What will be the value of pH of 0.01 `"mol dm"^(-3)CH_(3)COOH(K_(1)=1.74xx10^(-5))` ?

Options

A. 3.4
B. 3.6
C. 3.9
D. `3.0`

Answer

A

Explanation

`CH_(3)COOH+H_(2)OhArrCH_(3)COO^(-)+H_(3)O^(+)`
`K_(a)=([CH_(3)COO^(-)][H_(3)O^(+)])/([CH_(3)COOH])`
`K_(a)=([H_(3)O^(+)]^(2))/([CH_(3)COOH])" "(because[CH_(3)COO^(-)]=[H_(3)O^(+)])`
`[H_(3)O^(+)]^(2)=K_(a)[CH_(3)COOH]`
`[H_(3)O^(+)]=sqrt(K_(a)[CH_(3)COOH])`
Given : `K_(a)=1.74xx10^(-5),[CH_(3)COOH]=0.01"mol dm"^(-3)`
`[H_(3)O^(+)]=sqrt(1.74xx10^(-5)xx0.01)=sqrt(1.74xx10^(-7))`
`[H_(3)O^(+)]=4.17xx10^(-4)`
`pH=-log[H_(3)O^(+)]=-log(4.17xx10^(-4))=3.379~~3.4`