(C) For a very dilute solution of a strong acid like $1.0 \times 10^{-8} \ M \ HCl$,the concentration of $H^+$ ions contributed by the dissociation of

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The ionization of water cannot be assumed to be negligible in comparison with this low concentration of $HCl$

(C) For a very dilute solution of a strong acid like $1.0 \times 10^{-8} \ M \ HCl$,the concentration of $H^+$ ions contributed by the dissociation of water $(1.0 \times 10^{-7} \ M)$ cannot be ignored.Total $[H^+] = [H^+]_{HCl} + [H^+]_{H_2O} = 1.0 \times 10^{-8} + 1.0 \times 10^{-7} = 1.1 \times 10^{-7} \ M$.Therefore,$pH = -\log(1.1 \times 10^{-7}) \approx 6.96$,which is less than $7$ (and certainly less than $8$).

Class 11 Chemistry 2 Equilibrium: Ionic Equilibrium pH of strong Acids and strong Bases

Medium

The $pH$ value of $1.0 \times 10^{-8} \ M \ HCl$ solution is less than $8$ because

A
$HCl$ is completely ionised at this concentration
B
The ionization of water is negligible
C
The ionization of water cannot be assumed to be negligible in comparison with this low concentration of $HCl$
D
The $pH$ cannot be calculated at such a low concentration of $HCl$

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2 Equilibrium: Ionic Equilibrium

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pH of strong Acids and strong Bases

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The $pH$ value of $1.0 \times 10^{-8} \ M \ HCl$ solution is less than $8$ because

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$x \ mg$ of $Mg(OH)_2$ (molar mass $= 58 \ g/mol$) is required to be dissolved in $1.0 \ L$ of water to produce a $pH$ of $10.0$ at $298 \ K$. The value of $x$ is $........... mg$ ($Nearest$ $integer$). ($Given$: $Mg(OH)_2$ is assumed to dissociate completely in $H_2O$)

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