The hydrogen ion concentration of a `10^(-8)` M HCl aqueous solution at 298 K `(K_(w)=10^(-14))` is :

Options

A. `1.05xx10^(-7)M`
B. `9.525xx10^(-8)M`
C. `1.0xx10^(-8)M`
D. `1.0xx10^(-6)M`

Answer

A

Explanation

`[H^(+)]_("Total")=[H^(+)]_("acid")+[H^(+)]_("water")`
`[H^(+)]_(HCl)=10^(-8)M`
Let conc. of `H^(+)` from ionisation of water
`[H^(+)]_(H_(2)O)=[OH^(-)]_(H_(2)O)=x`
But `[H^(+)][OH^(-)=1xx10^(-14)`
`(1.0xx10^(-8)+x)(x)=1xx10^(-14)`
`x^(2)+10^(-8)x - 10^(-14)=0`
Solving for x, we get `x = 9.5xx10^(-8)`
`therefore [H^(+)]_("total")=1xxx10^(-8)+9.5xx10^(-8)`
`=1.05xx10^(-7)M`.