What is the pH of a 0.001M solution of H2SO4? HSO4- has a pKa =1.2*10^-2

Dear student
Please find the solution to the asked query:

There is no use of the second part in calculating the pH of 0.001M H2SO4 solution:
$$\begin{gathered} H_{2}S{O}_{4 }completely ionises as: \\ {H}_{2}S{O}_{4 }+2H_{2}O\to 2H_3{O}^{+}+ S{O_2}^{2-} \\ 1 molecule of sulphuric acid gives 2 molecules of hydronium ions \\ \left[H_3{O}^{+}\right] = 2\left[H_{2}S{O}_4\right] \\ As not mentioned in question 0.001 is considered as normality \\ Equivalent mass=49 \\ Molar mass of H_{2}S{O}_{4 }is 98 \\ {H}_{2}S{O}_4=0.001N\times 49g/L =\frac{0.001N\times 49g/L}{98}=0.005M \\ pH =-\log \left[H_3{O}^{+}\right]=-\log \left[{10}^{-3}\right] =3 \end{gathered}$$

Hope this information will clear your doubts regarding the topic. If you have any other doubts please ask here on the forum and our experts will try to solve them as soon as possible.

Regards