A `50 ml` solution of `pH=1` is mixed with a `50 ml` solution of `pH=2`. The `pH` of the mixture will be nearly

Answer

1.26

Explanation

In the mixture, `[H^(+)]=((10^(-1)+10^(-2)))/(2) = (0.11)/(2) = 0.055=5.5xx10^(-2)`
`pH = - log (5.5xx10^(-2))=2 - 0.74 = 1.26`