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50ml of 0.1M NH4OH ,25ml NH4Cl weere used to make a buffer . what is the PH is if Pkbis 4.8?Solution) The given buffer is a base buffer[NH4OH] =[(50*0.1)/75][NH4Cl] =[(25*2)/75]From the principle V1M1=V2M2For a base buffer pOH = pKb + log [Salt/Base]thus pOH 4.8+ log [{(25*2)/75}/{(50*0.1)/75}] = 4.8+ log (10) = 4.8+1 =5.8Therefore pH = (14-pOH) = (14-5.8) = 8.2
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