A student adds 5.00 mL of 0.775 M of oxalic acid to 6.00 mL of distilled water.

Question

A student adds 5.00 mL of 0.775 M of oxalic acid to 6.00 mL of distilled water. To this solution the student adds 1.00 mL of 0.130 M potassium permanganate. The total volume of the mixture is 12.0 mL. the elapsed time for the reaction is 240s. calculate initial concentration of oxalic acid in the reaction mixture. Calculate the initial concentration of potassium permanganate in the reaction mixture. Calculate the initial rate of the reaction. Explain the tones of color changes of thymol blue indicator produced by me addition of these solutions In each case give the approximate pH of the solution (Use the table on indicators in EQUL 397 for color change and pH range)) A few drops of thymol blue are added to HCl (a q). Small amount of sodium acetate (a basic salt) is added to a solution of (a) A small quantity (about 3-4 drops) of sodium hydroxide is added to solution (b). an additional larger quantity of sodium hydroxide is added to solution(C).

Explanation / Answer

a) Concentration of Oxalic Acid in reaction mixture-

Given- Volume of oxalic acid V1 = 5ml

M = 0.775 M oxalic acid

V2 = 6ml of water

Total volume = Vt

The volume is Vt = V1+V2 = 5 ml + 6 ml = 11 ml

Moles of Oxalic acid M1*V1 = M2*Vt

0.775 x 5ml = M2 x 11ml

So M2 = 0.775M x 5ml /11ml = 0.352M

Concentration of Oxalic Acid in reaction mixture = 0.352M

b) Concentration of KMnO4 in reaction mixture –

Given- Volume of oxalic acid V1 = 5ml

M = 0.130 KMnO4

V2 = 6ml of water

V3 = 1 ml KMnO4

Vt = V1+V2 + V3 =5 ml + 6 ml + 1 ml = 12 ml

M1*V1 = M2*Vt

0.130 x 1ml = M2 x 12ml

So M2 = 0.130 M x 5ml /12ml = 0.054M

Concentration of KMnO4 in reaction mixture is 0.054M

c) Initial rate of reaction-

3 H2C2O4 (aq) + 2 MnO4-(aq) --------à 6 CO2 (g) + 2 MnO2(s) + 2 OH-(aq) + 2 H2O

Rate of Reaction = -1/2 d [MnO4-]/dt

t = 240 s

Average Rate of Reaction = ½ [MnO4-]0/ time to completion

Average Rate of Reaction = ½[0.054M]/240s = 0.0001125 = 1.125 x 10^-4 M/s

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