Calculate the pH of `10^(-8)` M HCl solution .

Options

A. `9.525xx10^(-8)M`
B. `1.0xx10^(-8) M`
C. `1.0xx10^(-6) M`
D. `1.0525 xx 10^(-7)M`

Answer

D

Explanation

From acid, `[H^(+)]=10^(-8)M`
From `H_(2)O][H^(+)]=[OH^(-)]=x "mol" L^(-1)` (say)
`([H^(+)] or [OH^(-)]!=10^(-7)` M in presence of acid )
Thus, in the solution, `[H^(+)]=(10^(-8)+x) M and [OH^(-)]=xM`
But `[H^(+)] [OH^(-)]=K_(w)=10^(-14)`
`:. (10^(-8)+x)(x)=10^(-14)`
or `x^(2)+10^(-8)x-10^(-14)=0`
`:. x=(-10^(-8)pm sqrt(10^(-16)+4xx10^(-14)))/(2)`
`=9.51xx10^(-8), "i.e.," [OH^(-)]=9.51xx10^(-8)`.
Hence, `[H^(+)]=(10^(-14))/((9.51xx10^(-8)))=1.05xx10^(-7)M`
Alternatively, directly, the only concentration which will give pH close to 7 but less than 7 is (d).