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

Answer

B

Explanation

Let the volume of each acid = V
pH of first, second and third acids = 3,4 and 5 respectively
`[H^(+)]` of first acid `(M_(1))=1xx10^(-3)[because H^(+)=1xx10^(-)pH]`
`[H^(+)]` of second acid `(M_(2))=1xx10^(_4)`
`[H^(+)]` of third acid `(M_(3))=1xx10^(-5)`
Total `[H^(+)]` concentrated of mixture
`(M)=(M_(1)V_(1)+M_(2)V_(2)+M_(3)V_(3))/(V_(1)+V_(2)+V_(3))`
`=(1xx10^(-3)xxV+1xx10^(-4)xxV+1xx10^(-5)xxV)/(V+V+V)`
`=(1xx10^(-3)xx V(1+0.1+0.01))/(3V)`
`=(1.11xx10^(-3))/(3)=3.7xx10^(-4)M`