Problem 53 - Chemistry: Principles and Reactions

Question

Calcium ions in blood trigger clotting. To prevent that in donated blood, sodium oxalate, \(\mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\), is added to remove calcium ions according to the following equation. $$ \mathrm{C}_{2} \mathrm{O}_{4}^{2-}(a q)+\mathrm{Ca}^{2+}(a q) \longrightarrow \mathrm{CaC}_{2} \mathrm{O}_{4}(s) $$ Blood contains about \(0.10 \mathrm{mg} \mathrm{Ca}^{2+} / \mathrm{mL}\). If a 250.0-mL sample of donated blood is treated with an equal volume of \(0.160 \mathrm{M} \mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\), estimate \(\left[\mathrm{Ca}^{2+}\right]\) after precipitation. \(\left(K_{s p} \mathrm{CaC}_{2} \mathrm{O}_{4}=4 \times 10^{-9}\right)\)

Step-by-Step Solution

Answer

Answer: Approximately 2.0 x 10⁻⁵ M.

Step 1: Convert calcium concentration to moles

First, we need to convert the given concentration of \(\mathrm{Ca}^{2+}\) from mg/mL to moles per liter (M). The molar mass of \(\mathrm{Ca}^{2+}\) is 40.08 g/mol. $$ \frac{0.10 \, \text{mg}}{1 \, \text{mL}} \times \frac{1 \, \text{g}}{1000 \, \text{mg}} \times \frac{\text{1 mol} \, \mathrm{Ca}^{2+}}{40.08 \, \text{g}} \times \frac{1000 \, \text{mL}}{1 \, \text{L}} \approx 2.50 \times 10^{-3} \text{M} $$

Step 2: Find moles of \(\mathrm{Ca}^{2+}\) and \(\mathrm{C}_{2}\mathrm{O}_{4}^{2-}\)

Now, we will calculate moles of \(\mathrm{Ca}^{2+}\) in the 250 mL blood sample and moles of \(\mathrm{C}_{2}\mathrm{O}_{4}^{2-}\) in the 250 mL \(\mathrm{Na}_{2}\mathrm{C}_{2}\mathrm{O}_{4}\) solution. $$ \text{moles of } \mathrm{Ca}^{2+} = (2.50 \times 10^{-3}\text{M})\times(250\text{mL})\times \frac{1\text{L}}{1000\text{mL}} \approx 6.25\times10^{-4}\text{ mol} $$ $$ \text{moles of }\mathrm{C}_{2}\mathrm{O}_{4}^{2-} = (0.160\text{M})\times(250\text{mL})\times \frac{1\text{L}}{1000\text{mL}} =0.040\text{mol} $$

Step 3: Find the limiting reactant and moles of remaining \(\mathrm{Ca}^{2+}\)

As there is far more \(\mathrm{C}_{2}\mathrm{O}_{4}^{2-}\) than \(\mathrm{Ca}^{2+}\), \(\mathrm{Ca}^{2+}\) will be the limiting reactant. This means all \(\mathrm{Ca}^{2+}\) ions will react with \(\mathrm{C}_{2}\mathrm{O}_{4}^{2-}\) ions and some \(\mathrm{C}_{2}\mathrm{O}_{4}^{2-}\) ions will remain in the solution. $$ \text{moles of remaining } \mathrm{C}_{2}\mathrm{O}_{4}^{2-} = 0.040\text{mol} - 6.25\times10^{-4}\text{mol} \approx 0.0394\text{mol} $$

Step 4: Calculate solubility of \(\mathrm{CaC}_{2}\mathrm{O}_{4}\)

Now, we will use the \(K_{sp}\) value of \(\mathrm{CaC}_{2}\mathrm{O}_{4}\) to determine its solubility. $$ K_{sp} = 4 \times 10^{-9} = [\mathrm{Ca}^{2+}][\mathrm{C}_{2}\mathrm{O}_{4}^{2-}] $$ Since initial \(\mathrm{Ca}^{2+}\) ions precipitated completely, we can assume the solubility of \(\mathrm{CaC}_{2}\mathrm{O}_{4}\) will be equal to the concentration of the dissolved \(\mathrm{Ca}^{2+}\) ions. Thus, new \([\mathrm{Ca}^{2+}] = [\mathrm{C}_{2} \mathrm{O}_{4}^{2-}] = S\).

Step 5: Find the final concentration of \(\mathrm{Ca}^{2+}\) ions

Using the equation from Step 4, we can find the solubility, S, which is also the final concentration of \(\mathrm{Ca}^{2+}\) ions. $$ [\mathrm{Ca}^{2+}] = S = \sqrt{K_{sp}} = \sqrt{4 \times 10^{-9}} \approx 2.0 \times 10^{-5}\text{M} $$ Therefore, the final concentration of \(\mathrm{Ca}^{2+}\) ions will be approximately \(2.0 \times 10^{-5}\text{M}\) after precipitation.

Key Concepts

Solubility Product Constant

The solubility product constant, denoted as \( K_{sp} \), is a fundamental concept in chemistry that describes the extent to which a solid can dissolve in solution. It specifically pertains to the solubility of sparingly soluble salts.

In a chemical equation where a salt dissolves into its constituent ions, the solubility product is the product of the concentrations of these ions, each raised to the power of their stoichiometric coefficients. For the precipitate \( \text{CaC}_2\text{O}_4(s) \), we have the equilibrium:

• \( \text{CaC}_2\text{O}_4(s) \leftrightharpoons \text{Ca}^{2+}(aq) + \text{C}_2\text{O}_4^{2-}(aq) \)
• \( K_{sp} = [\text{Ca}^{2+}][\text{C}_2\text{O}_4^{2-}] \)

The value of the \( K_{sp} \) is critical to predict whether precipitation will occur if the concentrations of the ions in solution exceed this threshold.

This is used to find the solubility in mol/L by rearranging it. Understanding \( K_{sp} \) makes it easier to understand the solubility behavior of compounds in different solutions.

Molar Concentration Conversion

In many chemical reactions, concentrations are provided in different units like mg/mL, which is not convenient for calculations involving moles, the SI unit for amount of substance. Converting these units to molar concentration (M) is vital and involves several steps.

First, you need to convert the mass of the substance into moles using its molar mass. For calcium ions, \( \text{Ca}^{2+} \), the molar mass is 40.08 g/mol.

• Convert mg to g: \( 0.10 \, \text{mg/mL} = 0.0001 \, \text{g/mL} \)
• Convert grams to moles: \( \frac{0.0001 \, \text{g}}{40.08 \, \text{g/mol}} \approx 2.50 \times 10^{-3} \text{M} \)

After conversion, you can represent it in M (moles per liter), making it easier to use in stoichiometric calculations.

Understanding this process allows for accurate representation and manipulation of substances in various chemical reactions.

Limiting Reactant

The concept of the limiting reactant is pivotal in stoichiometry as it dictates the maximum amount of product that can be formed in a reaction. To identify the limiting reactant, compare the mole amounts of each reactant given in the problem.

In the reaction:

• \(\text{C}_2\text{O}_4^{2-}(aq) + \text{Ca}^{2+}(aq) \rightarrow \text{CaC}_2\text{O}_4(s)\)

We need to determine which reactant will be used up first. From the step-by-step solution, we know:

• Moles of \(\text{Ca}^{2+} = 6.25\times10^{-4} \text{ mol}\)
• Moles of \( \text{C}_2\text{O}_4^{2-} = 0.040 \text{ mol} \)

Given the larger amount of \(\text{C}_2\text{O}_4^{2-}\), \(\text{Ca}^{2+}\) is the limiting reactant.

This means all \(\text{Ca}^{2+}\) will react, leaving some \(\text{C}_2\text{O}_4^{2-}\) unreacted.

Recognizing the limiting reactant gives insight into which component in excess remains in solution, affecting the final concentrations post-reaction.

Precipitation Reaction

A precipitation reaction occurs when two solutions with soluble salts mix, and an insoluble solid, called a precipitate, forms. This process is governed by the solubility rules and involves exchange of ions.

In the given scenario, \(\text{C}_2\text{O}_4^{2-}\) ions react with \(\text{Ca}^{2+}\) ions to form \( \text{CaC}_2\text{O}_4(s) \). The solid form is a precipitate and falls out of solution as per the equation:

• \(\text{C}_2\text{O}_4^{2-}(aq) + \text{Ca}^{2+}(aq) \rightarrow \text{CaC}_2\text{O}_4(s) \)

This is a classic example of double displacement where ions swap partners to form an insoluble product.

In the presence of enough oxalate ions, most calcium is removed from the solution as a solid, significantly lowering its concentration.
This reaction is crucial in blood storage to prevent premature clotting by depleting soluble calcium ions, ensuring storable, safe blood. Understanding such reactions is key to predicting the outcomes of mixing different ionic solutions.