1.

500mL of 10^(-5)MNaOH is mixed with 500mL of 2.5xx10^(-5)M of Ba(OH_(2)),To the resulting solution ,99L water is added ,calculate pH of final solution.Take log0.303=-0.52.

Answer»

Solution :`[OH^(+)]_(f)=((500xx10^(-5))+(500xx2xx2.5xx10^(-5)))/(1000)=3XX10^(-5)M`
`V_(1)=1L&V_(f)=100L`
no of molesof `[OH^(+)]` in resulting solution =no. of moles of `[OH^(+)]` in FINAL
`3xx10^(-5)=[OH^(-)]_(f)xx100`
`:.[OH^(-)]_(f)=3xx10^(-7)M( lt 10^(-6)M)`
so `OH^(-)`ions coming form `H_(2)O` should also be considered.
`H_(2)OhArrunderset(x) H^(+)+underset(x+3xx10^(-7))(OH^(-))`
`K_(w)=x(x+3xx10^(-7))=10^(-14)`
`,.x=((sqrt(13)-3)/(2))xx10^(-7)M=[H^(+)]`
So `pH=7-log0.303=7.52`


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