If 9.8 g of sulfuric acid is dissolved in an excess quantity of water, it will yield _ moles of hydrogen ion ( H+) and _ mole of sulphate ions (SO4 -2).

Explanation

9.8 grams of sulphuric acid is 0.1 moles so there will be 0.2 moles of hydrogen ions ( two hydrogens per sulphuric acid molecule) H2SO4 dissociates in 2 steps H2SO4 → H+ + HSO4- Therefore 0.1 mol H2SO4 will produce 0.1 mol H+ HSO4- ↔ H+ + SO4- Ka for this reaction is : 1.2 x 10-2 Calculate [H+] using Ka equation Ka = [H+]² / [HSO4- ] 1.2*10^-2 = [H+]² / 0.1 [H+]² = 0.1( 1.2*10^-2) [H+]² = 1.2*10^-3 [H+] = 0.035 M Total moles of H+ ions = 0.1 + 0.035 = 0.135 moles