Question

Solutions of \(50.0 \mathrm{~mL}\) of \(0.10 M \mathrm{BaCl}_{2}\) and \(50.0 \mathrm{~mL}\) of \(0.15 M \mathrm{Na}_{2} \mathrm{CrO}_{4}\) are mixed, forming a precipitate of \(\mathrm{BaCrO}_{4}\). Calculate the concentration of \(\mathrm{Ba}^{2+}\) that remains in solution.

Short answer

The concentration of remaining \( \text{Ba}^{2+} \) is \( 4.8 \times 10^{-9} M. \)

Step-by-step solution

Determine the moles of each reactant

First, calculate the moles of each reactant using their concentrations and volumes. For \(\text{BaCl}_2\): \(\text{moles of Ba}^{2+} = 0.10 M \times 0.050 L = 0.005 \text{ moles}\). For \(\text{Na}_2 \text{CrO}_4\): \(\text{moles of CrO}_{4 }^{2-} = 0.15 M \times 0.050 L = 0.0075 \text{ moles}\).

Write the balanced reaction equation

The balanced equation for the reaction is \[ \text{Ba}^{2+} + \text{CrO}_{4}^{2-} \rightleftharpoons \text{BaCrO}_{4} \] which shows that one mole of \(\text{Ba}^{2+}\) reacts with one mole of \(\text{CrO}_{4}^{2-}\).

Determine the limiting reagent

Compare the moles of \(\text{Ba}^{2+}\) and \(\text{CrO}_{4}^{2-}\) to find the limiting reagent. Since \(\text{Ba}^{2+}\) has fewer moles (0.005 moles) compared to \(\text{CrO}_{4}^{2-}\) (0.0075 moles), \(\text{Ba}^{2+}\) is the limiting reagent and will completely react.

Calculate remaining moles of excess reagent

Subtract the moles of \(\text{Ba}^{2+}\) from the moles of \(\text{CrO}_{4}^{2-}\): \[ \text{Remaining moles of CrO}_{4}^{2-} = 0.0075 - 0.005 = 0.0025 \text{ moles} \].

Determine final volume of the solution

The total final volume after mixing is \(\text{50.0 mL} + \text{50.0 mL} = \text{100.0 mL} = 0.100 L\).

Apply the solubility product constant (Ksp)

Given \(\text{Ksp of BaCrO}_4 = 1.2 \times 10^{-10}\), the solubility equilibrium can be expressed as: \[ [\text{Ba}^{2+}][\text{CrO}_{4}^{2-}] = \text{Ksp} \].

Calculate the concentration of \(\text{Ba}^{2+}\) in solution

Using the remaining concentration of \(\text{CrO}_{4}^{2-}\): \[ [\text{CrO}_{4}^{2-}] = \frac{\text{Remaining moles}}{\text{Total volume}} = \frac{0.0025 \text{ mol}}{0.100 \text{ L}} = 0.025 M \]. Then, \([ \text{Ba}^{2+}] = \frac{\text{Ksp}}{0.025 M} = \frac{1.2 \times 10^{-10}}{0.025} = 4.8 \times 10^{-9} M \).

Key concepts

stoichiometry

Stoichiometry is a key concept in chemistry that helps us understand the quantitative relationships between reactants and products in a chemical reaction. This involves using the balanced chemical equation to determine the proportional amounts of each substance involved.

• In our problem, we start by figuring out the number of moles of each reactant. This is done through the formula: \( \text{moles} = \text{Molarity} \times \text{Volume} \)


• We found that we have 0.005 moles of \( \text{Ba}^{2+} \) and 0.0075 moles of \( \text{CrO}_{4}^{2-} \).

Understanding stoichiometry allows us to assess how much of each reactant is required and predict the amount of product obtained. It's like the recipe card for chemical reactions!

limiting reagent

The concept of limiting reagent is crucial in determining which reactant will be consumed first in a reaction, thus limiting the amount of products formed. Imagine you're baking cookies but run out of sugar - you can't make more cookies without it!

• In our example, we compare the moles of each reactant. Since there are only 0.005 moles of \( \text{Ba}^{2+} \), and 0.0075 moles of \( \text{CrO}_{4}^{2-} \), \( \text{Ba}^{2+} \) is the limiting reagent.

Without enough \( \text{Ba}^{2+} \), only a limited amount of \( \text{BaCrO}_{4} \) can form. This means some \( \text{CrO}_{4}^{2-} \) will be left over.

solubility product constant (Ksp)

Understanding the Solubility Product Constant (\text{Ksp}) helps us predict whether a precipitate will form in a solution. The \text{Ksp} value is a measure of the solubility of a compound.

• For \( \text{BaCrO}_4 \), the given \text{Ksp} is \( 1.2 \times 10^{-10} \).

To use this value, we set up the equilibrium expression:

\[ \text{Ba}^{2+} \times \text{CrO}_{4}^{2-} = \text{Ksp} \]

• When \( \text{BaCrO}_{4} \) forms, it slightly dissolves back into the solution. The product of the remaining ion concentrations should equal the \text{Ksp}.

molarity calculation

Molarity, the concentration of a solution, is a measure of moles of solute per liter of solution. Calculating molarity helps determine how much of the reactants and products are present at a given moment.

• In our problem, after identifying the limiting reagent and calculating remaining excess reagent, we focus on the concentration of \( \text{Ba}^{2+} \) in the remaining solution.


• Using the remaining moles of \( \text{CrO}_{4}^{2-} \) (0.0025 moles) and total volume (0.100 L), we find \( [ \text{CrO}_{4}^{2-}] = 0.025 M \).

• Then, we use \text{Ksp} to find \( [ \text{Ba}^{2+}] \): \( [ \text{Ba}^{2+}] = \frac{ \text{Ksp}}{ [ \text{CrO}_{4}^{2-}]} \).

This gives \( [ \text{Ba}^{2+}] = 4.8 \times 10^{-9} M \).
This calculation is crucial to understanding the concentrations and how they affect the formation of precipitates, crucial in predicting the behavior of chemical reactions in solution.