1.

What will be the value of pH of 0.01 mol dm^(-3) CH_3COOH (K_a=1.74xx10^(-5) ) ?

Answer»

3.4
3.6
3.9
`3.0`

Solution :If `K_a=1.74xx10^(-5)`
Concentration of `CH_3COOH`= 0.01 MOL `dm^(-3)`
`[H^+]=sqrt(K_a. C)`
`=sqrt(1.74xx10^(-5)xx0.01)=4.17xx10^(-4)`
`pH=-LOG[H^+]`
`=-log(4.17xx10^(-4))`=3.4


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