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Magazine Chapter 16: Problem 59
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
Expert verified
The concentration of remaining \( \text{Ba}^{2+} \) is \( 4.8 \times 10^{-9} M. \)
Step by step solution
01
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 \(\text{Na}_2 \text{CrO}_4\): \(\text{moles of CrO}_{4 }^{2-} = 0.15 M \times 0.050 L = 0.0075 \text{ moles}\).
02
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 \(\text{CrO}_{4}^{2-}\).
03
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.
04
Calculate remaining moles of excess reagent
Subtract the moles of \(\text{Ba}^{2+}\) from the moles of \(\text{CrO}_{4}^{2-}\): \[ \text{Remaining moles of CrO}_{4}^{2-} = 0.0075 - 0.005 = 0.0025 \text{ moles} \].
05
Determine final volume of the solution
The total final volume after mixing is \(\text{50.0 mL} + \text{50.0 mL} = \text{100.0 mL} = 0.100 L\).
06
Apply the solubility product constant (Ksp)
Given \(\text{Ksp of BaCrO}_4 = 1.2 \times 10^{-10}\), the solubility equilibrium can be expressed as: \[ [\text{Ba}^{2+}][\text{CrO}_{4}^{2-}] = \text{Ksp} \].
07
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, \([ \text{Ba}^{2+}] = \frac{\text{Ksp}}{0.025 M} = \frac{1.2 \times 10^{-10}}{0.025} = 4.8 \times 10^{-9} M \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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 \( \text{CrO}_{4}^{2-} \).
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!
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.
- 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.
\[ \text{Ba}^{2+} \times \text{CrO}_{4}^{2-} = \text{Ksp} \]
- For \( \text{BaCrO}_4 \), the given \text{Ksp} is \( 1.2 \times 10^{-10} \).
\[ \text{Ba}^{2+} \times \text{CrO}_{4}^{2-} = \text{Ksp} \]
- 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.
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.
- 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 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.
