The pH of a solution obtained by mixing 10 mL of `0.1MHCI` and 40mL of `0.2MH_(2)SO_(4)` is:

Options

A. `1.4865`
B. `0.4865`
C. `0.4685`
D. 3

Answer

C

Explanation

Milli-equivalent of `H^(+)` from `HCI=10xx0.1=1`
Milli-equivalent of `H^(+)` form `H_(2)SO_(4)`
`=40xx0.2xx2=16`
Total Meq. Of `H^(+)` in solution `= 1+16=17`
`:. [H^(+)]= (Meq.)/(V_(i nmL))=(17)/(50)=3.4xx10^(-1)`
`pH= -log[H^(+)= -log 0.34=0.4685`