What is (oh-) in a solution of 0.125m ch3nh2 and 0.130m ch3nh3cl? Ka = 4.4×10^-4

Explanation:

The solution which contains 0.125 M CH3NH2 and 0.130 M CH3NH3Cl acts as buffer solution and the pOH of the buffer solution use the following Henderson-

Hasselbalch equation:

pOH = pKo + log(BH+)/[B]

Given that

CH3NH2 (aq) + H2O(l) + CH3NH, (aq) + OH- (aq)

Ky = 4.4 x 10-4

Here, the value of Ky

=- 4.4 x 10-4 ,Therefore pK is 3

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