Equal volumes of three acid solutions of `pH 3, 4` and `5` are mixed in a vessel. What will be the `H^(+)` ion concentration in the mixture?

Options

A. `3.7xx10^(-3)M`
B. `1.11xx10^(-3)M`
C. `1.11xx10^(-4)M`
D. `3.7xx10^(-4)M`

Answer

D

Explanation

`pH=-log_(10)[H^(+)]`
`[H+]=10^(-pH)`
`[H+]" of solution 1"=10^(-3)`
`[H+]" of solution 2"=10^(-4)`
`[H+]" of solution 3"=10^(-5)`
Total concentration of `[H^(+)]`
The volume taken in each case is 1 L
`=10^(-3)(1+1xx10^(-1)+1xx10^(-2))`
`=10^(-3)((1)/(1)+(1)/(10)+(1)/(100))`
`=10^(-3)((111)/(100))=1.11xx10^(-3)`
Therefore, `H^(+)` ion concentration in mixture of equal volume of these acid solutions
`=(1.11xx10^(-3))/(3)=3.7xx10^(-4)N`