1.

Calculate the pH of 0.1 M acetic acid if its ionisation constant K_a=1.8xx10^(-5).

Answer»

Solution :Degree if dissociation `alpha=SQRT((K_a)(C )), C alpha = sqrt(K_a . C)`
`THEREFORE pH=log.(1)/(sqrt(K_a. C))=log.(1)/(sqrt(1.8xx10^(-5)xx.1))`
`1=log.(1)/(sqrt(1.8xx10^(-3)))=log.(10^3)/(sqrt(1.8))`
`=3-(1)/(2)log1.8`
`=3-(1)/(2)xx0.2553=3-0.1276=2.8724`
Alternation,
`[H^+]=sqrt(K_a xx C)`
`=sqrt(1.8xx10^(-5)xx0.1)=sqrt(1.8xx10^(-6))`
`=1.341xx10^(-3)`
`therefore pH=-log[H^+]=-log(1.341xx10^(-3))`
`=3-0.1277=2.8723`.


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