8. An 0.3130 g primary standard sodium oxalate, Na 2 C 2 O 4 (134.0 g/mol), was

Question

8. An 0.3130 g primary standard sodium oxalate, Na2C2O4 (134.0 g/mol), was dissolved in 75 mL of distilled water, acidified with sulfuric acid and titrated with a K2Cr2O7 solution. It took 20.503 g of the K2Cr2O7 solution to reach the equivalence point.

a.   Calculate the weight molarity (Mw) of the K2Cr2O7 solution.

Cr2O72– + 14 H+ +    3 C2O42–    2 Cr3+      +    6 CO2   +    7 H2O

b.   The above solution was used to determine the amount of sodium oxalate, Na2C2O4 (134.0 g/mol), in an impure sample. A 0.5650 g sample of the impure sodium oxalate was treated as above and titrated with the K2Cr2O7 solution. The titration required 5.0530 g of the K2Cr2O7 solution to reach the endpoint. Calculate the % Na2C2O4 in the impure sample.  

Cr2O72– + 14 H+ +    3 C2O42–    2 Cr3+      +    6 CO2   +    7 H2O

Explanation / Answer

Cr2O72-(aq) + 14H+(aq) + 3C2O42-(aq) ------> 2Cr3+(aq) + 6CO2 + 7H2O(l)

moles of sodium oxalate reacting = mass/molar mass = 0.313/134 = 0.00234

Thus, moles of K2Cr2O7 required = (1/3)*moles of sodium oxalate reacting = 0.00078

Let the density of K2Cr2O7 solution be 1 g/mole

Thus, molarity of the solution = moles of K2Cr2O7 /volume of solution in litres = 0.038 M

b) moles of K2Cr2O7 required = mass of K2Cr2O7 solution *molarity = 0.005053*0.038 = 0.000192 moles

Thus, moles of sodium oxalate reacting = 3*moles of K2Cr2O7 = 0.000576

mass of sodium oxalate reacting = moles*molar mass = 0.000576*134 = 0.077 g

thus, % purity = (mass of pure sodium oxalate/mass of sample)*100 = (0.077/0.565)*`100 = 13.653%

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