In aqueous solution, sulphuric acid dissociates into ions in 2 stages. The pKa for the first dissociation is -3. Calculate the pH of a 0.025 mol dm-3 solution of sulphuric acid using the pKa value of the 1st dissociation.
Ka = 10-pKa = 10--3 = 1000[H+] = √ Ka [0.025] = √1000 x 0.025 = 0.791pH = -log [H] = -log(0.791) = 0.102 = 0.1
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