Question

1a) A sample of a new cleaning product, "Joe's Famous Bleach Cleaner," with a mass of 59.0 g , was diluted with an acetic acid solution containing excess I−. A small amount of starch indicator solution was then added, turning the solution a deep bluish-purple. The solution was then titrated with 0.190 M sodium thiosulfate, Na2S2O3, containing the ion S2O32−. A volume of 45.0 mL of sodium thiosulfate, the titrant, was needed to turn the solution colorless. What is the percentage composition by mass of NaClO in the bleach product?

1b) A stainless steel alloy is to be analyzed for its chromium content. A 3.50 g sample of the steel is used to produce 250.0 mL of a solution containing Cr2O72−. A 10.0-mL portion of this solution is added to BaCl2(aq). When the pH of the solution is properly adjusted, 0.145 g BaCrO4(s) precipitates. What is the percent Cr, by mass, in the steel sample?

Answer 1

1a. The balanced chemical equation for the reactions are

ClO^- (aq) + 2 I^- (aq) + 2 H⁺ (aq) ----------> Cl^- (aq) + I₂ (aq) + H₂O (l) .......(1)

2 S₂O₃^2- (aq) + I₂ (aq) ----------> S₄O₆^2- (aq) + 2 I^- (aq) ........(2)

As per stoichiometric equation (2),

2 mols S₂O₃^2- = 1 mol I₂.

Millimol(s) S₂O₃^2- added = (45.0 mL)*(0.190 M) = 8.55 mmol.

Millimol(s) I₂ titrated = (8.55 mmol S₂O₃^2-)*(1 mol I₂)/(2 mols S₂O₃^2-) = 4.275 mmol

Again, as per stoichiometric equation (1),

1 mol I₂ = 1 mol ClO^-.

Therefore,

Millimol(s) ClO^- = millimol(s) NaClO in bleaching powder = 4.275.

The atomic masses are

Na: 22.99 g/mol

Cl: 35.45 g/mol

O: 15.99 g/mol

The gram molar mass of NaClO = (1*22.99 + 1*35.45 + 1*15.99) g/mol = 74.43 g/mol.

Mass of NaClO corresponding to 4.275 mmol = (4.275 mmol)*(74.43 g/mol)

= (4.275 mmol)*(1 mol)/(1000 mmol)*(74.43 g/mol)

= 0.318 g.

Percent NaClO in the sample of bleaching powder = (0.318 g)/(59.0 g)*100

= 0.5389

≈ 0.54 (ans).

1b. The balanced chemical equation for the reaction is given as

Cr₂O₇^2- (aq) + 2 BaCl₂ (aq) + 2 OH^- (aq) ---------> 2 BaCrO₄ (s) + 4 Cl^- (aq) + H₂O (l)

The atomic masses are

Ba: 137.327 g/mol

Cr: 51.996 g/mol

O: 15.999 g/mol

Gram molar mass of BaCrO₄ = (1*137.327 + 1*51.996 + 4*15.999) g/mol = 253.319 g/mol.

Mol(s) BaCrO₄ corresponding to 0.145 g = (0.145 g)/(253.319 g/mol)

= 5.7240*10^-4 mol.

As per the formulation of BaCrO₄,

1 mol BaCrO₄ = 1 mol Cr.

Therefore,

5.7240*10^-4 mol BaCrO₄ = 5.7240*10^-4 mol Cr.

Mass of Cr corresponding to 5.7240*10^-4 mol = (5.7240*10^-4 mol)*(51.996 g/mol) = 0.02976 g.

The above amount of Cr was present in 3.50 g sample of the stainless steel. Hence, the percentage of Cr in the sample = (0.02976 g)/(3.50 g)*100

= 0.8502

≈ 0.850 (ans).