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Solution

We can find the concentration of OH- from the expression
[OH-]=(Ka*C)^1/2
= (4*10^-4*0.01)1/2=2x10^-3
so pOH= -log [OH-]= -log[2*10^-3]=2.699
Hence pH of the solution is = 14-pOH= 14-2.699= 11.3 (approximately)

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