Slide presentation
Chemical Reactions in Solutions (Molarity)
General Chemistry • Chemical Reactions
Aqueous reactions
Chemical Reactions in Solutions: Molarity
Molarity connects the concentration of a solution to the number of moles available to react.
AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq)Learning target: use \(M = \frac{n}{V}\), volume in liters, and mole ratios from a balanced equation to solve solution stoichiometry problems.
Solution concentration
Molarity tells how many moles of solute are present in \(1\ \mathrm{L}\) of solution.
Moles react
Balanced equations compare substances in moles, not in milliliters.
Predict products
Use the limiting reactant to calculate precipitate or product amount.
Why it matters
Many reactions happen in water
In solution reactions, reactants are dissolved in water. Instead of measuring only solid mass, chemists often measure solution volume and concentration.
If the concentration and volume are known, the number of moles can be calculated. Then stoichiometry works the same way as in any balanced reaction.
Solution stoichiometry is used in
- acid-base titrations,
- precipitation reactions,
- water-quality testing,
- medicine and laboratory preparation,
- reaction yield predictions.
Core concept
Molarity converts volume of solution into moles of solute
Molarity is defined as moles of solute per liter of solution:
Rearranged for solution stoichiometry:
\[ n = M \times V \]The volume must be in liters. For example, \(25.0\ \mathrm{mL} = 0.0250\ \mathrm{L}\).
Vocabulary
Variables and units in solution reactions
| Quantity | Meaning | Formula or source | Unit |
|---|---|---|---|
| Molarity, \(M\) | Concentration of dissolved solute | \(M = \frac{n}{V}\) | mol/L or M |
| Moles, \(n\) | Amount of solute particles | \(n = M \times V\) | mol |
| Volume, \(V\) | Volume of solution used | Convert mL to L | L |
| Mole ratio | Relationship between reactants and products | Coefficients in balanced equation | mol/mol |
| Limiting reactant | Reactant that runs out first | Produces less product | mol or g |
Main relationship
Solution stoichiometry has two bridges
First convert solution information into moles. Then use the mole ratio from the balanced equation.
The dilution equation is used when the amount of solute stays constant while the solution volume changes.
Interactive simulation
Mix two solutions and predict the precipitate
Precipitation reaction
Stoichiometric prediction
NaCl is limiting because it provides fewer moles for the 1:1 reaction.
Static fallback: 25.0 mL of 0.200 M AgNO3 with 30.0 mL of 0.150 M NaCl gives 0.645 g AgCl.
Dynamic relationship
The smaller mole amount limits the precipitate
The reaction ratio is 1:1, so the solution that supplies fewer moles controls the amount of AgCl(s) that can form.
Interpretation: when the balanced ratio is 1:1, the smaller mole amount directly equals the moles of precipitate that can form.
Worked example
Find the mass of precipitate from two solutions
Problem: \(25.0\ \mathrm{mL}\) of \(0.200\ \mathrm{M}\) AgNO3 is mixed with \(30.0\ \mathrm{mL}\) of \(0.150\ \mathrm{M}\) NaCl. What mass of AgCl(s) forms?
- 1. Convert volumes to liters. \(25.0\ \mathrm{mL} = 0.0250\ \mathrm{L}\) \(30.0\ \mathrm{mL} = 0.0300\ \mathrm{L}\)
- 2. Calculate moles of each reactant. \[ n_{\mathrm{AgNO_3}} = 0.200\ \mathrm{mol/L} \times 0.0250\ \mathrm{L} = 0.00500\ \mathrm{mol} \] \[ n_{\mathrm{NaCl}} = 0.150\ \mathrm{mol/L} \times 0.0300\ \mathrm{L} = 0.00450\ \mathrm{mol} \]
- 3. Use the 1:1 mole ratio. NaCl is limiting, so \(0.00450\ \mathrm{mol}\) AgCl forms.
- 4. Convert moles AgCl to grams. \[ 0.00450\ \mathrm{mol\ AgCl} \times 143.32\ \mathrm{g/mol} = 0.645\ \mathrm{g\ AgCl} \]
Common mistake
Do not use milliliters directly in \(M = \frac{n}{V}\)
Incorrect
A student calculates \(0.200\ \mathrm{M} \times 25.0\ \mathrm{mL} = 5.00\ \mathrm{mol}\).
This is wrong because molarity uses liters, not milliliters.
Correct
Convert first:
\[ 25.0\ \mathrm{mL} = 0.0250\ \mathrm{L} \]
\[ 0.200\ \mathrm{mol/L} \times 0.0250\ \mathrm{L} = 0.00500\ \mathrm{mol} \]
Key idea: volume conversion is often the first step in solution stoichiometry.
Practice check
Use molarity and a balanced equation
Hydrochloric acid reacts with sodium hydroxide:
How many moles of NaOH are needed to react completely with \(40.0\ \mathrm{mL}\) of \(0.250\ \mathrm{M}\) HCl?
Show answer and reasoning
Convert volume to liters:
\[ 40.0\ \mathrm{mL} = 0.0400\ \mathrm{L} \]
Calculate moles of HCl:
\[ n_{\mathrm{HCl}} = 0.250\ \mathrm{mol/L} \times 0.0400\ \mathrm{L} = 0.0100\ \mathrm{mol} \]
The balanced equation has a 1:1 ratio between HCl and NaOH, so \(0.0100\ \mathrm{mol}\) NaOH is required.
Apply the topic
A reliable strategy for solution stoichiometry
Balance the equation
The coefficients determine the mole ratio between solutes.
Convert mL to L
Use liters before multiplying by molarity.
Find moles
Use \(n = M \times V\) for each solution.
Use stoichiometry
Apply mole ratios, identify limiting reactant, and calculate product amount.
For dilution, use \(M_1V_1 = M_2V_2\) when the amount of solute stays the same but the solution volume changes.
Summary
What to remember
Molarity means mol/L
\(M = \frac{n}{V}\), so \(n = M \times V\).
Volume must be in liters
Convert mL to L before using molarity.
Coefficients give mole ratios
Use the balanced equation to connect reactants and products.
Limiting reactant controls yield
The reactant with the smaller stoichiometric amount determines product formation.