Oxidation Reduction Pre-la Questions Titrations I: Determination of Oxalate Befo

Question

Oxidation Reduction Pre-la Questions Titrations I: Determination of Oxalate Before beginning this experiment in the laboratory, you should be able to answer the following questions. 1. If 0.4586 g of sodium oxalate. Nazczo4, requires 33.67 mL of a KMno. solution to reach the end point, what is the molarity of the KMnoa solution? 2. Titration of an oxalate sample gave the following percentages: 15.53%, 15.55%, and 15.56%. Calculate the average and the standard deviation 3. Why does the solution decolorize upon standing after the equivalence point has been reached? 4. why is the KMno solution filtered, and why should it not be stored in a rubber-stoppered bottle? s. What volume of 100 MKMno, would be required to titrate o.23 gof Kalcu o4): l 2Hzo? 6. Calculate the percent c o in each of the following H2C20a Na:C o and k, Alc os, 3H o.

Explanation / Answer

Using eqn ,M1*V1=M2V2

M1=molarity of Na2C2O4

V1=volume of Na2C2O4

Moles of Na2C2O4=M1*V1

M2=molarity of KMnO4=?

V2=volume of KMnO4=33.67 ml=0.03367L

M2=M1*V1/V2=0.0034/0.03367L=0.1 Mol/L

M2=0.1M

2)15.53%,15.55%,15.56%

Av=15.53+15.55+15.56/3=15.55%

Deviation1=15.53-mean=15.53-15.55=-0.02

Dev2=15.55-15.55=dev3=15.56-15.55=0.01

Std dev={[(dev1)^2+(dev2)^2+dev3^2]/3}^1/2=[(-0.02)^2+0+(0.01)^2/3]^1/2=0.013

3)MnO4- or permanganate ion are reduced to Mn2+ ions ,by oxalate so the chemical nature change causes decolorisation

4) To remove MnO2 precipitate that may have formed during titration it is filtered .As the titrant may reduce MnO2 and increase the concentration of KMno4.KMno4 is a strong oxidising agent ,so it may oxidize the rubber stopper

V2=M1*V1/M2=moles of complex/V2=4.0*10^-4 moles/0.1mol/L=0.0040 L=4 ml

Mass of H2C2O4=90.03 g/mol

Mass of C2O42-=88.019 g/mol

% C2O42-=88.019/90.03*100=97.7%

Na2C2O4=134g/mol

% C2O42=88.019/134*100=65.7%

K2C2O4=166.22 g/mol

% C2O42=88.019/166.22*100=52.95%

K3[Al(C2O4)3].3H2O=462.38g/mol

%C2O4=88.019*3/462.38*100=57.1%

7)5C2O42- (aq) +2MnO4- (aq) + 16H+ (aq) ------>   10CO2 (g) + 8H2O (l) + 2 Mn2+(aq)

C2O42- and MnO4- reacts in the ratio 5:2

moles of KMno4 required =molarity*volume=0.02 mol/L*28.7*10^-3 L=0.00057 moles=

moles of oxalate=5/2*0.00057=0.0014 moles

Moles of complex=0.25g/462.38g/mol=5.4*10^-4 moles=0.00054 moles

% oxalate=0.0014/0.00054*100=259.2 %

8) 1 mole complex has 3 moles oxalate

0.00054 moles complex has oxalate=3*0.00054 moles=0.00162 moles

But actual moles of oxalate=0.0014 moles

impurity moles=0.0002 moles*462.38 g/mol=0.0924 g(impurity)

% purity= amount of pure complex/total amount*100=(0.25-0.0924/0.25)*100=63%

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