Problem 71 - General Chemistry: Principles and Modern Applications

Question

When concentrated \(\mathrm{CaCl}_{2}(\mathrm{aq})\) is added to \(\mathrm{Na}_{2} \mathrm{HPO}_{4}(\mathrm{aq}),\) a white precipitate forms that is \(38.7 \%\) Ca by mass. Write a net ionic equation representing the probable reaction that occurs.

Step-by-Step Solution

Answer

The net ionic equation for the reaction is \(3Ca^{2+} + 2PO_4^{3-} \rightarrow Ca_3(PO_4)_2\)

Step 1: Identify the Chemical Reaction

First, determine the chemical reaction that would occur. The white precipitate formed implies that \( Ca^{2+} \) ions from \( CaCl_2 \) react with \( \text{PO}_4^{3-} \) ions from \( Na_2\text{HPO}_4 \) to form \( Ca_3(\text{PO}_4)_2 \), a white precipitate.

Step 2: Molar Mass Calculation

Next, we need to calculate the molar masses of Calcium and \( Ca_3(\text{PO}_4)_2 \). The molar mass of Calcium is about \( 40.08 \) g/mol and the molar mass of \( Ca_3(\text{PO}_4)_2 \) is calculated as \( 3 \times 40.08 \) g/mol for Calcium, \( 8 \times 16.00 = 128.00 \) g/mol for Oxygen, and \( 2 \times (1.007 \times 3 + 15.999) = 61.98 \) g/mol for Phosphate. Adding them up gives \( 310.18 \) g/mol.

Step 3: Verify the Mass Percent

Now, confirm that \( Ca_3(\text{PO}_4)_2 \) is \( 38.7 \% \) Ca by mass. This is done by taking the total mass of Calcium, dividing it by the total mass of the compound, and multiplying by 100. \(( 3 \times 40.08 / 310.18) \times 100 = 38.8 \% \), which is very close to our provided value.

Step 4: Write the balanced net ionic equation

Finally, write the balanced net ionic equation. The full equation is \( 3Ca^{2+} + 2PO_4^{3-} \rightarrow Ca_3(PO_4)_2 \). However, since \( Na^{+} \) and \( Cl^{-} \) ions are spectator ions (they do not participate directly in the reaction), the net ionic equation is the same as the full reaction.

Key Concepts

Precipitate Formation

In a chemical reaction, a precipitate is an insoluble solid that forms when two solutions containing soluble compounds are mixed. This solid is a result of ions joining together to form a compound that will not dissolve in the solvent—often water. When concentrated \( \text{CaCl}_2(\text{aq}) \) is mixed with \( \text{Na}_2\text{HPO}_4(\text{aq}) \), a white solid emerges. This indicates precipitate formation. This solid is calcium phosphate, \( \text{Ca}_3(\text{PO}_4)_2 \).

• The calcium ions, \( \text{Ca}^{2+} \), from \( \text{CaCl}_2 \) and the phosphate ions, \( \text{PO}_4^{3-} \), from \( \text{Na}_2\text{HPO}_4 \), come together in the reaction.
• The formation of the solid, non-soluble \( \text{Ca}_3(\text{PO}_4)_2 \), is the precipitate in question.

The formation of a precipitate is a key indication of a chemical reaction where new compounds form, highlighting the changes in solubility properties.

Chemical Reaction

A chemical reaction involves the transformation of substances, where reactants are turned into products through breaking and forming of bonds. In aqueous solutions, reactions often involve ions. For our exercise:
- **Reactants:** \( \text{CaCl}_2 \) and \( \text{Na}_2\text{HPO}_4 \).- **Products:** \( \text{Ca}_3(\text{PO}_4)_2 \) as a precipitate.- **Spectator Ions:** \( \text{Na}^{+} \) and \( \text{Cl}^{-} \) ions remain in the solution unchanged and do not form part of the net ionic equation.
The net ionic equation is central to understanding this process, as it strips away the non-reactive parts and highlights only the species that are involved in the actual reaction. This equation shows the formation of the precipitate:

• \( 3\text{Ca}^{2+} + 2\text{PO}_4^{3-} \rightarrow \text{Ca}_3(\text{PO}_4)_2 \)

Chemical reactions are fundamental in creating new materials and transforming compounds within different environments.

Mass Percent Calculation

The mass percent is calculated to understand the composition of a compound in terms of the masses of its elements. This is particularly helpful to verify given values or ascertain the accuracy of measured amounts in a chemical reaction—an accurate gauge of how much of each element makes up the compound. For calcium phosphate \( \text{Ca}_3(\text{PO}_4)_2 \):

• Total molar mass calculation: \( 3 \times 40.08 \ g/mol \) for calcium, alongside contributions from phosphate and oxygen, sums to 310.18 g/mol.
• Mass percent of calcium is derived by the formula: \[ \left(\frac{3 \times 40.08}{310.18}\right) \times 100 \approx 38.8\% \]

This calculated value closely approximates the provided mass percent, confirming its accuracy. Engaging in mass percentage calculations allows students to directly connect molecular compositions to measurable data, enhancing understanding of chemical formulation in real-world contexts.