago

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