A sample of oxalic acid is titrated with a standardized solution of KMNO4. A 25 mL sample of oxalic acid required 12.7 mL of 0.0206 M KMnO4 to achieve a pink colored solution. The balanced equation for this reaction is shown below:

6 H+ (aq) + 2 MnO4 - (aq) + 5 H2C2O4(aq) → 10 CO2(g) +8 H2O(l) + 2Mn2+(aq)

Required:
a. What does the pink color signify in this reaction?
b. What is the ratio of MnO4 - ions to H2C2O4 molecules in this reaction?
c. How many moles of MnO4 - ions reacted with the given amount of oxalic acid solution?
d. How many moles of H2C2O4 were present?
e. What was the molarity of the oxalic acid solution?
f. If the density of the oxalic acid solution was 1.00 g/mL, what was the percentage by mass of oxalic acid in the solution?

Its pink presence after full intake of oxalic acid with attachment to KMnO4 is suggested by the end-point of the process due to the small abundance of KMnO4, As just a self predictor, KMnO4 is used.

In point b:

[tex]H_2C_2O_4[/tex] molecules mole ratio to [tex]MnO_4^-[/tex] ions:

The equilibrium for both the oxalic acid and KMnO4 reaction is suggested:

[tex]6H+ (aq) + 2MnO_4- (aq) + 5H_2C_2O_4 (aq) \rightarrow 10CO_2 (g) + 8H_2O (l) + 2Mn_2+ (aq)[/tex]

The reaction of 5 mol of oxalic acid is 2 mol [tex]MnO_4^-[/tex] ions

[tex]H_2C_2O_4[/tex]: molecules mole proportion to [tex]MnO_4^-[/tex] ions:

[tex]5 H_2C_2O_4[/tex]: : [tex]2MnO_4^-[/tex]

In point c:

The Moles of [tex]MnO_4^-[/tex] ions reacted with the [tex]H_2C_2O_4[/tex]:

The molar mass of the solution is the number of solute moles in each volume of water

[tex]Molarity =\frac{moles}{Volume}\\\\Moles \ of\ KMnO_4 = Molarity \times volume[/tex]

Moles with ions reacted to mol with both the amount of : supplied.

In point d:

[tex]H_2C_2O_4[/tex] moles in the sample present:

[tex]H_2C_2O_4[/tex] moles = moles [tex]MnO_4^-[/tex] ions [tex]\times[/tex] mole ratio

[tex]H_2C_2O_4[/tex] moles in the sample = [tex]2.6162 \times 10^{-4}\ mol \times (\frac{5}{2})[/tex]

[tex]H_2C_2O_4[/tex] molecules = [tex]6,5405\times 10^{-4}[/tex] mol are present in the sample

In point e:

Oxalic acid molarity = [tex]\frac{mole}{volume}[/tex]

[tex]=\frac{ 6.54 \times 10^{-4} mol}{0.025\ L} \\\\ = 0.0260 \ M[/tex]

In point f:

Oxalic acid level by mass in the solution:

Oxalic acid mass calculation:

Oxalic acid molar weight = 90.0349 [tex]\frac{g}{mol}[/tex].

Oxalic acid mass per liter = oxalic acid moles per liter [tex]\times[/tex] molar mass

[tex]= 0.0260 \frac{mol}{L} \times 90.0349 \frac{g}{mol}\\\\= 2.3409 \frac{g}{L}\\\\ = 2.3409 \frac{g}{1000 \ mL}\\\\= 0.2409 \frac{g}{100 \ mL}[/tex]

When Oxalic acid solution density[tex]= 1.00 \ \frac{g}{mL}[/tex]

Mass oxalic acid percentage = [tex]0.2409 \%[/tex]

Oxalic acid mass proportion [tex]= 0.24\% \ \frac{W}{v} \ \ Mass[/tex]