1.2 L of 0.4 M HCl solution can be completely neutralized by

Options

A. `0.0099`
B. `0.099`
C. `0.99`
D. `9.9`

Answer

A

Explanation

Normality of NaOH solution `N_(b)=0.01xx1`
`N_(b)=0.01`
`V_(b)=19.85mL`.
`N_(b)xxV_(b)=19.85xx0.01`
`V_(a)=20mL`.
If `N_(a)` is the normality of HCl solution then for complete neutralization `N_(a)xxV_(a)=N_(b)xxV_(b)`
`N_(a)=(N_(b)xxV_(b))/(V_(a))`
`=(19.85xx0.01)/(20)=0.0099`
`N_(a)=M_(a)xx` basicity
`0.0099=M_(a)xx1`
`M_(a)=0.0099`.