1.

The pH of the a solution obtained by mixing 50 ml of 0.4 N HCl and 50 ml of 0.2 N NaOH is

Answer»

`-log 2`
`-log 0.2`
`1.0`
`2.0`

Solution :M. eq. of 50 ML 0.4 N HCL `= 0.4 xx 50 = 20`
M. eq. of 50 ml, 0.2 N NaOH `= 0.2 xx 50 = 10`
REMAINING M.eq. of HCl = 20 - 10 = 10
`[H^(+)]` in `HCl = (10)/(50+50) = 0.1 n = 10^(-1)`
`pH = - log 10^(-1) = 1`.


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