The pH of a 0.001 mol dm\(^{-3}\) of  H\(_2\)SO\(_4\) is  [Log\(_{10}\)2 = 0.3]

Explanation

H\(_2\)SO\(_4\) is a strong acid, so it will completely dissociate into corresponding ions.

On dissociation, H\(_2\)SO\(_4\)  → 2H\(^+\)  +  SO\(_4\)\(^{2-}\)

If the concentration of H\(_2\)SO\(_4\) is 0.001 M so, the concentration of 2H\(^+\) will be 0.002 M (0.001 x 2). This is because 1 molecule of H\(_2\)SO\(_4\) gives 2H\(^+\)  ions.

Recall that  pH = - Log\(_{10}\)[H\(^+\)]

                   pH =  - Log\(_{10}\)[2H\(^+\)]  (basicity is 2)

                        = - Log\(_{10}\)[2 x 0.001]

                        = - (Log\(_{10}\)2 +  Log\(_{10}\) 0.001)

                        = - (Log\(_{10}\)2 +  Log\(_{10}\) 10\(^{-3}\))

Also recall that Log\(_{10}\)2 = 0.3  and Log\(_{10}\)10 = 1

Therefore,  pH = - (0.3 - 3)

                        = - ( - 2.7)

                        = 2.7        - Option A

There is an explanation video available .