Calculate the pH of 0.005M H2SO4 which dissociated completely in water?
H2SO4 dissociates in water as follows:
H2SO4 -> 2H+ + SO4 2-
Since 1 mole of H2SO4 produces 2 moles of H+, the concentration of H+ ions is twice that of the initial H2SO4 concentration:
[H+] = 2 * 0.005M = 0.01M
To calculate the pH, we use the formula:
pH = -log[H+]
pH = -log(0.01) = -log(1 x 10^-2) = -(-2) = 2
Therefore, the pH of a 0.005M H2SO4 solution that dissociated completely in water is 2.