`K_(a)` of AcOH is `1.8 xx 10^(-5)` . What is `[H_(3)O^(+)]` in a solution which is 0.01 M AcOH and 0.005 M calcium acetate ?

Options

A. `1.8 xx 10^(-5)M`
B. `3.6 xx 10^(-5)M`
C. `0.9 xx 10^(-5)M`
D. `0.005 M`

Answer

A

Explanation

`AcOH +H_(2)O hArr AcO^(-) +H_(3)O^(+)`
`K_(a)=([AcO^(-)]\[H_(3)O^(+)])/([AcOH])`
`K_(a)=1.8 xx 10^(-5)`
`[AcOH]=0.01 M`
`[AcO^(-)]=2 xx [(CH_(3)COO)_(2)Ca]`
`=2 xx 0.005 =0.01M`
As the solution is very dilute, we assume complete ionisation of calcium acetate.
`1.8 xx 10^(-5)=((0.01)xx[H_(3)O^(+)])/(0.01)`
`[H_(3)O^(+)]=1.8 xx 10^(-5)M`