4.4 Solution Concentration and Stoichiometry

Solution Key Terms What type of mixture is also considered a solution? Give an example. – A homogeneous mixture. An example would be a saline solution, NaCl(aq). If I dissolve 2.5g NaCl in 25ml H 2 0, which substance is the solvent and which is the solute? – In an aqueous solution, water acts as the solvent (what is doing the dissolving, and sodium chloride acts as the solute (what is being dissolved). When two different liquids with similar structures are present, the majority component is the solvent and the minority component is the solute.

Solution Concentration If I have two sodium hydroxide solutions, we’ll label them A and B, and A was made by dissolving 1.0 g NaOH in 50 mL of H 2 O, and B was made by dissolving 2.5 g NaOH in 50 mL H 2 O, which would be considered dilute and which would be considered concentrated? (A) would be the dilute solution, and (B) would be considered concentrated. This is due to the fact that solution (A) was made using less solute relative to solvent.

Solution Concentration (continued) Molarity is the most common way to express solution concentration. mol solute Molarity(M) = L solution  (not L solvent)

Let’s Try a Sample Problem Calculate the molarity of a solution made by adding 45.4 g of NaNO 3 to a flask and dissolving it with water to create a total volume of 2.50 L. Step 1: convert g solute to mol solute 1.00 mol NaNO g NaNO 3 X = 5.34x10 -1 mol NaNO g NaNO 3 Step 2: calculate concentration 5.34x10 -1 mol NaNO 3 (s) M = = 2.14X10 -1 M NaNO 3 (aq) 2.50 L NaNO3(aq)

Using Molarity as a Conversion Factor A known molarity can also be used as a conversion factor between moles of solute and L of solution. Sample Problem 2: How many grams of sucrose (C 12 H 22 O 11 ) are in 1.55 L of M sucrose solution? Step 1: calculate moles of solute Moles solute = M x L = mol/L X 1.55 L = 1.17 mol C 12 H 22 O 11 Step 2: covert mol solute to g solute g C 12 H 22 O mol C 12 H 22 O 11 X = 401 g C 12 H 22 O mol C 12 H 22 O 11

Solution Dilution The easiest was to dilute a stock solution (concentrated solution), would be to use the dilution equation: M 1 V 1 = M 2 V 2 (M 1 and V 1 are the concentrated molarity and volume) (M 2 and V 2 are the dilute molarity and volume) F.Y.I.: When a solution is diluted, the number of solute particles remain the same since M(solution) V(solution) = mol solute.

Let’s Try a Practice Problem To what volume (in mL) should you dilute mL of a 5.00 M CaCl 2 solution to obtain a M CaCl 2 solution? (5.00 M)(100.0 mL) = (0.750M)(V 2 ) (5.00 M)(100.0 mL) V 2 = = 667 mL M

Solution Stoichiometry A balanced chemical equation can be used to make conversions between solution volumes and the amount of solute in moles using the molarities of the solutions. To do this type of problem, we usually use the following sequence of steps: V(solution A)  mol(solute in A)  mol(solute in B)  V(solution B)

Let’s Try a Practice Problem What volume (in mL) of a M HNO 3 solution will completely react with 35.7 mL of a M Na 2 CO 3 solution according to the following equation? Na 2 CO 3 (aq) + 2HNO 3 (aq)  2NaNO 3 + CO 2 (g) + H 2 O(l) Step 1: calculate moles of Na 2 CO 3 mol Na 2 CO 3 = (0.108 M)( L) = 3.86X10 -3 mol Na 2 CO 3 Step 2: Set up a mol:mol ratio 2 mol HNO X10 -3 mol Na 2 CO 3 X = 7.72X10 -3 mol HNO 3 1 mol Na 2 CO 3 Step 3: calculate volume of HNO 3 solution mol HNO X10 -3 mol HNO 3 L HNO 3 = = = 5.15X10 -2 L HNO 3 solution = 51.5 mL HNO 3 solution M HNO M HNO 3

Chapter 4 pg. 188 #’s 54, 56, 58 & 60 (just the a’s) Read pgs