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`

Explanation

`C_(H^(+))=10^(-pH)(mol)/(L)`
Thus, the acid solutions of pH 3, 4, and 5 have `C_(H^(+))` equal to `10^(-3)`, `10^(-4)` and `10^(-5)` mol `L^(-1)`, respectively. Suppose we use x liters of each to make the mixture. Number of moles of `H^(+)` = Molarity x Volume
Total number of moles of `H^(+)` ions
`=x10^(-3)+x10^(-4)+x10^(-5)`
Total volume of mixture = 3x
Final molarity of `H^(+)` ions `= ("Total moles")/("Total volume")`
`= (x10^(-3)+x10^(-4)+x10^(-5))/(3x)`
`=(10^(-3)+10^(-4)+10^(-5))/(3)`
`=(1)/(3)xx10^(-3)(1+10^(-1)+10^(-2))`
`=(1)/(3)xx10^(-3)(1.11)`
`=0.37xx10^(-3)`
`=3.7xx10^(-4)M`