Unit: Acids, Bases, and Solutions Day 2 - Notes Unit: Acids, Bases, and Solutions Units of Concentration: Molarity

After today you will be able to… Define molarity in terms of its mathematical formula Calculate moles, liters, or molarity of a given solution Explain how to make a solution

Recall, concentration is a measure of the amount of solute dissolved in a given quantity of solvent.

Molarity In chemistry, the most important unit of concentration is molarity. Molarity: (M) is the number of moles of solute dissolved in one liter of solution.

Molarity Molarity= Moles of solute Liters of solution Important: The volume is the total volume of resulting solution, not the solvent alone.

Steps for making a 0.5M solution Add 0.5mol of solute to a 1.0L volumetric flask half-filled with distilled water

Steps for making a 0.5M solution 2. Swirl the flask to dissolve the solute. 3. Fill the flask to the mark etched on the side of the flask (1-L mark).

Performing Calculations With Molarity Example: What is the molarity of a solution that contains 0.25 moles of NaCl in 0.75L of solution? M= mol= L= ? 0.25 mol NaCl 0.75L 0.25 mol NaCl M = 0.75L Molarity = 0.33 mol/L or 0.33M

Performing Calculations With Molarity Example: What volume of a 1.08M KI solution would contain 0.642 moles of KI? M= mol= L= 1.08 M 0.642 mol KI ? 0.642 mol KI 1.08 M = L Volume = 0.594L

Performing Calculations With Molarity Example: How many grams of CaBr2 are dissolved in 0.455L of a 0.39M CaBr2 solution? M= mol= L= 1 Ca=40.08 0.39 M CaBr2 2 Br=79.90(2) mol 199.88g 0.39 M = ? 0.455L 0.455L mol = 0.18 mol CaBr2 199.88gCaBr2 0.18 mol CaBr2 x = 35g CaBr2 1 mol CaBr2

Molality Molality (m) = moles solute/kg of solvent molality  molarity Example problem: What is the molality of a solution with 0.70 mol NaCl and 2.2 kg of water? m = mol solute = 0.70 mol kg solvent 2.2 kg answer = 0.32 m solution

Dilution is the procedure for preparing a less concentrated solution from a more concentrated solution. Dilution Add Solvent Moles of solute before dilution (1) after dilution (2) = M1V1 M2V2 =

How would you prepare 60.0 mL of 0.2 M HNO3 from a stock solution of 4.00 M HNO3? M1V1 = M2V2 M1 = 4.00 M2 = 0.200 V2 = 0.06 L V1 = ? L V1 = M2V2 M1 = 0.200 x 0.06 4.00 = 0.003 L = 3 mL 3 mL of acid + 57 mL of water = 60 mL of solution

Another Dilution Problem If 32 mL stock solution of 6.5 M H2SO4 is diluted to a volume of 500 mL, what would be the resulting concentration? M1*V1 = M2*V2 (6.5M) * (32 mL) = M2 * (500.0 mL) 6.5 M * 32 mL M2 = 500 mL M2 = 0.42 M

moles of solute before dilution = moles of solute after dilution Concentration of Solutions Dilution is the process of preparing a less concentrated solution from a more concentrated one. moles of solute before dilution = moles of solute after dilution

(2.00 M CuCl2)(Lc) = (0.100 M CuCl2)(0.2500 L) Concentration of Solutions In an experiment, a student needs 250.0 mL of a 0.100 M CuCl2 solution. A stock solution of 2.00 M CuCl2 is available. How much of the stock solution is needed? Solution: Use the relationship that moles of solute before dilution = moles of solute after dilution. (2.00 M CuCl2)(Lc) = (0.100 M CuCl2)(0.2500 L) Lc = 0.0125 L or 12.5 mL To make the solution: Pipet 12.5 mL of stock solution into a 250.0 mL volumetric flask. Carefully dilute to the calibration mark. Mc × Lc = Md × Ld

Mc × mLc = Md × mLd Concentration of Solutions Because most volumes measured in the laboratory are in milliliters rather than liters, it is worth pointing out that the equation can be written as Mc × mLc = Md × mLd