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

`0.365 g HCl =(0.365)/(36.5) "mole" = 0.01 "mole"`
`100 cm^(3) "of" 0.2 M NaOH=(0.2)/(1000)xx100`
`=0.02 ` mole
NaOH left unneutralized = 0.01 mole
Volume of solution = 100 ml
`:.` Molarity of NaOH in the solution
`=(0.01)/(100)xx1000=0.1M = 10^(-1)M`
`:. [H^(+)]=(10^(-14))/(10^(-1)M)=10^(-13)M :. pH=13`