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).
Correct answer: D. 0.2, 0.1
- A. 0.1, 0.2
- B. 0.1, 0.3
- C. 0.2, 0.4
- D. 0.2, 0.1
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
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About pH and pOH
pH measures hydrogen ion concentration and pOH measures hydroxide ion concentration on a logarithmic scale. Calculations use pH = minus log of hydrogen ion concentration, pOH = minus log of hydroxide ion concentration and pH plus pOH equals pKw, with separate treatment of strong and weak acids and bases.
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