What will the PH of a solution formed by mixing equal volumes of two solutions A and B having PH=6 and PH=4
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pH=-log[H+], so, [H+]= 10^-pH
[H+] of pH=6 =10^-6, [H+] of pH=4 =10^-4
[H+] of mixture= (10^-6+10^-4)/2= 5.05×10^-5
pH of mixture= -log([H+] of mixture)= -log(5.05×10^-5)=4.2967
[H+] of pH=6 =10^-6, [H+] of pH=4 =10^-4
[H+] of mixture= (10^-6+10^-4)/2= 5.05×10^-5
pH of mixture= -log([H+] of mixture)= -log(5.05×10^-5)=4.2967
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