1.

The pH of a solution obtained by mixing 100mL of a solutionpH of =3 with 400mL of a solution of pH =4 is

Answer»

`3- log 2.8`
`7- log 2.8`
`4- log 2.8`
`5-log 2.8 `

SOLUTION :`N_(1)V_(1)+N_(2)V_(2)=N_(3)(100+400)`
or `N_(3)=(0.1+0.04)/(500)=(0.14)/(500)=2.8 xx 10^(-4)M`
`pH=-log (2.8 xx 10^(-4))=4=log 2.8`


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