Example 20

The calcium salt of oxalic acid [Ca(O₂CCO₂)] is found in the sap and leaves of some vegetables, including spinach and rhubarb, and in many ornamental plants. Because oxalic acid and its salts are toxic, when a food such as rhubarb is processed commercially, the leaves must be removed, and the oxalate content carefully monitored.

The reaction of MnO₄⁻ with oxalic acid (HO₂CCO₂H) in acidic aqueous solution produces Mn²⁺ and CO₂:

$$\underset{\text{purple}}{{\text{MnO}}_{\text{4}}{}^{-}\text{(aq)}}+{\text{HO}}_{\text{2}}{\text{CCO}}_2\text{H(aq)}+{\text{H}}^{\text{ + }}\text{(aq)}\to \underset{\text{colorless}}{{\text{Mn}}_{}^{2+}\text{(aq)}}+{\text{CO}}_{\text{2}}\text{(g)}+{\text{H}}_{\text{2}}\text{O(l)}$$

Because this reaction is rapid and goes to completion, potassium permanganate (KMnO₄) is widely used as a reactant for determining the concentration of oxalic acid.

Suppose you stirred a 10.0 g sample of canned rhubarb with enough dilute H₂SO₄(aq) to obtain 127 mL of colorless solution. Because the added permanganate is rapidly consumed, adding small volumes of a 0.0247 M KMnO₄ solution, which has a deep purple color, to the rhubarb extract does not initially change the color of the extract. When 15.4 mL of the permanganate solution have been added, however, the solution becomes a faint purple due to the presence of a slight excess of permanganate (Figure 4.23 "The Titration of Oxalic Acid with Permanganate"). If we assume that oxalic acid is the only species in solution that reacts with permanganate, what percentage of the mass of the original sample was calcium oxalate?

Given: equation, mass of sample, volume of solution, and molarity and volume of titrant

Asked for: mass percentage of unknown in sample

Strategy:

A Balance the chemical equation for the reaction using oxidation states.

B Calculate the number of moles of permanganate consumed by multiplying the volume of the titrant by its molarity. Then calculate the number of moles of oxalate in the solution by multiplying by the ratio of the coefficients in the balanced chemical equation. Because calcium oxalate contains a 1:1 ratio of Ca²⁺:⁻O₂CCO₂⁻, the number of moles of oxalate in the solution is the same as the number of moles of calcium oxalate in the original sample.

C Find the mass of calcium oxalate by multiplying the number of moles of calcium oxalate in the sample by its molar mass. Divide the mass of calcium oxalate by the mass of the sample and convert to a percentage to calculate the percentage by mass of calcium oxalate in the original sample.

Solution:

A As in all other problems of this type, the first requirement is a balanced chemical equation for the reaction. Using oxidation states gives

2MnO₄⁻(aq) + 5HO₂CCO₂H(aq) + 6H⁺(aq) → 2Mn²⁺(aq) + 10CO₂(g) + 8H₂O(l)

Thus each mole of MnO₄⁻ added consumes 2.5 mol of oxalic acid.

B Because we know the concentration of permanganate (0.0247 M) and the volume of permanganate solution that was needed to consume all the oxalic acid (15.4 mL), we can calculate the number of moles of MnO₄⁻ consumed:

$$\text{15}\text{.4 }\cancel{\text{mL}}\left(\frac{\text{1 }\cancel{\text{L}}}{\text{1000 }\cancel{\text{mL}}}\right)\left(\frac{\text{0}{\text{.0247 mol MnO}}_4{}^{-}}{\text{1 }\cancel{\text{L}}}\right)=\text{ 3}\text{.80}\times {\text{10}}^{-\text{4}}{\text{ mol MnO}}_{\text{4}}{}^{-}$$

The number of moles of oxalic acid, and thus oxalate, present can be calculated from the mole ratio of the reactants in the balanced chemical equation:

$$\begin{array}{ll}{\text{moles HO}}_{\text{2}}{\text{CCO}}_{\text{2}}\text{H }\hfill & =\text{ 3}\text{.80}\times {\text{10}}^{-\text{4}}\cancel{{\text{mol MnO}}_{\text{4}}{}^{-}}\left(\frac{{\text{5 mol HO}}_{\text{2}}{\text{CCO}}_{\text{2}}\text{H}}{\text{2 }\cancel{{\text{mol MnO}}_{\text{4}}{}^{-}}}\right)\hfill \\ \hfill & =\text{ 9}\text{.50}\times {\text{10}}^{-\text{4}}{\text{ mol HO}}_{\text{2}}{\text{CCO}}_{\text{2}}\text{H}\hfill \end{array}$$

C The problem asks for the percentage of calcium oxalate by mass in the original 10.0 g sample of rhubarb, so we need to know the mass of calcium oxalate that produced 9.50 × 10⁻⁴ mol of oxalic acid. Because calcium oxalate is Ca(O₂CCO₂), 1 mol of calcium oxalate gave 1 mol of oxalic acid in the initial acid extraction:

Ca(O₂CCO₂)(s) + 2H⁺(aq) → Ca²⁺(aq) + HO₂CCO₂H(aq)

The mass of calcium oxalate originally present was thus

$$\begin{array}{ll}{\text{mass of CaC}}_{\text{2}}{\text{O}}_{\text{4}}\hfill & =\text{ 9}\text{.50}\times {\text{10}}^{-\text{4}}\cancel{{\text{mol HO}}_{\text{2}}{\text{CCO}}_{\text{2}}\text{H}}\left(\frac{\text{1 }\cancel{{\text{mol CaC}}_{\text{2}}{\text{O}}_{\text{4}}}}{\text{1 }\cancel{{\text{mol HO}}_{\text{2}}{\text{CCO}}_{\text{2}}\text{H}}}\right)\left(\frac{\text{128}{\text{.10 g CaC}}_{\text{2}}{\text{O}}_{\text{4}}}{\text{1 }\cancel{{\text{mol CaC}}_{\text{2}}{\text{O}}_{\text{4}}}}\right)\hfill \\ \hfill & =\text{ 0}{\text{.122 g CaC}}_{\text{2}}{\text{O}}_{\text{4}}\hfill \end{array}$$

The original sample contained 0.122 g of calcium oxalate per 10.0 g of rhubarb. The percentage of calcium oxalate by mass was thus

$${\text{\% CaC}}_{\text{2}}{\text{O}}_{\text{4}}=\frac{\text{0}\text{.122 g}}{\text{10}\text{.0 g}}\times \text{100 }=\text{ 1}\text{.22\%}$$

Because the problem asked for the percentage by mass of calcium oxalate in the original sample rather than for the concentration of oxalic acid in the extract, we do not need to know the volume of the oxalic acid solution for this calculation.

Exercise

Glutathione is a low-molecular-weight compound found in living cells that is produced naturally by the liver. Health-care providers give glutathione intravenously to prevent side effects of chemotherapy and to prevent kidney problems after heart bypass surgery. Its structure is as follows:

Glutathione is found in two forms: one abbreviated as GSH (indicating the presence of an –SH group) and the other as GSSG (the disulfide form, in which an S–S bond links two glutathione units). The GSH form is easily oxidized to GSSG by elemental iodine:

2GSH(aq) + I₂(aq) → GSSG(aq) + 2HI(aq)

A small amount of soluble starch is added as an indicator. Because starch reacts with excess I₂ to give an intense blue color, the appearance of a blue color indicates that the equivalence point of the reaction has been reached.

Adding small volumes of a 0.0031 M aqueous solution of I₂ to 194 mL of a solution that contains glutathione and a trace of soluble starch initially causes no change. After 16.3 mL of iodine solution have been added, however, a permanent pale blue color appears because of the formation of the starch-iodine complex. What is the concentration of glutathione in the original solution?

Answer: 5.2 × 10⁻⁴ M

Example 21

The structure of vitamin C (ascorbic acid, a monoprotic acid) is as follows:

An absence of vitamin C in the diet leads to the disease known as scurvy, a breakdown of connective tissue throughout the body and of dentin in the teeth. Because fresh fruits and vegetables rich in vitamin C are readily available in developed countries today, scurvy is not a major problem. In the days of slow voyages in wooden ships, however, scurvy was common. Ferdinand Magellan, the first person to sail around the world, lost more than 90% of his crew, many to scurvy. Although a diet rich in fruits and vegetables contains more than enough vitamin C to prevent scurvy, many people take supplemental doses of vitamin C, hoping that the extra amounts will help prevent colds and other illness.

Suppose a tablet advertised as containing 500 mg of vitamin C is dissolved in 100.0 mL of distilled water that contains a small amount of the acid–base indicator bromothymol blue, an indicator that is yellow in acid solution and blue in basic solution, to give a yellow solution. The addition of 53.5 mL of a 0.0520 M solution of NaOH results in a change to green at the endpoint, due to a mixture of the blue and yellow forms of the indicator (Figure 4.24 "The Titration of Ascorbic Acid with a Solution of NaOH"). What is the actual mass of vitamin C in the tablet? (The molar mass of ascorbic acid is 176.13 g/mol.)

Given: reactant, volume of sample solution, and volume and molarity of titrant

Asked for: mass of unknown

Strategy:

A Write the balanced chemical equation for the reaction and calculate the number of moles of base needed to neutralize the ascorbic acid.

B Using mole ratios, determine the amount of ascorbic acid consumed. Calculate the mass of vitamin C by multiplying the number of moles of ascorbic acid by its molar mass.

Solution:

A Because ascorbic acid acts as a monoprotic acid, we can write the balanced chemical equation for the reaction as

HAsc(aq) + OH⁻(aq) → Asc⁻(aq) + H₂O(l)

where HAsc is ascorbic acid and Asc⁻ is ascorbate. The number of moles of OH⁻ ions needed to neutralize the ascorbic acid is

$${\text{moles OH}}^{-}=\text{ 53}\text{.5 }\cancel{\text{mL}}\left(\frac{\cancel{\text{1 L}}}{\text{1000 }\cancel{\text{mL}}}\right)\left(\frac{\text{0}{\text{.0520 mol OH}}^{-}}{\cancel{\text{1 L}}}\right)=\text{ 2}\text{.78}\times {\text{10}}^{-\text{3}}{\text{ mol OH}}^{{}^{-}}$$

B The mole ratio of the base added to the acid consumed is 1:1, so the number of moles of OH⁻ added equals the number of moles of ascorbic acid present in the tablet:

$$\begin{array}{cc}\text{mass ascorbic acid}& =\text{ 2}\text{.78}\times {\text{10}}^{-\text{3}}\cancel{\text{mol HAsc}}\end{array}\left(\frac{\text{176}\text{.13 g HAsc}}{\text{1 }\cancel{\text{mol HAsc}}}\right)=\text{ 0}\text{.490 g HAsc}$$

Because 0.490 g equals 490 mg, the tablet contains about 2% less vitamin C than advertised.

Exercise

Vinegar is essentially a dilute solution of acetic acid in water. Vinegar is usually produced in a concentrated form and then diluted with water to give a final concentration of 4%–7% acetic acid; that is, a 4% m/v solution contains 4.00 g of acetic acid per 100 mL of solution. If a drop of bromothymol blue indicator is added to 50.0 mL of concentrated vinegar stock and 31.0 mL of 2.51 M NaOH are needed to turn the solution from yellow to green, what is the percentage of acetic acid in the vinegar stock? (Assume that the density of the vinegar solution is 1.00 g/mL.)

Answer: 9.35%