Question

The pH of 0.005 molar aqueous solution of sulphuric acid is approximately
(A) 0.005
(B) 1
(C) 0.1
(D) 2.0

Answer 1

Hint: pH of a solution means that negative logarithm \[\left[ {base{\text{ }}10} \right]\] of molar concentration of \[{H_{3}}O^ + \] ion in the dilute solution.
2. Sulphuric acid is acidic as it gives protonium ion \[({H_{3}}O^ + )\] in aqueous solution.
3. At \[{25^o}C\] temperature pH of an acid is less than seven.

Complete answer:
We have to know that, pH of a solution means negative logarithm \[\left[ {base{\text{ }}10} \right]\] of molar concentration of \[{H_{3}}O^ + \] ion in the dilute which expressed as-
\[pH = - lo{g_{10}}[{H_{3}}O^ + ]\]
From the pH value we can easily find out that maybe the solution is acidic or basic. We will match the value from the pH scale which has ranged from zero to fourteen at \[ - {25^o}C\] temperature.
At this temperature, the pH value of the acidic solution is less than seven and the basic solution is greater than seven.
Here, given that the concentration of sulphuric acid is 0.005 molar means that 0.005 mole sulphuric acid is present in 1-liter solution.
Sulphuric acid \[[{H_2}S{O_4}]\] gives two \[{H_3}{O^ + }\] ions in aqueous solution.
\[{H_2}S{O_4} + 2{H_2}O \rightleftharpoons 2{H_3}{O^ + } + S{O_4}^{2 - }\]
From the equation we see that 1 mole \[{H_2}S{O_4}\] give 2-mole \[{H_3}{O^ + }\] ion are the stoichiometric coefficient so, \[0.005\] mole \[{H_2}S{O_4}\] gives \[(0.005 \times 2) = 0.01\,\,mole\,{H_3}{O^ + }\,ion\]
Therefore the concentration of \[{H_3}{O^ + }\] ion in the given solution is 0.01 molar.
Putting this data in expression of \[pH\] we get
\[pH = - lo{g_{10}}[{H_{3}}O^ + ]\]
\[= - lo{g_{10}}[0.01]\]
\[= -lo{g_{10}[{10}}^{-2}]\]
= 2
Therefore the pH value of a 0.005 molar aqueous solution of sulphuric acid is approximately 2.0.

So, the correct answer is (D).

Note: Care must be taken while calculating the concentration of $H_3O^+$ ions in the solution because here 1 molecule of $H_2SO_4$ is giving 2 moles of $H_3O^+$ ions in the solution so the concentration will become doubled.