Q2. Calculate the pH of the solution formed after 50.0 cm^3 of 0.0108 mol/dm^3 aqueous sodium hydroxide are added to beaker B. Give your answer to 2 decimal places
A2. First thing to work out is the moles of NaOH added, since moles = concentration x volume, = 0.0108 mol/dm^3 x 50ml, but need volume to be in units of Litres to match with the units of dm^3 so the formula becomes 0.0108 mol/dm^3 x (50 x 10^-3)L to give the number of moles of NaOH = 5.4 x 10^-4 Since we know that beaker B also contains 0.0125 mol/dm^3 of nitric acid in 100ml, and so the OH- will react with the H+, we have to account for the number of moles of HNO3 So, (0.0125 x (100 x 10^-3)) - (5.4 x 10^-4) = (moles of H+) - (moles of OH-) = moles of excess H+ = 7.1 x 10^-4 To get [H+] we need to do moles/volume = 7.1 x 10^-4 mol/dm^3 / 150 x 10^-3 L = 4.73 x 10^-3 pH = -log[H+] = 2.32
TD
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