Question

Q1)

lead ions can be precipitated from aqueous solutions by the addition of aqueous iodide: Pb2+(aq) + 2I-(aq) → PbI2(s)

Lead iodide is virtually insoluble in water so that the reaction appears to go to completion. How many milliliters of 1.180 M HI(aq) must be added to a solution containing 0.200 mol of Pb(NO3)2(aq) to completely precipitate the lead?

Q2) What is the molarity of a NaOH solution if 15.5 mL of a 0.220 M H2SO4 solution is required to neutralize a 25.0 mL sample of the NaOH solution?

Q3) A Sample of clay of mass 10.750g prepared to make ceramics was analyzed to determine its iron content.

The clay was washed with hydrochloric acid, and the iron was converted into iron(II) ions. The resulting solution was titrated with cerium(IV) sulfate solution.

Fe2+(aq) + Ce4+(aq) Fe3+(aq) + Ce3+(aq)
In the titration, 13.45mL of 1.340M Ce2SO4 was needed to reach the stoichiometric point. What is the mass percentage of iron in the clay?

Answer 1

1) Pb2+(aq) + 2I-(aq) ----> PbI2(s)

So, 1 mole of Pb2+ react with 2 moles of I- to form 1 mole of PbI2

Given, mols of Pb(NO3)2 = 0.200 mol

So, mols of HI = 2 x 0.200 mol = 0.400 mol

Molarity of HI solution = 1.180 M

Volume of HI solution required for comletely precipitate Pb2+ = moles/molarity

= 0.400/1.180 = 0.339 L = 339.0 mL

2) We have, 2NaOH + H2SO4 ---> 2NaCl + 2H2O  

So, 2 mole of NaOH reacts with 1 mole of H2SO4

moles of H2SO4 = molarity x volume = 0.220 M x 0.0155 L = 0.00341 mols

So, moles of NaOH = 2 x 0.00341 = 0.00682 mol

Volume of NaOH used = 25 mL = 0.025 L

moalrity = moles/L = 0.00682/0.025 = 0.273 M will be the molarity of original NaOH solution

3) 1 mole of Ce2SO4 reacts with 1 mole of Fe2+ to form 1 mole of Fe3+

moles of Ce2SO4 = molarity x volume = 1.340 M x 0.01345 L = 0.018 mols

So, mols of Fe2+ = 0.018 mols

mass of iron = moles x molar mass of Fe = 0.018 mols x 55.845 g/mol = 1.00521 g

So, mass % of iron in clay = (1.00521/10.750) x 100 = 9.35%