Given the reaction: H2SO4 + NaOH --> ? + H2O. (a). Work out the salt produced (?) and (b). calculate the pH of the remaining solution when 1.2 g of NaOH and 4.41 g of H2SO4 were added in a 500 ml solution. Of the unreacted H2SO4 95% dissociated.
(a). Firstly, to work out the product, take 2Na+ + SO2- --> 2Na2SO4.(b). To calculate the pH, the equation: pH = -log10([H+]) must be used. Therefore we need to find the concentration of hydrogen ions in the resulting solution, [H+].Step 1: Work out the molar quantities of the reactants.Looking in a periodic table, the molar mass of S = 32 g mol-1, H = 1 g mol-1, Na = 23 g mol-1, O = 16 g mol-1. Therefore:H2SO4 = 2(1) + 32 + 4(16) = 98 g mol-1. NaOH = 23 + 16 + 1 = 40 g mol-1. To work out the molar quantities look at the units or the equation: moles = mass/Mr where Mr is relative molecular mass.for H2SO4 = 4.41 g / 98 g mol-1 = 0.045 mol, for NaOH = 1.2 g / 40 g mol-1 = 0.030 mol.Step 2: Balance the chemical equation.To do this, count the amount of each atom on each side of the reaction equation then add in the appropriate mole fraction to make sure there is the same amount of each atom on both sides. With the above equation on the LHS: 3 x H, 1 x Na, 5 x O, 1 x S. On the RHS: 2 x H, 2 x Na, 5 x O, 1 x S. Therefore the balanced equation should be:H2SO4 + 2NaOH --> Na2SO4 + 2H2O Counting atoms: LHS: 4 x H, 2 x Na, 6 x O, 1 x S. RHS: 4 x H, 2 x Na, 6 x O, 1 x S therefore balanced.Step 3: Calculate concentration of H+ ions remaining.Since the mole ratio of H2SO4 : NaOH = 1:2, this means that the H2SO4 is in excess as only 0.030 mols/2 = 0.015 mols of H2SO4 has reacted. Therefore to calculate how many moles of H2SO4 are left: 0.045 - 0.015 = 0.030 mols. Since only 95% of the unreacted H2SO4 dissociated: 0.030 mols * 0.95 = 0.0285 mols dissociated. Each mol of H2SO4 donates 2 mols of H+ ions, therefore the mols of H+ ions = 0.0285 mols * 2 = 0.057 mols.Therefore the concentration of H+ ions in a 500 ml = 0.5 dm3 solution, [H+] = 0.057 mols/0.5 dm3 = 0.114 mol dm-3. Step 4: Calculating the pH of the solution.Using the equation for pH shown above: pH = -log10([H+]) we can now work out the pH. pH = -log10(0.114 mol dm-3) = 0.94
