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Question

Calculate the pH of a 0.02M aqueous solution of NH4OH.
Given, pKNH4OH=4.73

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Solution

NH4OH is weak base so
pOH=0.5(pKblogC)
pOH=0.5(4.73log(0.02))
pOH=0.5(4.73+1.69)
pOH=3.21
As we know pH+pOH=14
So pH=9.79

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