100 ml,0.2M ch3cooh is mixed with 100 ml,0.1M naoh find out ph of resulting solution.(given: Ka(ch3cooh)=2*10^-5)?

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The above reactants form buffer sol. of CH3COOH + CH3COONa Hence" Ph= PKà + log(salt÷acid). Here conc. of salt=0.05M. conc. of acid=0.05M. In solving. Ph = -log{2×10^(-5)÷3} + log(.05÷.05) = 5+0.48-0.3 "Ph=5.18.".