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

Options

A. 13
B. 8
C. 7
D. 9

Answer

A

Explanation

`0.365 g HCl=(0.365)/(36.5)=0.01" moles of HCl"`
0.2 M NaOH in `100 cm^3=(0.2 xx 100)/(1000)="0.02 moles of NaOH"`
0.01 moles of HCl get neutralized with 0.01 moles of NaOH, thus resulting solution contains 0.01 moles of NaOH in 100 `cm^3` solution.
Concentration of `NaOH=(0.01)/(100) xx 1000=0.1M`
`[OH^-]=0.1 M =10^(-1)M, [H^+] [OH^-]=10^(-14)`
`[H^+]=(10^(-14))/(10^(-1))=10^(-13) M, pH =-log 10^(-13)=13`