0.365g of HCl gas was passed through 100 `cm^(3)` of 0.2 M NaOH solution. The pH of the resulting solution would be

Options

A. 1
B. 5
C. 8
D. 13

Answer

d

Explanation

Moles of HCl `=(0.365 )/(36.5 ) =10^(-2) =0.01 ` mol
Moles of NaOH `=(0.2 xx 100)/(1000)=0.02 mol`
Moles of NaOH left unreacted `=0.01 mol`
Volume of solution `=100 cm^(3) =0.1 L`
`:. [NaOH]=(0.01 )/(0.1)=10^(-1)M`
`[OH^(-)]=10^(-1)M`
`[H^(+)]=(K_(w))/([OH^(-)])=(10^(-14))/(10^(-1))=10^(-13)M`
`pH-13`.