1.

Calculate the pH of the solution obtained by mixing 100 `cm^(3)` of solution with pH = 3 with `400 cm^(3)` of solution with pH = 4 .

Answer» `100 cm^(3) ` of solution with pH = 3 contains `H^(+)=(10^(-3))/(1000)xx100=10^(-4)`mole
`400cm^(3) ` of solution with pH = 4 contains `H^(+)=(10^(_4))/(1000)xx400=4xx10^(-5) ` mole
Total `H^(+)=10^(-4)+4xx10^(-5)=10^(-4)(1+0.4)=1.4xx10^(4)`.
Total volume = `500 cm^(3)`
`:. [H^(+)]=(1.4xx10^(-4))/(500) xx1000M = 2.8xx10^(-4)M`
`pH = - log (2.8xx10^(-4))=4-0.4472 ~~ 3.55`


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