Calculate the ph of a solution formed by mixing equal volumes of two solutions A and B of a strong acids having `ph=6" and "ph=4` respectively.

Explanation

pH of solution A=6
`therefore [H_3O^+]` of solution A=`10^(-6)` M
pH of solution B=4
`therefore [H_3O^+]` of solution `B=10^(-4)`
On mixing one litre of each solution, total volume =1L+1L=2L
`[H_3O^+]` in mixture =`(10^(-4)+10^(-6))/2`
`=(10^(-4)(1+0.01))/2=(1.01xx10^(-4))/2=5xx10^(-5)`
`pH=-log (5xx10^(-5))`=4.3