how to find the pH of 0.10 M H2SO4 ??

Well since sulfuric acid is a strong acid it dissociate completely so you just do the -log(0.10) =pH

To add a bit of information here, H2SO4 has TWO H atoms. The first one is a strong acid and (H^+) is due largely to the ionization of the first acid. The second H is not a strong one and it has a k2 but it is relatively large as k values go (about 0.012). The second H contributes a small amount. I've worked through to see how much and the answer is about 0.0098 so the final concn H^+ is 0.1+0.0098 = about 1.1M. The main thing I want to point out hee is that it is NOT 2 x 0.1 BUT the second H does contribute a small amount.

To find the pH of a solution, you need to use the equation:

pH = -log[H+]

In the case of strong acids like H2SO4, they completely dissociate in water to release H+ ions. H2SO4 is a diprotic acid, meaning it can donate two protons (H+) per molecule. Therefore, its dissociation can be represented as:

H2SO4 → 2H+ + SO4^2-

Since the concentration of H2SO4 is given as 0.10 M, the concentration of H+ ions is also 0.10 M.

Now, substitute the value of [H+] into the pH equation and calculate:

pH = -log(0.10)

Using a scientific calculator, you can find the value of the logarithm and obtain the pH.