Sodium oxalate, Na_2C_2O_4, in solution is oxidized to CO_2 (9) by MnO_4 which i

Question

Sodium oxalate, Na_2C_2O_4, in solution is oxidized to CO_2 (9) by MnO_4 which is reduced to Mn^2+. A 50.1 - mL volume of a solution of Mn04 is required to titrate a 0.338 - g sample of sodium oxalate. This solution of Mn04 is thon used to analyze uranium-containing samples. A 4 65 - g sample of a uranium-containing material requires 33.0 mL of the solution for titration The oxidation of the uranium can be represented by the change UO^2+ rightarrow UO^2+_2. Calculate the percentage of uranium in the sample.

Explanation / Answer

Solution

Step 1: Balance the equation

2 MnO4- + 5 Na2C2O4 10 CO2 + 2 Mn2+

Step 2:Find out he molar masses of the reactants and subsequently the amount of 2 MnO4- and 5 Na2C2O4 needed for the redox reaction

Molar mass of MnO4- =55+(4x160)=119g

Molar mass of Na2C2O4 =(2x23)+(2x12+(4x16)=134g

                                    2 MnO4-        +        5 Na2C2O4                   10 CO2       +        2 Mn2+

                                    (2x119)g                  5(134)g

                                     238g                        670g

In this problem we have to standardize the MnO4- solution used to oxidize Na2C2O4 (given in g=0.338g)

From the equation we know that 238g of MnO4- is needed to oxidize 670g of Na2C2O4 .Now we can find out how many grams of    MnO4- is needed to oxidize 0.338g of Na2C2O4

670g of Na2C2O4 is oxidized by = 238g of MnO4-

Therefore 0.338g of Na2C2O4 is oxidized by = 238g x 0.338g /670g= 0.120g of MnO4-

0.120g of MnO4- is present in 50.1 mL of the solution. Now we can find out the Molarity 9concentration) of the solution using the formula

M=(number of moles of MnO4- x1000mL)/50.1mL

Number of moles=0.120g/119g= 0.001

M=(0.001 x1000mL)/50.1mL = 0.01996 or 0.02M

The sample is UO2+ .We can calculate the % of uranium in the sample as follows

Molecular mass of UO2+(238.03g+16g)=244.03g

244.03g of UO2+ contains = 238.03g of U

Therefore 4.65g UO2+ contains = (238.03gx 4.65g )/244.03=4.5356g or 4.54gof U

4.65g of UO2+ contains = 4.54gof U

Therefore 100g of UO2+ contains = (4.54gx100g)/ 4.65= 97.63g of U

The % of uranium in the sample of 4.65g UO2+=97.63%

Only this part is needed. If the question is read carefully we will understand that standardization of MnO4- is not needed as the question is only about the % of uranium in the sample

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