the ph of a mixture of 0.01M hcl and 0.1M ch3cooh is approximately

0.01M Hcl and 0.1M CH3COOH is Given 
So H+  in HCl will be 10-2 and also in CH​3COOH is 10-1 
Let volume of HCl and CH3COOH be V 
According to formula H+ =N1V1 + N2V2/ total volume 
here N1=1* 10-2 and N2=1*10-1 
so by putting in formula,
H+= 10-2* V +10-1* V/2V


H+= 5.5 * 10-2

So, PH= - log[H+] 

Therefore  PH= - log[5.5 * 10-2]

PH=2- log 5.5
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