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

Options

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

Answer

D

Explanation

HCI is a strong monobasic acid. Thus, `C_(H)^(+)` from HCl is equal to `C_(HCl)`, i.e., `10^(-8)` M. If we just focus on HCl, then pH of the solution will be 8:
`C_(H^(+))=10^(-pH) mol L^(-1)`
This is not acceptable as the solution is acidic. Its pH must be below 7. For this purpose, we must consider the contribution of `H^(+)` ions from water. In the absence of HCl, `C_(H^(+))` from water (at `25^(@)C`) is `10^(-7)` M. But due to the addition of HCl, the water equilibrium is distrubed and shifts backwards. Let the concentration of `H^(+)` ions be now x mol `L^(-1)`
`H_(2)OhArr underset(x)(H^(+))+underset(x)(H^(-))`
Total concentration of `H^(+)` ions will then be `(10^(-8)+x)` M. Thus,
`K_(w)=C_(H^(+))C_(OH^(-))`
`10^(-14)=(10^(-8)+x)(x)`
or `x^(2)+10^(-8)x-10^(-14)=0`
It is a quardatic equation `(ax^(2)+bx+c=0)` having the solution
`x=(-b+-sqrt(b^(2)-4ac))/(2a)`
`:. x= (-10^(-8)+-sqrt(10^(-16)+4xx10^(-14)))/(2)`
`=9.51xx10^(-8)`
This implies that
`C_(OH^(-))=9.51xx10^(-8)`
`:. C_(H^(+))=(10^(-14))/(C_(OH^(-)))=(10^(-14))/(9.51xx10^(-8))`
`=1.05xx10^(-7)M`
Also note that the only concentration which gives pH close to 7 but less than 7is `1.05xx10^(-7)`M.