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

Hint: $ 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^ - } $