Question

What is the pH value of 1 M \[{H_2}S{O_4}\] ?

Answer 1

Hint: The values of ${H_3}{O^ + }$ and $O{H^ - }$ concentrations can vary widely in aqueous solutions depending on the kinds and quantities of solutes present in it. To cover the range more conveniently a scale was introduced.

Complete step by step solution:
Sorrensen introduced a logarithmic scale which depicted,
$pH = - \log [{a_{{H_3}{O^ + }(aq)}}]$
The pH of the solution is the negative logarithm of the concentration in moles per litre of hydrogen ions which is contained in the solution. this question evolves the concept of determination of pH of strong acid, since sulphuric acid is a strong acid and we know that they ionize almost completely on in water no matter what is their concentration and this ionization is hardly affected by the presence/absence of other solutes present in the same solution. therefore dissociation occurs like,
${H_2}S{O_4} \to 2{H^ + } + SO_4^{(2 - )}$
$
  [{H_{(2)}}S{O_4}] = 1M \\
  [{H^ + }] = 2[{H_2}S{O_4}] = 2 \times 1 = 2M
 $
(Since one mole of ${H_2}S{O_4}$ gives 2 mole of ${H^ + }$ ions therefore its concentration will be double the concentration of ${H_2}S{O_4}$)
We know that pH is defined as the negative logarithm of hydrogen ion concentration.. By the pH scale we can find out the acidity or basicity of an aqueous solution. Acidic solutions are measured to have lower pH values than basic or alkaline solutions. The first electronic method for measuring pH was invented by Arnold Orville Beckman.
pH is defined as the decimal logarithm of.
$pH = - log[{H^ + }] $
Now let us substitute the value of concentration in the above equation to calculate the pH we get,
$pH = - log[{H^ + }] = - log2 = - 0.3010$

Hence this is the value of pH = -0.301.

Note:
Pure water is neutral. pH will be less than 7 when acid is dissolved in the water and when a base will be dissolved in the water the pH will be greater than 7. Negative ph values are completely possible as in this case.