Question

The hydrogen ion concentration of a 10-8 M HCl aqueous solution at 298K is (Kw = 10-14)

A) 1.0 x 10-8 M B) 1.0 x 10-6 M C) 1.0525 x 10-7 M D) 9.525 x 10-8 M

Answer 1

Answer:

Explanation:

The computation is expressed below,

HCl→. H+ + Cl−

10^−8 10^−8

Kw=[H+][OH−]

10^−14=(10^−8+a)×a

a=0.95×10^−7[H+]

=10^−8+(0.95×10^−7)

=1.05×10^−7M

Hence the answer is Option (C) 1.0525 x 10-7

In aqueous solution of HCl we have to calculate H+ ion from both water and the acid.

Answer 2

Answer: C) $1.1\times 10^{-7}M$

Explanation:

$10^{-8}MHCl$

$HCl\rightarrow H^+Cl^-$

1 mole of $HCl$ gives 1 mole of $H^+$

$10^{-8}$ moles of $HCl$ gives $10^{-8}$ mole of $H^+$

$H_2O\rightarrow H^++OH^-$

1 mole of $H_2O$ gives 1 mole of $H^+$ and 1 mole of $OH^-$                                                                       $K_w=[H^+][OH^-]$

$10^{-14}=[H^+][OH^-]$

$[H^+]=10^{-7}$

Thus total concentration of $H^+$ is $10^{-8}+10^{-7}=10^{-7}(0.1+1)=1.0525\times 10^{-7}M$

The hydrogen ion concentration will be $1.0525\times 10^{-7}M$