The pH of a solution obtained by mixing 50 mL of 1N HCl and 30 mL of 1N NaOH is `[log 2.5 =0.3979]`

Options

A. 3.979
B. 0.6021
C. 12.042
D. 1.2042

Answer

B

Explanation

1N HCl`=` 1 MHCl, 1 NNaOH `=` 1 M NaOH
Moles of HCl`=Mxx V =50 xx 10^(-3) xx 1`
`=5 xx 10^(-2)`
Moles of NaOH `=M xx V =30 xx 10^(-3) xx 1`
`= 3 xx 10^(-2)`
As number of moles of HCl are more than NaOH, The resulting solution `(30+50 =80 mL)` will contain excess of HCl
`=5 xx 10^(-2) -3 xx 10^(-2) =2 xx 10^(-2) mol`
`:. H^(+)` ( after mixing ) `=(2 xx 10^(-2)mol)/( 80 xx 10^(-3)L)=(1)/(4)M`
`pH=-log H^(+)=-log. (1)/(4) =log 4=0.6021`