Question

An impure sample of lime weighing 0.150g was reacted with 50.00cm3 of 0.8mol/L hydrochloric acid. After the reaction was complete the solution was transferred to a 250cm3 volumetric flask and made up to the mark with distilled water. A 25.00cm3 portion of this solution was then titrated against 0.25mol/L sodium hydroxide using methyl orange as the indicator. A 15.00cm3 volume of the sodium hydroxide was consumed. Calculate the percentage purity of lime sample. Unbalanced equations are as follows;

CaCO3(s) + HCl(aq) -> CaCl2(aq) + CO2(g) + H2O(l)

HCl(aq) + NaOH(aq) -> NaCl(aq) + H2O(l)

Answer 1

The reactions are CaCO3(S)+2HCl(aq)======CaCl2(aq)+CO2(g)+H2O(l)

                          HCl(aq)+NaOH(aq)======NaCl(aq)+H2O(l)

so 1 mol of CaCO3 reacts with 2 mol of HCl & 1 mol of HCl reacts with 1 mol of NaOH

for 25.00 cm3 of the solution 15.00 cm3 of 0.25mol/L NaOH solution required

mol of NaOH = 0.015*0.25 = 0.00375 mol

so mol of HCl in the 25.00 cm3 solution = 0.00375 mol

total mol of HCl in the 250 cm3 flask = 0.0375 mol

mol of HCl in 50.00 cm3 0.8 mol/L HCl = 0.05*0.8 = 0.04 mol

mol of HCl reacts with CaCO3 = 0.04-0.0375 = 0.0025 mol

mol of CaCO3 = 0.0025/2 =0.00125 mol

weight of CaCO3 = 0.125 g

% of CaCO3 in the lime sample = 0.125/0.150*100 = 83.33%

so the impure lime sample is of 83.33% purity