1.

Calculate the ph of a solution formed by mixing equal volumes of two solutions A and B of a strong acids having `ph=6" and "ph=4` respectively.

Answer» ph of solution `A=6,[H^(+)]` "of solution" `[A]=10^(-6)mol L^(-1)`.
ph of solution `B=4,[H^(+)]` "of solution" `[B]=10^(-4)mol L^(-1)`.
Total `[H^(+)]` of the solution `=(10^(6)+10^(-4)) mol L^(-1)`
Volme of the solution upon mixing `=(1+1)=2L`
Total `[H^(+)]` on mixing solution `=(10^(-6)+10^(-4))/(2)mol L^(-1)= (10^(-4)(0.01+1))/(2) mol L^(-1)`
`=(0.01xx10^(-4))/(2)mol L^(-1)=0.5xx10^(-4)mol L ^(-1)`
`=5xx10^(-5)mol L^(-1)`
ph of solution `=-log[H^(+)]=-[5xx10^(-5)]`
`=[5-log5]=[5-0.6990]=4.3010=4.3`


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