Question

The ${[OH]^ - }$ of 0.005 M ${H_2}S{O_4}$ is:
A. $2 \times {10^{ - 2}}M$
B. $5 \times {10^{ - 3}}M$
C. ${10^{ - 2}}M$
D. ${10^{ - 12}}M$

Answer 1

The sulphuric acid dissociates to give two hydrogen ion and one sulphate ion. To determine the concentration of hydroxide ion, first find out the concentration of hydrogen ion present in the sulphuric acid. The product of hydrogen ion and hydroxide is equal to the$1.0 \times {10^{ - 14}}$.

Complete step by step answer:
The concentration of ${H_2}S{O_4}$is 0.005 M.
The sulphuric acid is a strong acid which on dissolving in water completely dissociates into its constituent ions.
The dissociation of sulphuric acid is shown below.
${H_2}S{O_4} \to 2{H^ + } + SO_4^ -$
In this reaction, one mole of sulphuric acid dissociates to give two hydrogen ions and one sulphate ion.
As two hydrogen ions are present in the sulphuric acid then the concentration of hydrogen ion is given as shown below.
$\Rightarrow 2 \times 0.005 = 0.01M$
Thus, the concentration of hydrogen ion ${H^ + }$ is 0.01 M.
The pH of an acid is defined as the negative logarithm of hydrogen ion concentration.
The pH of the acid is calculated by the formula as shown below.
$pH = - \log [{H^ + }]$
The pOH of a base is defined as the negative logarithm of hydroxide ion concentration.
$pOH = - \log [O{H^ - }]$
The product of hydrogen ion and hydroxide ion is equal to $1.0 \times {10^{ - 14}}$.
The equation is shown below
$[{H^ + }][O{H^ - }] = {10^{ - 14}}$
Substitute the value of hydrogen ion concentration in the above equation.$\Rightarrow [{10^{ - 2}}][O{H^ - }] = {10^{ - 14}}$
$\Rightarrow [O{H^ - }] = \dfrac{{{{10}^{ - 14}}}}{{{{10}^{ - 2}}}}$
$\Rightarrow [O{H^ - }] = {10^{ - 2}}$M.
Thus, the concentration of ${[OH]^ - }$ of 0.005 M ${H_2}S{O_4}$ is ${10^{ - 2}}M$.
Therefore, the correct option is C.

Note:
There is a relationship between the pH of the acid and pOH of base. The equation to convert the pH to pOH and vice versa is shown below.
$pH + pOH = 14$