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An experimenter makes up a solution of \(0.375 \mathrm{~mol}\) \(\mathrm{Na}_{2} \mathrm{CO}_{3}, 0.125 \mathrm{~mol} \mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\), and \(0.200 \mathrm{~mol} \mathrm{AgNO}_{3}\) in \(2.000 \mathrm{~L}\) of water solution. Note any precipitations that occur, writing a balanced equation for each. Then, calculate the molarities of each ion in the solution.

Short Answer

Expert verified
\(\text{CaCO}_3(s)\) and \(\text{Ag}_2\text{CO}_3(s)\) precipitate, resulting in \([\text{CO}_3^{2-}] = 0.075\,\text{M}\), \([\text{Ca}^{2+}] = 0\), \([\text{Ag}^+] = 0\), \([\text{Na}^+] = 0.3750\,\text{M}\), \([\text{NO}_3^-] = 0.1625\,\text{M}\).

Step by step solution

01

Identify Possible Precipitations

Start by determining if any of the ions in the solution will form a precipitate when they combine. The ions present are \( ext{Na}^+, \ ext{CO}_3^{2-}, \ ext{Ca}^{2+}, \ ext{NO}_3^-,\ \text{Ag}^+\), and \(\text{NO}_3^-\) again from silver nitrate. Use solubility rules: \(\text{Na}_2\text{CO}_3\) is soluble, but \(\text{CaCO}_3\) and \(\text{Ag}_2\text{CO}_3\) are not soluble. \(\text{Ca(NO}_3)_2\) and \(\text{AgNO}_3\) are soluble. Hence, \(\text{CaCO}_3\) and \(\text{Ag}_2\text{CO}_3\) could precipitate.
02

Write Balanced Equations for Precipitations

Write down the balanced equations for the substances that precipitate: 1. \(\text{Ca}^{2+} + \text{CO}_3^{2-} \rightarrow \text{CaCO}_3(s) \)2. \(2\text{Ag}^+ + \text{CO}_3^{2-} \rightarrow \text{Ag}_2\text{CO}_3(s) \)
03

Calculate Initial Concentrations

To calculate the initial molar concentrations, divide the moles of each substance by the total volume of the solution (2 L):- \([\text{Na}_2\text{CO}_3] = \frac{0.375 \text{ mol}}{2.000 \text{ L}} = 0.1875\,\text{M}\)- \([\text{Ca(NO}_3)_2] = \frac{0.125 \text{ mol}}{2.000 \text{ L}} = 0.0625\,\text{M}\)- \([\text{AgNO}_3] = \frac{0.200 \text{ mol}}{2.000 \text{ L}} = 0.1000\,\text{M}\)
04

Calculate Molarities After Precipitation

1. \(\text{CaCO}_3\) precipitates until Ca or CO2 is depleted from reaction: - For \(\text{CO}_3^{2-}\) left: \([\text{CO}_3^{2-}] = 0.1875 \text{ M} - 0.0625 \text{ M} = 0.1250\,\text{M}\) - [Ca\(^{2+}\)] = 0 (Total consumed by CaCO3 precipitation)2. \(\text{Ag}_2\text{CO}_3\) precipitation: - \([\text{CO}_3^{2-}] = 0.1250 \text{ M} - \frac{0.1000}{2} = 0.075\,\text{M}\) - [Ag\(^+\)] = 0 (Total consumed by Ag2CO3 precipitation)
05

Calculate Molarity of Non-Participating Ions

Ion not involved in precipitation retains its initial concentration:- \([\text{Na}^+] = 2 \times 0.1875 = 0.3750 \,\text{M}\) (Since Na remains unreacted) - \([\text{NO}_3^-] = 0.0625 + 0.1000 = 0.1625 \,\text{M}\) (Sum from both salts containing NO3)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Precipitation Reactions
In chemistry, precipitation reactions occur when two aqueous solutions are mixed together and an insoluble substance, or "precipitate," forms and settles out of the solution. When identifying possible precipitates, you rely on solubility rules to predict which combinations of ions will result in an insoluble compound. In the given exercise, upon mixing solutions of sodium carbonate (Na鈧侰O鈧), calcium nitrate (Ca(NO鈧)鈧), and silver nitrate (AgNO鈧), two potential precipitates are identified:
  • Calcium carbonate (CaCO鈧) forms when calcium ions (Ca^{2+}) react with carbonate ions (CO_3^{2-}).
  • Silver carbonate (Ag鈧侰O鈧) forms when silver ions (Ag^+) react with carbonate ions (CO_3^{2-}).
The balanced chemical reactions for the precipitation of each substance are essential. For calcium carbonate:\[ \text{Ca}^{2+} + \text{CO}_3^{2-} \rightarrow \text{CaCO}_3(s) \]And for silver carbonate:\[ 2\text{Ag}^+ + \text{CO}_3^{2-} \rightarrow \text{Ag}_2\text{CO}_3(s) \]These reactions demonstrate how ions in solution can form an insoluble compound based on their solubility characteristics.
Molarity Calculations
Molarity is a way to express the concentration of a solution, defined as the number of moles of solute per liter of solution. Calculating molarity allows you to determine how concentrated a solution is. To calculate molarity, use the formula:\[ \text{Molarity (M)} = \frac{\text{moles of solute}}{\text{volume of solution in liters}}\]In the provided exercise, the experimenter starts with known amounts of Na鈧侰O鈧, Ca(NO鈧)鈧, and AgNO鈧 in 2 liters of water:
  • The molarity of Na鈧侰O鈧 is found by dividing its moles (0.375) by 2 L, giving 0.1875 M.
  • Ca(NO鈧)鈧 has 0.125 moles, resulting in a molarity of 0.0625 M when divided by 2 L.
  • The molarity of AgNO鈧 is calculated as 0.1000 M from 0.200 moles in the solution.
These calculations provide the initial concentrations, which are needed before considering any changes due to precipitation. After precipitation, certain ions are used up and molarity must be recalculated to reflect the amount remaining in solution.
Solubility Rules
Understanding solubility rules is crucial for predicting whether a compound will dissolve in water. These rules guide us in determining whether certain combinations of cations and anions will result in soluble or insoluble compounds. Some general solubility guidelines include:
  • All sodium ( Na鈦), potassium ( K鈦), and ammonium ( NH鈧勨伜) salts are soluble.
  • All nitrates ( NO鈧冣伝) and most chlorides ( Cl鈦) are soluble, except for those of silver ( Ag鈦), lead ( Pb虏鈦), and mercury ( Hg鈧偮测伜).
  • Most carbonates ( CO鈧兟测伝), phosphates ( PO鈧劼斥伝), and hydroxides ( OH鈦) are insoluble, except when paired with Na鈦 or K鈦.
In this exercise, these rules help determine that CaCO鈧 and Ag鈧侰O鈧 are insoluble and will precipitate when formed. These insights allow experimenters to predict and write balanced reactions for observed chemical behavior, facilitating a deeper understanding of the interaction between solutes in solution.

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