# Back Titration Powerpoint

Pages: 3 (657 words) Published: March 18, 2013
Back Titrations

Question 1

A 10.0 g piece of rusty steel wool is dissolved in 200.00 ml of 1.00 M sulfuric acid. The excess sulfuric acid is determined by titration with a 0.500 M sodium hydroxide solution. 300.00 mL of sodium hydroxide is required to neutralise the acid. What was the % purity of iron in steel wool?

Question 2

A 3.145 g sample of a certain lead ore containing lead(II) carbonate, PbCO3, was heated gently with 25.00 mL of nitric acid of concentration 0.6293 mol/L.

The equation for this reaction is:
PbCO3(s) + 2HNO3(aq) ( Pb(NO3)2(aq) + H2O(l) + CO2(g)

When the reaction was complete and no more CO2 was evolved, the mixture was cooled and the excess nitric acid was titrated against NaOH solution of concentration 0.1423 mol/L. The volume of NaOH required was 23.41 mL.

The equation for the titration is:
HNO3(aq) + NaOH(aq) ( NaNO3(aq) + H2O(l)

a Calculate the percentage by mass of PbCO3 in the original ore. b Predict the effect on the final answer if:
i the ore contained some limestone, CaCO3, in addition to PbCO3 ii not all the CO2 was evolved during the heating stage

Question 3
Three aspirin tablets, which were claimed by the manufacturer to each contain 320 mg of aspirin, C6H4(OCOCH3)COOH, were heated gently in 50.00 mL of NaOH solution of concentration 0.5190 mol/ L.

The aspirin tablets reacted according to the equation:
C6H4(OCOCH3)COOH(s) + 2NaOH(aq) ( C6H4(OH)COONa(aq) + CH3COONa(aq) + H2O(l)

After cooling, the solution was transferred to a 100.00 mL volumetric flask and the volume was made up to exactly the 100.00 mL mark. Aliquots of 20.00 mL of this solution were then titrated against HCl of concentration 0.1232 mol/L. The mean titre was 24.19 mL.

The equation for the titration is:
HCl(aq) + NaOH(aq) ( NaCl(aq) + H2O(l)

a Draw a flow chart to illustrate this procedure, ( it’s a back titration). b...