### Redox Titration

Problem #1: 0.2640 g of sodium oxalate is dissolved in a flask and requires 30.74 mL of potassium permanganate (from a buret) to titrate it and cause it to turn pink (the end point).

The equation for this reaction is:

5Na2C2O4(aq) + 2KMnO4(aq) + 8H2SO4(aq) ---> 2MnSO4(aq) + K2SO4(aq) + 5Na2SO4(aq) + 10CO2(g) + 8H2O(ℓ)

(a) How many moles of sodium oxalate are present in the flask?
(b) How many moles of potassium permanganate have been titrated into the flask to reach the end point?
(c) What is the molarity of the potassium permanganate?

Solution to (a):

0.2640 g / 134.00 g/mol = 0.001970149 mol

to four sig figs, this would be 0.001970 mol

Solution to (b):

From the balanced equation, the oxalate-permanganate molar ratio is five to two.

0.001970149 mol oxalate times (2 mol permanganate / 5 mol oxalate) = 0.00078806 mol permanganate

to four sig figs, this is 0.0007881 mol

Solution to (c):

0.00078806 mol / 0.03074 L = 0.02564 M (to four sig figs)

Problem #2: Potassium dichromate is used to titrate a sample containing an unknown percentage of iron. The sample is dissolved in H3PO4/H2SO4 mixture to reduce all of the iron to Fe2+ ions. The solution is then titrated with 0.01625 M K2Cr2O7, producing Fe3+ and Cr3+ ions in acidic solution. The titration requires 32.26 mL of K2Cr2O7 for 1.2765 g of the sample.

(a) Balance the net ionic equation using the half-reaction method.
(b) Determine the percent iron in the sample.
(c) Is the sample ferrous iodate, ferrous phosphate, or ferrous acetate?

Solution to (a):

6Fe2+ + Cr2O72¯ + 14H+ ---> 6 Fe3+ + 2Cr3+ + 7H2O

The mechanics of balancing is left to the reader.

The key point is the ferrous ion-dichromate molar ratio, which is six to one.

Solution to (b):

1) Determine Fe(II) in solution:

(0.01625 mol/L) (0.03226 L) = 0.000524225 mol dichromate

0.000524225 mol times (6 mol Fe(II) / 1 mol dichromate) = 0.00314535 mol Fe(II)

0.00314535 mol Fe(II) times 55.845 g/mol = 0.175652 g

2) Determine percent of iron in the sample:

0.175652 g / 1.2765 g = 13.76%

Solution to (c):

Which compound contains 13.76% iron? The only way to determine this is to calculate the percent composition of the three sbstances.

Ferrous iodate:

Fe(IO3)2

% Fe: 55.845 g/mol divided by 405.67 g/mol = 13.77%

I think we have a winner!

Here's the percent composition calculator I used to calculate the three percent compositions. Here are the other two substances in the question:

Fe3(PO4)2 = 46.87%

Fe(C2H3O2)2 = 32.11 %

Problem #3: A solution contains both iron(II) and iron(III) ions. A 50.0 mL sample of the solution is titrated with 35.0 mL of 0.00280 M KMnO4, which oxidizes Fe2+ to Fe3+. The permangante ion is reduced to manganese(II) ion. Another 50.0 mL sample of solution is treated with zinc metal, which reduces all the Fe3+ to Fe2+. The resulting solution is again titrated with 0.00280 M KMnO4, this time 48.0 mL is required. What are the concentrations of Fe2+ and Fe3+ in the solution?

Solution:

1) The chemical equation of interest is this:

5Fe2+ + 8H+ + MnO4¯ ---> 5Fe3+ + Mn2+ + 4H2O

The stoichiometric relationship of Fe2+ to permangante is five to one.

2) Calculate moles of Fe2+ reacted:

(0.00280 mol / L) (0.0350 L) = 0.000098 mol of MnO4¯

(0.0000980 mol MnO4¯) (5 mol Fe / 1 mol MnO4¯) = 0.000490 mol Fe(II)

3) Determine the TOTAL iron content. For this, we will use the second 50.0 mL sample after it has been treated with zinc. Since all the Fe3+ has been reduced to Fe2+, we can determine how much iron was in the second 50.0 mL sample.

(0.00280 mol / L) (0.0480 L) = 0.0001344 mol of MnO4¯

(0.0001344 mol MnO4¯) (5 mol Fe / 1 mol MnO4¯) = 0.000672 mol of total Fe

4) Determine Fe(III) in solution and its molarity:

0.000672 mol - 0.000490 mol = 0.000182 mol

0.000182 mol / 0.050 L = 0.00364 M

5) Determine molarity of Fe(II):

0.000490 mol / 0.050 L = 0.0098 M

Problem #4: The quantity of antimony in a sample can be determined by an oxidation-reduction titration with an oxidizing agent. A 9.62 g sample of stibnite, an ore of antimony, is dissolved in hot, concentrated HCl(aq) and passed over a reducing agent so that all the antimony is in the form Sb3+. The Sb3+(aq) is completely oxidized by 43.70 mL of a 0.1250 M solution of KBrO3. Calculate the amount of antimony in the sample and its percentage in the ore.

Solution:

1) Balance the equation (in acidic solution, remember the HCl from the problem statement):

The unbalanced equation for the reaction is:
BrO3¯(aq) + Sb3+(aq) ---> Br¯(aq) + Sb5+(aq)

The balanced equation is:

6H+(aq) + BrO3¯(aq) + 3Sb3+(aq) ---> Br¯(aq) + 3Sb5+(aq) + 3H2O(ℓ)

The key point is the bromate to Sb(III) molar ratio, which is one to three.

2) Determine the bromate used:

(0.1250 mol/L) (0.04370 L) = 0.0054625 mol bromate

3) Determine the Sb(III) that reacted:

0.0054625 mol bromate times (3 mol Sb(III) / 1 mol bromate) = 0.0163875 mol Sb(III)

4) Determine grams of antimony:

0.0163875 mol times 121.760 g/mol = 1.995342 g

to three sig figs, this is 2.00 g

5) Calculate percent of antimony in the sample:

1.995342 g / 9.62 g = 20.7416 %

to three sig figs, this is 20.7%

Problem #5: A rock sample is to be assayed for its tin content by an oxidation-reduction titration with I3¯(aq). A 10.00 g sample of the rock is crushed, dissolved in sulfuric acid, and passed over a reducing agent so that all the tin is in the form Sn2+. The Sn2+(aq) is completely oxidized by 34.60 mL of a 0.5560 M solution of NaI3. The balanced equation for the reaction is

I3¯(aq) + Sn2+(aq) ---> Sn4+(aq) + 3I¯(aq)

Calculate the amount of tin in the sample and its mass percentage in the rock.

Solution:

1) Determine the amount of triiodide used:

(0.5560 mol/L) (0.03460 L) = 0.0192376 mol of I3¯ reacted

2) Determine the Sn(II) that reacted:

the key point is the 1:1 molar ratio between triiodide and the stannous ion

therefore, 0.0192376 mol of Sn(II) reacted

3) Calculate grams of tin(II) ion:

0.0192376 mol times 118.710 g/mol = 2.284 g (to four sig figs)

4) Calculate mass percentage:

2.284 g / 10.00 g = 22.84%

Problem #6: The amount of I3¯(aq) in a solution can be determined by titration with a solution containing a known concentration of S2O32¯(aq) (thiosulfate ion). The determination is based on the balanced equation:

I3¯(aq) + 2S2O32¯(aq) ---> 3I¯(aq) + S4O62¯(aq)

Given that it requires 36.40 mL of 0.3300 M Na2S2O3(aq) to titrate the I3¯(aq) in a 15.00 mL sample, calculate the molarity of I3¯(aq) in the solution.

Solution:

1) Determine the amount of thiosulfate used:

(0.3300 mol/L) (0.03640 L) = 0.012012 mol of thiosulfate

2) Determine triiodide that reacts:

from the balanced equation, the molar ratio between triiodide and thiosulfate is:
one to two
therefore:
0.012012 mol of thiosulfate times (1 mol triiodide / 2 mol thiosulfate) = 0.006006 mol triiodide

3) Determine molarity:

0.006006 mol / 0.01500 L = 0.4004 M (to 4 sig figs)

Problem #7: The amount of Fe2+(aq) in an FeSO4(aq) solution can be determined by titration with a solution containing a known concentration of Ce4+(aq). The determination is based on the reaction

Fe2+(aq) + Ce4+(aq) ---> Fe3+(aq) + Ce3+(aq)

Given that it requires 37.50 mL of 0.09650 M Ce4+(aq) to oxidize the Fe2+(aq) in a 35.00 mL sample to Fe3+(aq), calculate the molarity of Fe2+(aq) and the number of milligrams of iron in the sample.

Solution:

1) Determine the amount of Ce(IV) that reacts:

(0.09650 mol/L) (0.03750 L) = 0.00361875 mol

2) Determine amount of iron(II) ion that reacts:

There is a 1:1 molar ratio between Fe(II) and Ce(IV), therefore:

0.00361875 mol of Fe(II) reacts

3) Determine molarity:

0.00361875 mol / 0.03500 L = 0.1034 M (to four sig figs)

4) Determine milligrams:

0.00361875 mol times 55.845 g/mol = 0.2021 g = 202.1 mg

Problem #8: A solution of I3¯(aq) can be standardized by using it to titrate As4O6(aq). The titration of 0.1021 g of As4O6(s) dissolved in 30.00 mL of water requires 36.55 mL of I3¯(aq). Calculate the molarity of the I3¯(aq) solution. The unbalanced equation is

As4O6(s) + I3¯(aq) ---> As4O10(s) + I¯(aq)

Solution:

1) Balance the equation:

4H2O + As4O6(s) + 4I3¯(aq) ---> As4O10(s) + 12I¯(aq) + 8H+

See a discussion concerning this equation at the end of the solution.

2) Determine how many moles of As4O6 reacted:

0.1021 g divided by 395.6828 g/mol = 0.000258035 mol

3) Determine moles of triiodide that react:

the molar ratio between As4O6 and I3¯ is:
one to four
therefore:
0.000258035 mol As4O6 times (4 mol I3¯ / 1 mol As4O6) = 0.00103214 mol

4) Calculate the molarity:

0.00103214 mol / 0.03655 L = 0.02824 M (to four sig figs)

Comment: the actual reactions are different from the equation presented above. They can be viewed here.

Please note that four H3AsO4 would be produced from one As4O6. However, to produce one As4O10 would require four H3AsO4. So the above equation, while not showing the actual chemical reactions, does accurately represent the ratio (which is 1:4) between As4O6 and triiodide.

Problem #9: A 32.15 mL sample of a solution of MoO42¯(aq) was passed through a Jones reductor (a column of zinc powder) in order to convert all the MoO42¯(aq) to Mo3+(aq). The filtrate required 20.85 mL of 0.09550 M KMnO4-(aq) for the reaction given by

MnO4¯(aq) + Mo3+(aq) ---> Mn2+(aq) + MoO22+(aq)

Balance this equation and then calculate the concentration of the original MoO42¯(aq) solution.

Solution:

1) Balance the equation:

the balanced half-reactions:
5e¯ + 8H+ + MnO4¯(aq) ---> Mn2+(aq) + 4H2O(ℓ)
2H2O + Mo3+(aq) ---> MoO22+(aq) + 4H+ + 3e¯

the balanced equation:

4H+ + 3MnO4¯(aq) + 5Mo3+(aq) ---> 3Mn2+(aq) + 5MoO22+(aq) + 2H2O

2) Calculate the moles of permanganate required:

(0.09550 mol/L) (0.02085 L) = 0.001991175 mol

3) Calculate moles of the molybdate:

the key point is the molar ratio between permanganate and molybdate, which is:
three to five

therefore:

0.001991175 mol permanganate times (5 mol molybdate / 3 mol permanganate = 0.003318625 mol molybdate

4) Calculate the concentration:

0.003318625 mol / 0.03215 L = 0.1032 M (to four sig figs)

Problem #10: An ore is to be analyzed for its iron content by an oxidation-reduction titration with permanganate ion. A 4.230 g sample of the ore is dissolved in hydrochloric acid and passed over a reducing agent so that all the iron is in the form Fe2+. The Fe2+(aq) is completely oxidized by 31.60 mL of a 0.05120 M solution of KMnO4. The unbalanced equation for the reaction is

KMnO4(aq) + HCl(aq) + FeCl2(aq) ---> MnCl2(aq) + FeCl3(aq) + H2O(ℓ)

Calculate the amount of iron in the sample and its mass percentage in the ore.

Solution:

1) The unbalanced, net ionic equation is:

MnO4¯ + Fe2+ ---> Mn2+ + Fe3+
the balanced equation is:
8H+ + MnO4¯ + 5Fe2+ ---> Mn2+ + 5Fe3+ + 4H2O

2) Calculate moles of permanganate:

(0.05120 mol/L) (0.03160 L) = 0.00161792 mol

3) Determine the amount of Fe(II) that reacted:

the molar ratio for the reaction of permanganate and Fe(II) is:
one to five
therefore:
0.00161792 mol permanganate times (5 mol iron(II) / 1 mol permanganate) = 0.0080896 mol iron(II)

4) Calculate grams of iron in sample, then its mass percentage:

0.0080896 mol iron(II) times 55.845 g/mol = 0.4518 g (to four sig figs)

0.4517637 g / 4.230 g = 10.68%

Problem #11: 10.0 cm3 of 0.10 mol dm-3 aqueous iron(II) sulfate is titrated against 0.025 mol dm-3 aqueous potassium manganate(VII) in the presence of an excess of both hydrogen ions and fluoride ions. It is found that exactly 10.0 cm3 of the manganate(VII) solution is required to reach end point. What is the oxidation number of the manganese at the end point? (Warning from the ChemTeam: be sure to read commentary at the end of the solution.)

Solution:

1) Determine how much Fe2+ is present:

(0.10 mol/L) (0.0100 L) = 0.0010 mol

2) Determine how much MnO4¯ reacted:

(0.025 mol/L) (0.0100 L) = 0.00025 mol

3) Calculate the number of Fe2+ oxidized per one MnO4¯ reduced:

0.0010 mol / 0.00025 mol = 4

4) Conclusion:

Mn is reduced from +7 to +3

Comment: this is a problem with a fake answer. Under the chemical circumstances specified, manganese would be reduced to Mn2+. This problem is deliberately set up to catch a "quickie" answer by the student who knows that, in the real world, the Fe goes +2 to +3 and the Mn goes from +7 to +2.

Part of the real world underlying this problem is the knowledge that Mn, being at its highest oxidation state, can only be reduced. Therefore, the Fe must be oxidized and, based on real world knowledge, it can only go from +2 to +3.

Consequently, we can reason that the Mn goes down 4 in its oxidation number.

Problem #12: A 2.50 g sample of bronze, an alloy of copper and tin, was dissolved in sulfuric acid. The copper in the alloy reacts with sulfuric acid as shown by the following balanced reaction:

Cu + 2H2SO4 ---> Cu2+ + SO2 + 2H2O + SO42¯

Adding KI produces CuI and triodide, I3¯

2Cu2+ + 5I¯ ---> 2CuI + I3¯

Finally, titrating the I3¯ with S2O32¯

I3¯ + 2S2O32¯ ---> 3I¯ + S4O62¯

provides an indirect method for determining the amount of Cu in the original sample. Calculate the percentage, by mass, of copper in a sample of bronze if 31.50 mL of 1.000 M S2O32¯ is consumed in the titration.

Solution:

1) Determine the moles of S2O32¯ consumed:

(1.000 mol/L) (0.03150 L) = 0.03150 mol

2) S2O32¯ reacts with triiodide in 2:1 molar ratio. How much triiodide was consumed?

2 is to 0.03150 mol as 1 is to x

x = 0.01575 mol

3) Triiodide is in a 2:1 molar ratio with Cu2+ ion. For every 1 triiodide produced, two copper ions were consumed. How much Cu2+ was consumed?

1 is to 0.01575 mol as 2 is to x

x = 0.03150 mol of Cu2+

4) In the reaction between Cu and sulfuric acid, the molar ratio between Cu and Cu2+ is 1:1. This means 0.03150 mol of Cu was consumed. How many grams of Cu is this?

0.03150 mol times 63.546 g/mol = 2.00 g (to three sig figs)

5) The percentage of copper in the bronze is:

(2.00 g / 3.50 g) * 100 = 57.1%

6) A slight variation would be to combined chemical equations. Take the last equation in the question and rearrange it as follows:

I3¯ ---> 3I¯ + S4O62¯ - 2S2O32¯

Notice how I subtracted the 2S2O32¯.

7) Now, substitute for the I3¯ in the second equation in the question:

2Cu2+ + 5I¯ ---> 2CuI + 3I¯ + S4O62¯ - 2S2O32¯

The thing to notice in the above equation is the 1:1 molar ratio between the S2O32¯ and the Cu2+. That means that 0.03150 mol of S2O32¯ reacts with 0.03150 mol of Cu2+. We can skip the calculation involving the triiodide.

Sometimes what I did can be done. Sometimes, it's too complex, so you suck it up and just do all the backwards calculations one step at a time.