Announcement: Chemistry 6A F2007. Dr. J.A. Mack 11/9/07. Molarity: The ratio of moles of solvent to liters of solute. Moles/Liters and Molarity:

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1 Chemistry 6A F007 Dr. J.A. Mack Announcement: This weeks experiment (Atomic Spectra/Flame Test) is due next week, even though there is no lab scheduled for the next two weeks. Monday s Lab must turn in the lab by Tuesday (11/1) Tuesday s Lab must turn in the lab by Tuesday (11/1) Thursday s Lab must turn in the lab by Thursday (11/15) Late labs will have points deducted. 11/9/07 11/9/07 Dr. Mack. CSUS 1 11/9/07 Dr. Mack. CSUS The relative amounts of solute in the solution is expressed by a concentration. Molarity: The ratio of moles of solvent to liters of solute Moles/Liters and Molarity: Molarity relates mols and volume (L) Molarity knowing corners, you can find the rd molarity (M) = moles of solute L of solution moles Liters If you know moles & L, you know Molarity (M) 11/9/07 Dr. Mack. CSUS 11/9/07 Dr. Mack. CSUS 4

if you know Molarity and volume, you know moles!

L = moles if you know mols and molarity, you know volume!

that is accurately calibrated to a precise know volume at a

2 if you know Molarity and volume, you know moles! Molarity Volume = moles Preparation of a 1 molar solution mols L L = moles if you know mols and molarity, you know volume! 1 moles = Volume M mol L = L mol 11/9/07 Dr. Mack. CSUS 5 11/9/07 Dr. Mack. CSUS 6 A volumetric flask is a piece of laboratory glassware that is accurately calibrated to a precise know volume at a known temperature.. Fill the volumetric flask partially with the solvent to dissolve the solute. One must be sure that the solute (if it is in solid form) is completely dissolved before the addition of more solvent. Solute 11/9/07 Dr. Mack. CSUS 7 11/9/07 Dr. Mack. CSUS 8

3 . Fill the volumetric flask partially with the solvent to dissolve the solute.. Fill the volumetric to the calibration mark using a bottle, then a dropper. The bottom of the curved portion of the meniscus must be even with the calibration mark. 11/9/07 Dr. Mack. CSUS 9 Knowing the volume of the flask and the moles of solute, one can determine the molarity of the solution! 11/9/07 Dr. Mack. CSUS 10 A student adds 5.15 g of sodium sulfide into a 500.0mL volumetric flask then fills the solution to the calibration mark with water. What is the molarity of this solution. M(Na S) = molarity (M) = moles of solute L of solution g Na S mols Na S M(Na S) 5.15g Na S 1 mol Na S 78.05g Na S 500.0mL = M Na S 11/9/07 Dr. Mack. CSUS 11 How many grams of sodium sulfide are there in 5.1mL of a M sodium sulfide solution? 5.1mL Volume (L) molarity(mol/l) = moles moles molar mass = grams ml L moles grams mol Na S 78.05g Na S mol Na S 11/9/07 Dr. Mack. CSUS 1 =1.6g Na S

4 What is the sodium ion concentration of this solution? Na S(aq) Na (aq) S - (aq) When the salt dissolves in solution, moles of sodium ion result for every one mole of the sodium sulfide salt. How many moles of sodium ion are there in 5.1mL of a M sodium sulfide solution? Volume (L) molarity(mol/l) = moles The moles of Na are found by: M Na S mol Na = 1.89M Na 1mol Na S 5.1mL mol Na S mol Na = 1mol Na S The concentration of ions depends upon the molar ratios in the salt mols Na 11/9/07 Dr. Mack. CSUS 1 11/9/07 Dr. Mack. CSUS 14 Suppose one wants to prepare 50.0 ml of a M solution of AgNO. How would this be done? Volume (L) molarity(mol/l) = moles moles molar mass = grams volume mols grams (that need to be added to 0.50 L) 50.0mL 0.105mol AgNO g AgNO 1 mol = 4.46g AgNO add 4.46g of AgNO to a 50.0mL volumetric flask and dilute to the calibration mark with water. 11/9/07 Dr. Mack. CSUS 15 A 5.7 % solution of potassium nitrate has a density of 1.11 g/ml. Calculate the molarity of the solution. 5.7% = 5.7 g KNO g solution 5.7 g KNO 1.11 g solution g solution 1.00 ml solution 11/9/07 Dr. Mack. CSUS 17 d = 1 mol KNO 1L 1.11 g solution 1.00 ml solution g KNO = = mol KNO /L

5 Solution Stoichiometry: When solutions mix, a chemical reaction may result. One can calculate the concentrations of products and reactants based on the stoichiometry of the reaction. Consider solutions of nitric acid and potassium hydroxide: HNO (aq) and KOH(aq) When they mix: acid base make a salt water HNO (aq) KOH(aq) KNO (aq) H O(l) How many ml of 0.15M nitric acid are needed to completely neutralize 5.1mL of 0.105M potassium hydroxide? HNO (aq) KOH(aq) KNO (aq) H O(l) ml KOH mols KOH mols HNO ml HNO 5.1mL 0.105mol KOH 1mol HNO 1mol KOH = 1.1mL 0.15mol HNO 1L 11/9/07 Dr. Mack. CSUS 18 11/9/07 Dr. Mack. CSUS 19 How many grams of silver chloride will result form the mixing of 5.1 ml of 0.15M NaCl and 15.6 ml of 0.105M AgNO? Step 1: Write the balanced chemical reaction. AgNO (aq) NaCl(aq) AgCl(s) NaNO (aq) Step : Recognize that moles can be found form Molarity and volume Volume (L) molarity(mol/l) = moles How many grams of silver chloride will result form the mixing of 5.1 ml of 0.15M NaCl and 15.6 ml of 0.105M AgNO? L 0.15mol NaCl 5.1mL L L 0.105mol AgNO 15.6mL L 1 mol AgCl 1 mol NaCl 1 mol AgCl 1 mol AgNO 14.4g AgCl mol AgCl = 0.450g AgCl 14.4g AgCl mol AgCl Step : Recognize that this is a limiting reactant problem; The LR will determine the maximum amount of product that can form. AgNO limits! = 0.5g AgCl 11/9/07 Dr. Mack. CSUS 0 11/9/07 Dr. Mack. CSUS 1

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