Chemistry, 04.01.2021 02:00 vazquezemmy8

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?

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Answer from: levicorey846

Following are the solution to the given points:

Explanation:

Oxalic acid volume $= 25.00 \ mL = 0.0250 \ litres$

KMnO4 volume $= 12.70 \ ml = 0.0127 \ litres$

KMnO4 molarity $= 0.0206\ M = 0.0206 \ \frac{mol}{l}$

In point a:

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:

H_2C_2O_4 molecules mole ratio to MnO_4^- ions:

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

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

The reaction of 5 mol of oxalic acid is 2 mol MnO_4^- ions

H_2C_2O_4 : molecules mole proportion to MnO_4^- ions:

5 H_2C_2O_4 : : 2MnO_4^-

In point c:

The Moles of MnO_4^- ions reacted with the H_2C_2O_4 :

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

$Molarity =\frac{moles}{Volume}\\\\Moles \ of\ KMnO_4 = Molarity \times volume$

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

In point d:

H_2C_2O_4 moles in the sample present:

H_2C_2O_4 moles = moles MnO_4^- ions $\times$ mole ratio

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

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

In point e:

Oxalic acid molarity = $\frac{mole}{volume}$

$=\frac{ 6.54 \times 10^{-4} mol}{0.025\ L} \\\\ = 0.0260 \ M$

In point f:

Oxalic acid level by mass in the solution:

Oxalic acid mass calculation:

Oxalic acid molar weight = 90.0349 $\frac{g}{mol}$ .

Oxalic acid mass per liter = oxalic acid moles per liter $\times$ molar mass

$= 0.0260 \frac{mol}{L} \times 90.0349 \frac{g}{mol}\\\\= 2.3409 \frac{g}{L}\\\\ = 2.3409 \frac{g}{1000 \ mL}\\\\= 0.2409 \frac{g}{100 \ mL}$

When Oxalic acid solution density $= 1.00 \ \frac{g}{mL}$

Mass oxalic acid percentage = $0.2409 \%$

Oxalic acid mass proportion $= 0.24\% \ \frac{W}{v} \ \ Mass$

Answer from: Quest

Answerthis solution will not form a precipitateexplanationsolubility equilbrium: pbf₂(s) ⇄ pb²⁺(aq) + 2 f⁻(aq)the solubility product expression is $k_{sp} = [\text{pb}^{2+}][\text{f}^- ]^2 = 4 \times 10^{-8}$ (we exclude solids from the equilibrium expression).now suppose we have the trial ion product q, which we can define as $q = [\text{pb}^{2+}]_{\text{initial}} \cdot [\text{f}^- ]^2_{\text{initial}}$ then, we can interpret q as the product of the concentration of the ions that we have and ksp as the product of the ion concentrations that are needed to form a saturated solution.if q > ksp, then that implies that there will be a surplus of ions that exceed the amount that is required to form a saturated solution. a precipitate will then form (the precipitate continually forms until the surplus ions are removed and q is decreased to ksp's value).if q = ksp, the saturated solution that forms is barely saturated and the precipitate will be minimal.if q < ksp, then the ion concentration will not be enough to form a saturated solution (so no precipitate).using our information, we calculate the trial ion product: $\begin{aligned} q & = [\text{pb}^{2+}]_{\text{initial}} \cdot [\text{f}^- ]^2_{\text{initial}} \\ & = (1.9 \times 10^{-3} ) (3.8\times 10^{3})^2 \\ & = 2.7 \times 10^{-8} \end{aligned}$ since q = 2.7 × 10⁻⁸ < ksp = 4 × 10⁻⁸, a precipitate will not form.

Answer from: Quest

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