Question

Calcium oxalate, ${\mathrm{CaC}}_2{\mathrm{O}}_4,$ is very insoluble in water. What mass of sodium oxalate, ${\mathrm{Na}}_2{\mathrm{C}}_2{\mathrm{O}}_4,$ is required to precipitate the calcium ion from $37.5\mathrm{mL}$ of $0.104M{\mathrm{CaCl}}_2$ solution?

Short answer

The mass of sodium oxalate, ${\mathrm{Na}}_2{\mathrm{C}}_2{\mathrm{O}}_4$, required to precipitate the calcium ions from $37.5\mathrm{mL}$ of $0.104M{\mathrm{CaCl}}_2$ solution is approximately $0.5226\mathrm{g}$.

Step-by-step solution

Calculate the moles of ${\mathrm{Ca}}^{2+}$ ions in the solution

Moles of ${\mathrm{Ca}}^{2+}$ ions can be calculated using the given volume and molarity: Number of moles = Volume × Molarity n(\mathrm{Ca^{2+}}) = (37.5 \times 10^{-3} \ \mathrm{L}) (0.104 \ \mathrm{mol/L}) n(\mathrm{Ca^{2+}}) = 3.9 \times 10^{-3} \ \mathrm{mol}

Find the moles of sodium oxalate needed for precipitation

Since the moles of sodium oxalate required for precipitation are equal to the moles of ${\mathrm{Ca}}^{2+}$ ions present, we can directly use the value: n(\mathrm{Na}_{2} \mathrm{C}_{2}\mathrm{O}_{4}) = n(\mathrm{Ca^{2+}}) n(\mathrm{Na}_{2} \mathrm{C}_{2}\mathrm{O}_{4}) = 3.9 \times 10^{-3} \ \mathrm{mol}

Convert moles of sodium oxalate to mass

Now we can find the mass of sodium oxalate required to precipitate the calcium ions using the molar mass of sodium oxalate, which is $134.0\mathrm{g}/\mathrm{mol}$: Mass = moles × molar mass m(\mathrm{Na}_{2} \mathrm{C}_{2}\mathrm{O}_{4}) = (3.9 \times 10^{-3} \ \mathrm{mol}) (134.0 \ \mathrm{g/mol}) m(\mathrm{Na}_{2} \mathrm{C}_{2}\mathrm{O}_{4}) = 0.5226 \ \mathrm{g} Thus, $0.5226g$ of sodium oxalate is required to precipitate the calcium ions from the given solution.

Key concepts

Precipitation Reaction

In a precipitation reaction, two soluble salts in solution react to form an insoluble solid, called the precipitate, alongside a free ion in solution. This is a key concept in chemistry, especially during processes that involve removing ions from a solution.
In our problem, calcium chloride ($${\mathrm{CaCl}}_2$$ ) in the solution reacts with sodium oxalate ($${\mathrm{Na}}_2{\mathrm{C}}_2{\mathrm{O}}_4$$ ) to form calcium oxalate, an insoluble compound.

• Calcium ions ($${\mathrm{Ca}}^{2+}$$ ) react with oxalate ions ($${\mathrm{C}}_2{\mathrm{O}}_4^{2-}$$ ) to form calcium oxalate ($${\mathrm{CaC}}_2{\mathrm{O}}_4$$ ).
• This reaction removes the calcium ions from the solution as a solid, thereby forming a precipitate.

Precipitation reactions such as these are common when performing ionic reactions in aqueous solutions. They are important for various applications, including water softening and treatment, analytical chemistry, and even forming certain minerals in nature.

Molarity Calculation

Molarity (M) is the number of moles of solute per liter of solution. It is a convenient way to express the concentration of a solution in stoichiometry problems.
To find the molarity, the formula is:$$\text{Molarity (M)}=\frac{\text{moles of solute}}{\text{liters of solution}}$$
In the given problem, we used this relation to calculate the moles of calcium ions present in the solution.

• First, convert the volume from milliliters (mL) to liters (L) by multiplying by $${10}^{-3}$$.
• Given as 37.5 mL, this becomes 0.0375 L.
• Then, multiply the volume in liters by the molarity to find moles ($$0.104\mathrm{mol}/\mathrm{L}\times 0.0375\mathrm{L}=3.9\times {10}^{-3}\mathrm{mol}$$).

This calculated value is critical when working with stoichiometric conversions, allowing us to determine how much reactants are needed in a chemical reaction.

Mole Concept

The mole concept is fundamental in chemistry, providing a bridge between the atomic and macroscopic worlds.
A mole contains $$6.022\times {10}^{23}$$ entities, known as Avogadro's number. This concept allows chemists to count atoms and molecules by weighing them.
In stoichiometry, the mole plays a key role in balancing chemical reactions, allowing chemists to relate quantities and convert between masses and moles.

• Our problem involved converting the moles of sodium oxalate to mass by using its molar mass.
• The molar mass of sodium oxalate is known to be 134.0 g/mol.
• By multiplying the number of moles by the molar mass ($$3.9\times {10}^{-3}\text{mol}\times 134.0\text{g/mol}=0.5226\text{g}$$), we determine the weight required for the precipitation.

Understanding the mole concept allows accurate measurements and calculations, crucial for scientific experimentation and chemical production.