ph of a buffer solution containing 0.15 moles of NH4OHand 0.4 moles of NH4Cl . Kb for NH4OH is 1.98×10^-5 . Then the buffer solution is approximately equals to
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Applying the equation
pOH=log[Salt]/[Base]−logKb
= log(0.4/0.15)-log(1.98×10^-5)
= 5.129
Therefore, pH of buffer solution is (14 - 5.129) = 8.871
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