Core Concept

Solution stoichiometry is the study of quantitative relationships in chemical reactions that occur in solution. It combines the principles of stoichiometry with solution concentrations to calculate the amounts of reactants and products in reactions involving dissolved substances.

Key Concepts

1. Molarity (M):

   • Molarity is the concentration of a solution, defined as the moles of solute per liter of solution:

     $M = \frac{\text{moles of solute}}{\text{liters of solution}}$

   • Units are in moles per liter (mol/L \text{mol/L}mol/L), also written as MMM.

2. Mole Ratios:

   • As in standard stoichiometry, the coefficients in a balanced equation provide mole ratios, which relate the moles of reactants to the moles of products.

   • These mole ratios are essential for converting between moles of different substances in a reaction.

3. Volume Relationships:

   • Solution stoichiometry problems often involve calculating volumes of solutions required to react with a given amount of a substance, or the volume of solution needed to achieve a desired concentration.

4. Dilution:

   • In many stoichiometry problems, solutions are diluted to a lower concentration.

   • The dilution formula is: M1V1=M2V2M_1 V_1 = M_2 V_2M1​V1​=M2​V2​

   • Where M1M_1M1​ and V1V_1V1​ are the initial molarity and volume, and M2M_2M2​ and V2V_2V2​ are the final molarity and volume.

Steps for Solution Stoichiometry Problems

1. Write and Balance the Chemical Equation:

   • Ensure that the reaction is balanced so that you can use accurate mole ratios.

2. Convert Volumes and Concentrations to Moles:

   • Use molarity to convert between moles and volume of a solution: moles=M×V\text{moles} = M \times Vmoles=M×V

   • Ensure volume is in liters; if given in milliliters, convert to liters by dividing by 1000.

3. Use Mole Ratios to Find Unknown Quantities:

   • Use the balanced equation’s mole ratios to relate the moles of the known substance to the moles of the unknown substance.

4. Convert Moles Back to Desired Units:

   • If needed, convert moles back to volume using the molarity, or to grams using molar mass.

Example Problem: Solution Stoichiometry in a Neutralization Reaction

Problem: How many milliliters of 0.500 M0.500 \, M0.500M HCl\text{HCl}HCl are needed to react completely with 25.0 mL25.0 \, \text{mL}25.0mL of 0.250 M0.250 \, M0.250M NaOH\text{NaOH}NaOH?

HCl+NaOH→NaCl+H2O\text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O}HCl+NaOH→NaCl+H2​O

Solution:

1. Write and Balance the Equation:

   • The equation is already balanced as written.

2. Calculate Moles of NaOH\text{NaOH}NaOH:

   • Molarity of NaOH=0.250 M\text{NaOH} = 0.250 \, MNaOH=0.250M, volume = 25.0 mL (0.0250 L0.0250 \, L0.0250L). moles of NaOH=0.250 M×0.0250 L=0.00625 mol\text{moles of NaOH} = 0.250 \, M \times 0.0250 \, L = 0.00625 \, \text{mol}moles of NaOH=0.250M×0.0250L=0.00625mol

3. Use Mole Ratio to Find Moles of HCl\text{HCl}HCl:

   • The balanced equation shows a 1:11:11:1 mole ratio between HCl\text{HCl}HCl and NaOH\text{NaOH}NaOH. moles of HCl=0.00625 mol\text{moles of HCl} = 0.00625 \, \text{mol}moles of HCl=0.00625mol

4. Convert Moles of HCl\text{HCl}HCl to Volume:

   • Molarity of HCl=0.500 M\text{HCl} = 0.500 \, MHCl=0.500M. Volume of HCl=moles of HClmolarity of HCl=0.00625 mol0.500 M=0.0125 L=12.5 mL\text{Volume of HCl} = \frac{\text{moles of HCl}}{\text{molarity of HCl}} = \frac{0.00625 \, \text{mol}}{0.500 \, M} = 0.0125 \, \text{L} = 12.5 \, \text{mL}Volume of HCl=molarity of HClmoles of HCl​=0.500M0.00625mol​=0.0125L=12.5mL

Answer: 12.5 mL of 0.500 M0.500 \, M0.500M HCl\text{HCl}HCl is needed.

Example Problem: Precipitation Reaction in Solution Stoichiometry

Problem: What mass of AgCl\text{AgCl}AgCl precipitate will form when 50.0 mL50.0 \, \text{mL}50.0mL of 0.100 M0.100 \, M0.100M AgNO3\text{AgNO}_3AgNO3​ is mixed with excess NaCl\text{NaCl}NaCl solution?

AgNO3+NaCl→AgCl+NaNO3\text{AgNO}_3 + \text{NaCl} \rightarrow \text{AgCl} + \text{NaNO}_3AgNO3​+NaCl→AgCl+NaNO3​

Solution:

1. Write and Balance the Equation:

   • The equation is balanced as written.

2. Calculate Moles of AgNO3\text{AgNO}_3AgNO3​:

   • Molarity of AgNO3=0.100 M\text{AgNO}_3 = 0.100 \, MAgNO3​=0.100M, volume = 50.0 mL (0.0500 L0.0500 \, L0.0500L). moles of AgNO3=0.100 M×0.0500 L=0.00500 mol\text{moles of AgNO}_3 = 0.100 \, M \times 0.0500 \, L = 0.00500 \, \text{mol}moles of AgNO3​=0.100M×0.0500L=0.00500mol

3. Use Mole Ratio to Find Moles of AgCl\text{AgCl}AgCl:

   • The balanced equation shows a 1:11:11:1 mole ratio between AgNO3\text{AgNO}_3AgNO3​ and AgCl\text{AgCl}AgCl. moles of AgCl=0.00500 mol\text{moles of AgCl} = 0.00500 \, \text{mol}moles of AgCl=0.00500mol

4. Convert Moles of AgCl\text{AgCl}AgCl to Grams:

   • Molar mass of AgCl=143.32 g/mol\text{AgCl} = 143.32 \, \text{g/mol}AgCl=143.32g/mol. mass of AgCl=0.00500 mol×143.32 g/mol=0.7166 g\text{mass of AgCl} = 0.00500 \, \text{mol} \times 143.32 \, \text{g/mol} = 0.7166 \, \text{g}mass of AgCl=0.00500mol×143.32g/mol=0.7166g

Answer: 0.7166 g of AgCl\text{AgCl}AgCl will form.

Key Tips for Solution Stoichiometry

• Always Balance the Equation First: Correct mole ratios are essential for solution stoichiometry.

• Convert Volumes to Liters: Ensure all volumes are in liters when using molarity.

• Use Molarity for Mole Conversions: Molarity is the key to finding moles from solution volumes.

• Be Mindful of Units: Watch for conversions between liters, milliliters, grams, and moles.

• Practice with Different Reaction Types: Solution stoichiometry applies to neutralization, precipitation, and redox reactions.