Determine the pH of a 0.85 M CH3NH2 solution
Kb of CH3NH2 = 4.4*10^-4
CH3NH2 + H2O <-----> CH3NH3+ + OH-
0.85 0 0 (initial)
0.85-x x x (at equilibrium)
Kb = x*x/ (0.85-x)
since Kb is small, x will be small and it be ignored as compared to
0.85
4.4*10^-4 = x^2 / (0.85)
x = 0.019 M
[OH-]= x = 0.019 M
pOH = -log[OH-]
= -log (0.019)
= 1.71
pH = 14 - pOH
= 14- 1.71
= 12.29
Answer: 12.29
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