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?

Explanation

For first solution,
`pH =3 or - log [H^(+)] =3 or log [H^(+)] =- 3=overset(-)(3).0`
` [H^(+)] = " Antilog "[H^(+)] = " Antilog "(overset(-)(3).0) = 10^(-3)M`
Similarly ,for second solution, " "`pH=4 and [H^(+)] =10^(-4) M`
For third solution " "`pH =5 and [H^(+)] =10^(-5) M`
Total `[H^(+)]` in the solution `=(10^(-3)+ 10^(-4) +10^(-5))=10^(-3) (1+10^(-1) +10^(-2))`
`=10^(-3)(1+0.1 +0.01) = 1.11 xx 10^(-3)M`
Since equal volumes of the three solutions have been mixed. `[H^(+)]` of the resulting solutin
`=((1.11 xx 10^(-3)M))/(3) = 3.7 xx 10^(-4) M`