Molarity, Solubility, and Precipitation Reactions

Fundamentals of Solutions

Definition of solutions
- A solution is a homogeneous mixture formed when a substance (like table salt) is mixed with a solvent (like water).
- The solute appears to disappear or become a liquid, but it remains present in the mixture, as evidenced by taste or by boiling away the solvent to recover the solid.
- Components of a Solution:
  - Solute: The component that changes its state during the forming of the solution. If both components are in the same state, the solute is the minor component.
  - Solvent: The component that maintains its original physical state. If both components start in the same state, the solvent is the major component.

Describing and Quantifying Solutions

Composition Variability
- Unlike pure substances, which have a constant composition across all samples, solutions are mixtures with varying compositions.
- Example: Saltwater samples collected from different seas or lakes will contain varying amounts of dissolved salt.
- To accurately describe a solution, one must specify the amount of each component present.
Qualitative Descriptions
- Dilute Solution: Contains a relatively small amount of solute compared to the volume of the solvent.
- Concentrated Solution: Contains a relatively large amount of solute compared to the volume of the solvent.
Quantitative Measurement: Molarity ( $M$ )
- Concentration refers to the quantitative relative amount of solute in a solution.
- Molarity Definition: The number of moles of solute per $1\,\text{liter}$ of solution.
- Formula: $M = \frac{\text{moles solute}}{\text{L solution}}$
- This unit is used because it describes the exact number of solute molecules present in each liter of solution.

Molarity Calculations and Examples

Example 1: Finding Molarity from Mass
- Given: $25.5\,g$ of $KBr$ dissolved in $1.75\,L$ of solution.
- Goal: Find the molarity ( $M$ ).
- Relationships & Constants:
  - Molar mass of $KBr = 119.00\,g/mol$ .
  - $M = \frac{\text{moles}}{L}$ .
- Concept Plan: $g\,KBr \rightarrow mol\,KBr \rightarrow M$ .
- Solution:
  - Convert grams to moles: $\frac{25.5\,g}{119.00\,g/mol} = 0.21428\,mol\,KBr$ .
  - Divide by volume: $\frac{0.21428\,mol}{1.75\,L} \approx 0.122\,M$ .
- Check: Most solutions range between $0$ and $18\,M$ , so this result is reasonable.
Using Molarity as a Conversion Factor
- Molarity defines the relationship between moles of solute and liters of solution.
- Example: A $2.0\,M$ sugar solution means $1\,L$ of solution contains $2.0\,moles$ of sugar.
  - $2\,L = 4.0\,moles$ sugar.
  - $0.5\,L = 1.0\,mole$ sugar.
Example 2: Finding Volume from Moles and Molarity
- Given: $0.125\,M\,NaOH$ and $0.255\,mol\,NaOH$ .
- Goal: Find the volume in liters ( $L$ ).
- Concept Plan: $mol\,NaOH \rightarrow L\,\text{solution}$ .
- Solution:
  - Relationship: $0.125\,mol\,NaOH = 1\,L\,\text{solution}$ .
  - $0.255\,mol \times \frac{1\,L}{0.125\,mol} = 2.04\,L$ .
- Check: Since each liter has only $0.125\,mol$ , requiring slightly more than $2\,L$ to hold $0.255\,mol$ is logically consistent.

Dilution of Solutions

Concept
- Concentrated "stock solutions" are often stored and then diluted to lower concentrations for use.
- Dilution is achieved by adding more solvent.
- Crucial Rule: The amount of solute (moles) does not change during dilution; only the volume of the solution increases. $\text{moles solute in solution 1} = \text{moles solute in solution 2}$
Dilution Formula
- The concentrations and volumes of the stock and the new solution are inversely proportional: $M_1 \times V_1 = M_2 \times V_2$
Example 3: Dilution Calculation
- Given: Stock solution $V_1 = 0.200\,L$ , stock concentration $M_1 = 15.0\,M\,NaOH$ . Target concentration $M_2 = 3.00\,M\,NaOH$ .
- Goal: Find final volume $V_2$ .
- Solution:
  - $V_2 = \frac{M_1 \times V_1}{M_2}$
  - $V_2 = \frac{15.0\,M \times 0.200\,L}{3.00\,M} = 1.00\,L$ .
- Check: The solution is diluted by a factor of $5$ (from $15$ to $3$ ), so the volume must increase by a factor of 5 ( $0.200 \times 5 = 1.00$ ).

Solution Stoichiometry

Molarity in Chemical Reactions
- Because molarity relates moles to liters, it serves as a bridge to convert between volumes of solutions and the amounts of reactants or products in a balanced chemical equation.
Example 4: Stoichiometry Calculation
- Reaction: $2\,KCl(aq) + Pb(NO_3)_2(aq) \rightarrow PbCl_2(s) + 2\,KNO_3(aq)$
- Given: $0.150\,L$ of $0.175\,M\,Pb(NO_3)_2$ . The second reactant is $0.150\,M\,KCl$ .
- Goal: Find the volume of $KCl$ required for complete reaction.
- Concept Plan: $L\,Pb(NO_3)_2 \rightarrow mol\,Pb(NO_3)_2 \rightarrow mol\,KCl \rightarrow L\,KCl$ .
- Relationships:
  - $1\,L\,Pb(NO_3)_2 = 0.175\,mol$
  - $1\,L\,KCl = 0.150\,mol$
  - Stoichiometric ratio: $1\,mol\,Pb(NO_3)_2 : 2\,mol\,KCl$
- Check logic: We need twice the moles of $KCl$ as $Pb(NO_3)_2$ . Since the molarity of $Pb(NO_3)_2$ is higher than that of $KCl$ , the volume of $KCl$ should be more than double the volume of the lead solution.

The Dissolution Process

Microscopic Interactions
- There are attractive forces holding solute particles together.
- There are attractive forces between solvent molecules.
- When mixing, attractive forces develop between solute particles and solvent molecules.
- Condition for Dissolving: A solute will dissolve only if the attractions between the solute and the solvent are strong enough to overcome the internal attractions within the solute and within the solvent.

Solubility of Ionic Compounds

Solubility Trends
- Soluble: Compounds that dissolve significantly in a solvent (e.g., $NaCl$ in water).
- Insoluble: Compounds that hardly dissolve at all (e.g., $AgCl$ in water).
- Nuances:
  - The degree of solubility depends on temperature.
  - Even "insoluble" compounds dissolve slightly, but not to a meaningful or measurable degree in standard contexts.
The Empirical Method
- Predicting solubility is difficult. Scientists use the empirical method: performing experiments to see what dissolves and then developing rules based on those results.

Solubility Rules

Compounds that are Generally Soluble
- Ions with NO Exceptions:
  - $Li^+, Na^+, K^+, NH_4^+$
  - $NO_3^-, C_2H_3O_2^-$
- Ions with Exceptions:
  - $Cl^-, Br^-, I^-$ : Generally soluble except when paired with $Ag^+, Hg_2^{2+},$ or $Pb^{2+}$ .
  - $SO_4^{2-}$ : Generally soluble except when paired with $Ag^+, Ca^{2+}, Sr^{2+}, Ba^{2+},$ or $Pb^{2+}.$
Compounds that are Generally Insoluble
- $OH^-$ and $S^{2-}$ :
  - Exceptions: Become soluble when paired with $Li^+, Na^+, K^+,$ or $NH_4^+.$
  - Slightly Soluble Exceptions: When paired with $Ca^{2+}, Sr^{2+},$ or $Ba^{2+}.$
- $CO_3^{2-}$ and $PO_4^{3-}$ :
  - Exceptions: Become soluble when paired with $Li^+, Na^+, K^+,$ or $NH_4^+.$

Precipitation Reactions

Definition
- A precipitation reaction occurs when two aqueous solutions of ionic compounds react to produce an insoluble ionic compound.
- The insoluble product is called the precipitate.
Example of Precipitation
- $2\,KI(aq) + Pb(NO_3)_2(aq) \rightarrow PbI_2(s) + 2\,KNO_3(aq)$
- In this reaction, $PbI_2$ is the insoluble precipitate (s), while $KNO_3$ remains soluble (aq).
No Reaction Scenario
- If combining two aqueous solutions results in all possible products being soluble, no precipitate forms.
- Example: $KI(aq) + NaCl(aq) \rightarrow KCl(aq) + NaI(aq)$ .
- In this case, all ions remain floating in the water; therefore, no chemical reaction has occurred.

Process for Predicting Precipitation Products

Identify Reactant Ions: Determine which ions are present in each aqueous reactant.
Determine Potential Products:
- Exchange the ions (the cation of one reactant pairs with the anion of the other).
- Balance the charges of the new pairs to write correct chemical formulas.
Check Solubility: Use the solubility rules to determine the state of each potential product.
- If a product is insoluble or slightly soluble, it will precipitate (s).
- If a product is soluble, it remains aqueous (aq).
Evaluate for Reaction:
- If neither product precipitates, write "no reaction."
- If at least one product is insoluble, write the full balanced equation.
Finalize the Equation:
- Write (s) for the precipitate and (aq) for soluble products.
- Balance the final equation.

Worked Example: Precipitation Reaction (Example 5)

Task: Write the equation for the reaction between potassium carbonate and nickel(II) chloride.
Step 1: Reactants:
- Potassium carbonate: $K_2CO_3(aq)$
- Nickel(II) chloride: $NiCl_2(aq)$
Step 2: Ion List & Exchange:
- Ions: $(K^+ + CO_3^{2-})$ and $(Ni^{2+} + Cl^-)$ .
- Exchange: $(K^+ + Cl^-)$ and $(Ni^{2+} + CO_3^{2-})$ .
Step 3: Product Formulas:
- $KCl$ and $NiCO_3$ .
Step 4: Solubility Rules:
- $KCl$ : Contains $K^+$ , so it is soluble (aq).
- $NiCO_3$ : Carbonates are generally insoluble, and $Ni^{2+}$ is not an exception. Therefore, it is insoluble (s).
Step 5: Final Equation:
- Unbalanced: $K_2CO_3(aq) + NiCl_2(aq) \rightarrow KCl(aq) + NiCO_3(s)$
- Balanced: $K_2CO_3(aq) + NiCl_2(aq) \rightarrow 2\,KCl(aq) + NiCO_3(s)$