By adding `20 ml 0.1 N HCl` to `20 ml 0.001 N KOH`, the `pH` of the obtained solution will be

Options

A. `2`
B. `1.3`
C. `0`
D. `7`

Answer

B

Explanation

`20ml` of `0.1 NHCl=(0.1)/(1000)xx20g eq=2xx10^(-3)g eq.`
`20 ml` of `0.001 KOH=(0.001)/(1000)xx20g eq`
`=2xx10^(-5) g eq.`
`:. HCl` left unneutralised `=2(10^(-3)-10^(-5))`
`= 2xx10^(-3)(1-0.01)=2xx0.99xx10^(-3)=1.98xx10^(-3)g eq.`
Volume of solution `=40 ml`
`:. [HCl]=(1.98xx10^(-3))/(40)xx1000M=4.95xx10^(-2)`
`:. pH=2-log 4.95=2-0.7=1.3`