pH of the mixture containing 200ml of 0.1M CH3COOH and 100ml of 0.15M KOH is pka of  CH3COOH=4.8

Number of milli equivalents of CH3COOH = 200(0.1M)1 = 20Number of milli equivalents of KOH = 100(0.15M)1 = 15Excess WA + limited SB acts like an acidic bufferWA15  meq20  meq5  meqexcess + SB15  meq → salt15  meq + water  The solution contains 5 meq of WA and  15 meq of Salt. ∴pH=pka+logVsNsVaNapH=4.8+log155pH=4.8+log3