9.4 Solution Concentrations and Reactions Suppose we prepare a solution from 8.00 g of KI (solute) and 42.00 g of water solvent). Learning Goal Calculate the concentration of a solute in a solution; use concentration units to calculate the amount of solute or solution. Given the volume and concentration of a solution, calculate the amount of another reactant or product in a reaction.

Solution Concentrations The amount of a solute may be expressed in units of grams, milliliters, or moles. The amount of a solution may be expressed in units of grams, milliliters, or liters.

Mass Percent (m/m) Concentration Mass percent (m/m) is the concentration by mass of solute in mass of solution. the grams of solute in 100 grams of solution. × ×

Mass of Solute − Mass Solution When 42.00 g of water is added to 8.00 g of KCl, the mass percent concentration is 16.0% (m/m). × ×

Guide to Calculating Solution Concentration

Calculating Mass Percent What is the mass percent of NaOH in a solution prepared by dissolving 30.0 g of NaOH in 120.0 g of H2O? STEP 1 State the given and needed quantities. ANALYZE Given Needed THE 30.0 g NaOH solute mass percent PROBLEM 30.0 g NaOH + 120.0 g H2O = (m/m) 150.0 g of NaOH solution

Calculating Mass Percent What is the mass percent of NaOH in a solution prepared by dissolving 30.0 g of NaOH in 120.0 g of H2O? STEP 2 Write the concentration expression. STEP 3 Substitute solute and solution quantities into the expression and calculate. × ×

Study Check A solution is prepared by mixing 15.0 g of Na2CO3 and 235 g of H2O. Calculate the mass percent (m/m) of the solution. A. 15.0% (m/m) Na2CO3 solution B. 6.38% (m/m) Na2CO3 solution C. 6.00% (m/m) Na2CO3 solution

Solution A solution is prepared by mixing 15.0 g of Na2CO3 and 235 g of H2O. Calculate the mass percent (m/m) of the solution. STEP 1 State the given and needed quantities. STEP 2 Write the concentration expression. ANALYZE Given Need THE 15.0 g Na2CO3 solute mass percent (m/m) PROBLEM 15.0 g Na2CO3 + 235 g H2O = 250. g solution ×

Solution A solution is prepared by mixing 15.0 g of Na2CO3 and 235 g of H2O. Calculate the mass percent (m/m) of the solution. STEP 3 Substitute solute and solution quantities into the expression and calculate. The answer is C, 6.00% Na2CO3 solution. ×

Volume Percent (v/v) Concentration The volume percent (v/v) is the percent volume (mL) of solute (liquid) to volume (mL) of solution. volume of solute (mL) in 100 mL of solution (conversion factor for volume percent) ×

Mass/Volume Percent The mass/volume percent (m/v) is the percent mass (g) of solute to volume (mL) of solution. mass of solute (g) in 100 mL of solution. (conversion factor for mass/volume percent) × Core Chemistry Skill Calculating Concentration

Study Check Write two conversion factors for each solution. A. 8.50% (m/m) NaOH B. 5.75% (v/v) ethanol C. 4.8% (m/v) HCl

Solution A. 8.50% (m/m) NaOH B. 5.75% (v/v) ethanol C. 4.8% (m/v) HCl

Molarity Molarity (moles of solute/liter of solution) is defined as the moles of solute per volume (L) of solution. A 1.0 M solution of NaCl is defined as

Molarity Calculations What is the molarity of 0.500 L of NaOH solution if it contains 6.00 g of NaOH? STEP 1 State the given and needed quantities. 1 mole NaOH = 40.00 g NaOH ANALYZE Given Need THE 6.00 g NaOH solute molarity (mole/L) PROBLEM 0.500 L of NaOH solution ×

Molarity Calculations What is the molarity of 0.500 L of NaOH solution if it contains 6.00 g of NaOH? STEP 2 Write the concentration expression. STEP 3 Substitute solute and solution quantities into the expression and calculate.

Study Check What is the molarity of 0.225 L of a KNO3 solution containing 34.8 g of KNO3? A. 0.344 M B. 1.53 M C. 15.5 M

Solution What is the molarity of 0.225 L of a KNO3 solution containing 34.8 g of KNO3? STEP 1 State the given and needed quantities. 1 mole of KNO3 = 101.11 g KNO3 ANALYZE Given Need THE 34.8 g KNO3 solute molarity (mole/L) PROBLEM 0.225 L of KNO3 solution ×

Solution What is the molarity of 0.225 L of a KNO3 solution containing 34.8 g of KNO3? STEP 2 Write the concentration expression. STEP 3 Substitute solute and solution quantities into the expression and calculate. The answer is B, 1.53 M KNO3.

Molarity as a Conversion Factor The units of molarity are used as conversion factors in calculations with solutions. Molarity Equality 3.5 M HCl 1 L solution = 3.5 moles of HCl Written as Conversion Factors

Conversion Factors, Concentrations Core Chemistry Skill Using Concentration as a Conversion Factor

Guide to Using Concentration to Calculate Mass or Volume General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C. Timberlake © 2016 Pearson Education, Inc.

Study Check How many grams of NaOH are needed to prepare 75.0 g of 14.0% (m/m) NaOH solution? A. 10.5 g of NaOH B. 75.0 g of NaOH C. 536 g of NaOH

Solution How many grams of NaOH are needed to prepare 75.0 g of 14.0% (m/m) NaOH solution? STEP 1 State the given and needed quantities. STEP 2 Write a plan to calculate mass or volume. grams of grams of solution solute ANALYZE Given Need THE 75.0 grams of 14.0% grams NaOH PROBLEM (m/m) NaOH solution Percent (m/m)

Solution How many grams of NaOH are needed to prepare 75.0 g of 14.0% (m/m) NaOH solution? STEP 3 Write equalities and conversion factors. 14.0 g NaOH = 100 g of solution STEP 4 Set up the problem to calculate mass or volume. The answer is A, 10.5 g NaOH. ×

Study Check How many grams of AlCl3 are needed to prepare 125 mL of a 0.150 M solution? A. 20.0 g of AlCl3 B. 16.7 g of AlCl3 C. 2.50 g of AlCl3

Solution How many grams of AlCl3 are needed to prepare 125 mL of a 0.150 M solution? STEP 1 State the given and needed quantities. STEP 2 Write a plan to calculate mass or volume. liters of grams of solution solute ANALYZE Given Need THE 125 mL (0.125 L) solution grams AlCl3 PROBLEM 0.150 M solution Molarity

Solution How many grams of AlCl3 are needed to prepare 125 mL of a 0.150 M solution? STEP 3 Write equalities and conversion factors. 0.150 moles AlCl3 = 1 L of solution 1 mole AlCl3 = 133.33 g AlCl3

Solution How many grams of AlCl3 are needed to prepare 125 mL of a 0.150 M solution? STEP 4 Set up the problem to calculate mass or volume. Answer is C, 2.50 g AlCl3. × ×

Chemical Reactions in Solution When chemical reactions involve aqueous solutions, we use the balanced chemical equation, the molarity, and the volume to determine the moles or grams of the reactants or products. Core Chemistry Skill Calculating the Quantity of a Reactant or Product for a Chemical Reaction in Solution

Guide to Calculations Involving Solutions in Chemical Reactions General, Organic, and Biological Chemistry: Structures of Life, 5/e Karen C. Timberlake © 2016 Pearson Education, Inc.

Study Check Zinc reacts with HCl to produce hydrogen gas, H2 , and ZnCl2. Zn(s) + 2HCl(aq) H2(g) + ZnCl2(aq) How many liters of a 1.50 M HCl solution completely react with 5.32 g of zinc?

Solution Zinc reacts with HCl to produce hydrogen gas, H2 , and ZnCl2. Zn(s) + 2HCl(aq) H2(g) + ZnCl2(aq) How many liters of a 1.50 M HCl solution completely react with 5.32 g of zinc? STEP 1 State the given and needed quantities. ANALYZE Given Need THE 5.32 g Zn PROBLEM 1.50 M HCl solution liters of HCl Equation solution Zn(s) + 2HCl(aq) H2(g) + ZnCl2(aq)

Solution Zinc reacts with HCl to produce hydrogen gas, H2 , and ZnCl2. Zn(s) + 2HCl(aq) H2(g) + ZnCl2(aq) How many liters of a 1.50 M HCl solution completely react with 5.32 g of zinc? STEP 2 Write a plan to calculate the needed quantity. moles moles volume of of of zinc HCl HCl Mole–mole factor Molarity

Solution Zinc reacts with HCl to produce hydrogen gas, H2, and ZnCl2. Zn(s) + 2HCl(aq) H2(g) + ZnCl2(aq) How many liters of a 1.50 M HCl solution completely react with 5.32 g of zinc? STEP 3 Write equalities and conversion factors including mole–mole and concentration factors.

Solution Zinc reacts with HCl to produce hydrogen gas, H2, and ZnCl2. Zn(s) + 2HCl(aq) H2(g) + ZnCl2(aq) How many liters of a 1.50 M HCl solution completely react with 5.32 g of zinc? STEP 4 Set up the problem to calculate needed quantity. × × ×

Concept Map for Chemical Equations