A solution of KMnO 4 is standarized against pure sodium oxalate. 0.250 g of sodi

Question

A solution of KMnO4 is standarized against pure sodium oxalate. 0.250 g of sodium oxalate is dissolved in water and then strongly acidified using 4 M H2SO4 solution. The solution is then heated to 80 degrees celcius. The titration is required 28.00 mL of KMnO4 solution according to the following reaction:

5 Na2C2O4 + 2 KMnO4 + 8 H2SO4 --> 5 Na2SO4 + 10 CO2 + K2SO4 + 8 H2O + MnSO4

a. determine the concentration of the permanganate solution?

b. This solution is used to determine the concentration of an unknown commercial peroxide solution. The volume of permanganate required to titrate 10.0 mL of peroxide solution is 15.90 mL. What is the concentration of the peroxide solution?

5H2O2 + 2 KMnO4 + 3 H2SO4 --> 2 MnSO4 + 5 O2 + K2SO4 + 8 H2O

Explanation / Answer

m = 0.25 g of NaOx

MW of NaOX = 134 g/mol

mol of NaOx = mass/MW = 0.25/134 = 0.00186 mol of NaOX

dissolved in water and acidified in H2SO4

V = 28 mL of KMnO4 required

so

a)

Concentration of KMnO4

mol of NaOx = 0.00186 mol

ratio is 5 mol of NaOx = 2 mol of KMnO4

so

0.00186 mol of NaOx = 2/5*0.00186 = 0.000744 mol of KMnO4

[KMnO4] = mmol/V = 0.000744 / (28*10^-3) = 0.026571 M

b)

V = 10 mL of peroxide required

V = 15. 90 mL of Permangante

the reaction is

5:2 ratio as well

so

mol of KMnO4 = MV = 0.026571*15.90 = 0.42247 mmol of KMnO4

then

ratio for H2O2

mmol of H2O2 = 5/2*0.42247 = 1.056175 mmol of H2O2

[H2O2] = mmol/V = 1.056175/(10)= 0.1056175 M of H2O2

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