Question

What will be the $pH$ of $1 \times {10^{ - 4}}M$ ${H_2}S{O_4}$ solution?
$A.$ $10.4$
$B.$ $03.70$
$C.$ $3$
$D.$ $13$

Answer 1

$pH$ is a measure of how acidic and basic solution is. The range of $pH$ is $0 - 14$ . The neutral solution has $pH$ equal to $7$ , $pH$ of acidic solution is always less than $7$ and $pH$ of basic solution is always greater than $7.$

Formula used:
$pH = - \log \left[ {{H^ + }} \right]$
Where $\left[ {{H^ + }} \right]$ is the concentration of hydronium ion.

Complete step by step solution:
In the given question we have given acid is sulphuric acid ${H_2}S{O_4}$ . ${H_2}S{O_4}$ is a dibasic acid. Thus it gives two ${H^ + }$ ions its solution. As we know it is a strong acid it dissociate completely as
${H_2}S{O_4} \rightleftharpoons 2{H^ + } + SO_4^{2 - }$
Now, we can see sulphuric acid give two ${H^ + }$ ion in solution. Now the concentration of ${H^ + }$ .
Concentration of ${\left[ H \right]^ + } = 2 \times {10^{ - 4}}M$
We will have to multiply concentration by $2$ because one ${H_2}S{O_4}$ gives two ${H^ + }$ .
Now using the above formula we can calculate the $pH$ of the solution.
$pH = - \log \left[ {{H^ + }} \right]$
By putting the value of $\left[ {{H^ + }} \right]$ , we will get the $pH$ of ${H_2}S{O_4}$ solution.
$pH = - \log \left[ {2 \times {{10}^{ - 4}}} \right]$
$\Rightarrow$ $pH = 4 - 0.3$
$\Rightarrow$ $pH = 03.70$
Hence $pH$ of $1 \times {10^{ - 4}}M$ ${H_2}S{O_4}$ solution is $03.70$
So, the correct option is $B$ .

Additional information:
$pH$ scales: $pH$ scales measured the concentration of hydrogen ion in solution. The more hydrogen ions, the more acidic solution, the range of $pH$ scale is $0 - 14$ . The $pH$ scale was invented by the Danish chemist Soren Sorenson to measure the acidity of beer in a brewery.

Note:
It is to be noted that we can calculate the $pH$ of solution if concentration of $\left[ {O{H^ - }} \right]$ is known by some modification in above formula just replace the concentration of $\left[ {{H^ + }} \right]$ by concentration of $\left[ {O{H^ - }} \right]$ . Now the new formula is $pOH = - \log \left[ {O{H^ - }} \right]$ and we know $pH = 14 - pOH$ so we can calculate the $pH$ in this way.
Here is the table which summarizes the $pH$ of solution, Nature of solution and concentration of ions

$pH$ of solution	Nature of solution	Concentration of ${H^ + }$ and $O{H^ - }$ ions
Less than $7$	Acidic	${H^ + } > O{H^ - }$
Greater than $7$	Basic	$O{H^ - } > {H^ + }$
Equal to $7$	Neutral	${H^ + } = O{H^ - }$