Question

A Fajans titration of a 0.7908-g total sample weight required 45.32 mL of 0.1046 M AgNO3. Express the results of this analysis in terms of the percentage of: (a) Cl–, or (b) BaCl2.H2O.

A. Write balanced net ionic equations for the following reactions supply Ht and/or H20 as needed. Mn0,2- +H2SO, Mn2+ + SO2 Balanced Hal Reactions Balanced overal reaction Fed+ + Sn2+ - Fe2+ + Sn++ Balanced half reactions Balanced overall reaction

Answer 1

Ans: Fajans titration

Molecular Weights

Silver Nitrate    = 169.87 g/mole

Cl- (chloride)    = 35.5 g/mole

BaCl2.2H2O     = 244.26 g/mole

The precipitation takes place as follows:

(1)        AgNO3                                         +          Cl-                         →        AgCl (s)

            169.87 g/mole                          35.5 g/mole

(2)        2AgNO3                         +                       BaCl2.2H2O       →        2AgCl (s)

            2 *169.87 g/mole                                 244.26 g/mole

Need to calculate the weight of silver nitrate present in 45.32 mL of 0.1046 M AgNO3

1 M silver nitrate solution has 169.87 g dissolved in 1000 mL

The 0.1046 M AgNO3 will have = 169.87 * 0.1046 = 17.7684 g dissolved in 1000 mL

So, 45.32 mL of 0.1046 M AgNO3 will have = (17.7684 g / 1000 mL) * 45.32 m                

= 0.8053 g of AgNO3

(a)Calculation of percent of chloride ions:

Since, 169.87 g of silver nitrate reacts completely with 35.5 g of chloride ions to precipitate AgCl.

So, 0.8053 g of AgNO3 will react = (35.5 g / 169.87 g) * 0.8053 g = 0.1683 g of Cl-

The 0.7908 g of the above sample contains 0.1683 g of Cl- ions

Percentage of chloride ions = (0.1683 / 0.7908) * 100 = 21.28 %

(b) Calculation of percent of BaCl2.2H2O in the sample:

Since, 339.74 g (2*169.87 g) of silver nitrate reacts completely with 244.26 g of barium chloride to precipitate 2.AgCl.

So, 0.8053 g of AgNO3 will react = (244.26 g / 339.74 g) * 0.8053 g = 0.5790 g of BaCl2.2H2O

The 0.7908 g of the above sample contains 0.5790 g of BaCl2.2H2O

Percentage of BaCl2.2H2O = (0.5790 / 0.7908) * 100 = 73.21 %

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Ans:

A (i)     Half Reaction:

2 MnVII + 10 e- → 2 MnII (reduction)

5 SIV - 10 e- → 5 SVI (oxidation)

Balanced ionic equation:

2MnO4- (aq) + 10 H+ (aq) + 5SO32- (aq) → 2Mn2+ + 5SO42- (aq) + 3H2O (l) + 4H+ (aq)

                       

Net ionic equation:

2MnO4- (aq) + 6H+ (aq) + 5SO32- (aq) → 2Mn2+ + 5SO42- (aq) + 3H2O (l)

A (ii)    Half Reactions:

2 Fe3+ + 2e- → 2Fe2+

Sn2+ → Sn4+ + 2e -

Balanced ionic equation:

2 Fe3+ (aq) + Sn2+ (aq) → 2 Fe2+ (aq) + Sn4+ (aq)

Net Ionic Reaction:

2 Fe3+ (aq) + Sn2+ (aq) → 2 Fe2+ (aq) + Sn4+ (aq)