1. 1 Solution Stoichiometry The concentration of a solution is the amount of solute present in a given quantity of solvent or solution. M = molarity = moles of solute liters of solution

2. M = moles of solute liters of solution Molarity (M) Molality (m) m = moles of solute mass of solvent (kg) Because density (volume) can change with temperature it is helpful to express solvent by mass when sample undergoes temperature changes Includes solute volume Excludes solute mass No volumetric measurements needed; all mass Measures of Concentration The concentration of a solution is the amount of solute present in a given quantity of solvent or solution.

3. 3 M = moles of solute liters of solution (grams) (mL) L = moles of solute Molarity moles = Liters x Molarity

4. Calculate the Molarity and Molality of a H 2 SO 4 solution containing 24.4 g of sulfuric acid in 198 g of water at 70°C to produce a 204 mL solution. M = moles L solution M = 0.249 mol 0.204 L = 1.22 Molar H 2 SO 4 Molarity/Molality Problem 24.4 g H 2 SO 4 98.1 g H 2 SO 4 1 mol H 2 SO 4 = 0.249 mol H 2 SO 4 m = moles kg solvent m = 0.249 mol 0.198 kg = 1.26 molal H 2 SO 4

5. 5 What is the Molar concentration of the Potassium ion [K + ] when 9.85 g of K 2 CO 3 are dissolved in a 250 mL solution? 9.85 g K 2 CO 3 138.2 g K 2 CO 3 = 0.143 mol K + M = 0.143 mol 0.250 L = 0.57 M K + Ion Molarity Problem 1 mol K 2 CO 3 2 mol K + K 2 CO 3(s) → 2K + (aq) + CO 3 -2 (aq) 1 mol K 2 CO 3

6. Preparing a Solution of Known Concentration from solids Volumetric FlaskMix till dissolved Bring to desired volume

7. 7 What is the Molar concentration of the Sodium ion [Na + ] when 23.4 g of NaCl and 34.1 g of Na 2 O are dissolved in 0.60 L H 2 O? 23.4 g NaCl 1 mol 58.5 g NaCl = 0.40 mol NaCl= 0.40 mol Na + 34.1 g Na 2 O 1 mol Na 2 O 62.8 g Na 2 O = 1.08 mol Na + 0.40 mol Na + + 1.08 mol Na + = 1.48 mol Na + M = 1.48 mol 0.60 L = 2.5 M Na + (doubles from subscript) Molarity Problem 1 mol Na 2 O 2 mol Na +

8. How many grams of potassium dichromate (K 2 Cr 2 O 7 ) are required to prepare a 250-mL solution whose concentration is 2.16 M? M = mol L mol = M  L Molarity Problem

9. 9 Dilution is the procedure for preparing a less concentrated solution from a more concentrated solution (stock). Dilution Add Solvent Moles of solute before dilution (1) Moles of solute after dilution (2) = M1V1M1V1 M2V2M2V2 = Number of moles does not change

10. 4.9 Dilution Practice Describe how you would prepare 500 mL of a 1.75 M H 2 SO 4 solution, starting with an 8.61 M stock solution of H 2 SO 4. Keep in mind that in dilution, the concentration of the solution decreases but the number of moles of the solute remains the same. M 1 V 1 = M 2 V 2

11. 4.9 Solution Solution We prepare for the calculation by tabulating our data: M 1 = 8.61 M M 2 = 1.75 M V 1 = ? V 2 = 500 mL Thus, we must dilute 102 mL of the 8.61 M H 2 SO 4 solution with water to give a final volume of 500 mL

12. Describe how you would prepare 300 mL of a 0.4 M H 3 PO 4 solution, starting with an 1.5 M stock solution of H 3 PO 4. Dilution Problem M 1 V 1 = M 2 V 2

13. You have 250 mL of a 3.0 M Ba(OH) 2 solution. What is the concentration if we add 150 mL of water to the solution? Bell Ringer M 1 = 3.0 M V 1 = 250 mL M 2 = ? V 2 = 250 mL + 150 mL = 400 mL Crash Course: Water and Solutions for Dirty Laundry www.youtube.com/watch?v=AN4KifV12DA

14. 14 2) What is the Molar concentration of the Sodium ion [Na + ] when 2.8 g of Na 3 PO 4 and 4.5 g of Na 2 CO 3 are dissolved in 85 mL? 1) How many grams of solid NaNO 3 are needed to produce 125 mL of a 0.85 M NaNO 3 solution? 3) How would you prepare 250 mL of a 0.65 M H 2 SO 4 solution from a stock of 6.5 M H 2 SO 4 solution? 4) You have 50 mL of 6.0 M NaF and 450 mL water are added, what is the new Molarity? Bell Ringer

15. 15 Titrations In a titration, a solution of accurately known concentration is added gradually to another solution of unknown concentration until thechemical reaction between the two solutions is complete. Standard solution – solution with known concentration to be precisely added for comparison Equivalence point – the point at which the reaction is complete example) 1 mol H 2 SO 4  2 mol NaOH ( obtained from balanced equation )

16. Titrations Indicator – substance that changes color at (or near) the equivalence point Slowly add base to unknown acid UNTIL the indicator changes color

17. 17 Titrations can be used in the analysis of: Acid-base reactions Redox reactions H 2 SO 4 + 2NaOH 2H 2 O + Na 2 SO 4 5Fe 2+ + MnO 4 - + 8H + Mn 2+ + 5Fe 3+ + 4H 2 O *Can involve color changes without indicator

18. 1)Write Balanced equation for stoichiometry (mole-to-mole ratio) Titration Steps 2) Determine moles of Standard solution used. (mol = M x L) 3) With stoichiometry, convert moles standard to moles unknown (train tracks) 4) Determine unknown Molarity using given volume (L) (M = mol/L) Alternative Equation: M s  V s = M u  V u Coefficient #

19. It takes 32 mL of 2.0M HCl standard to neutralize a 500. mL solution of Ba(OH) 2. What is the concentration of Ba(OH) 2 ? Titration Problem #1 1) 1Ba(OH) 2 + 2HCl → BaCl 2 + 2H 2 O (Reacts 1:2) 2) mol HCl = 2.0M x 0.032 L = 0.064 mol HCl 3) 0.064 mol HCl 1 mol Ba(OH) 2 4) M Ba(OH) 2 = 0.032 moles 0.500 L = 0.064 M Ba(OH) 2 2 mol HCl = 0.032 mol Ba(OH) 2 M H  V H = M OH  V OH Coefficient # Alternative: 2.0  32 = M OH  500 1 2 M OH = 0.064

20. Titration Problem #2 How many milliliters (mL) of a 0.610 M NaOH solution are needed to neutralize 20.0 mL of a 0.245 M H 2 SO 4 solution? NaOH + H 2 SO 4 H 2 O + Na 2 SO 4 2 2 1) For every 2 moles base added, it neutralizes 1 mole acid

21. 2) Next we calculate the number of moles of H 2 SO 4 in a 20.0 mL solution: 3) From the Balanced Equation: 1 mol H 2 SO 4  2 mol NaOH. = 9.80 × 10 -3 mol NaOH 4.9 x 10 -3 mol H 2 SO 4 2 mol NaOH 1 mol H 2 SO 4 4) L = 9.80 x 10 -3 mol 0.61 M NaOH = 0.0161 L or 16.1 mL NaOH Titration #2 Solution Moles = M x L 0.245 M x 0.0200 L = 0.00490 mol H 2 SO 4

22. Titration: Finding the Molar Mass of an Unknown Lauric Acid is a short-chain fatty acid that is solid at room temperature and monoprotic. We dissolve 0.022 grams into 500. mL of water and then titrate it with 0.010 M NaOH. If it takes 11.0 mL of NaOH to neutralize the fatty acid, what is the Molar Mass of Lauric Acid? Molar mass has the units grams/mole. We weighed out the mass of the solid acid in grams. Titration can tell us how many moles of acid are present in the same sample. Because it is monoprotic, it will react 1:1 with NaOH. Moles H + = ? Molar mass 0.022 grams

23. Molar Mass Titration Solution 1) Given information states it reacts 1:1 3) 1.1 x 10 -4 mol NaOH 1 mol Lauric acid 4) Molar Mass = = 200 g/mol 1 mol NaOH = 1.1 x 10 -4 mol Lauric Acid 2) (0.010 M NaOH) x (0.0110 L) = 1.1 x 10 -4 mol NaOH 1.1 x 10 -4 moles 0.022 grams Large component of coconut oil ~ 3-6% of milk C 11 H 23 COOH

24. Example: Redox Titration A 16.42-mL volume of 0.1327 M KMnO 4 solution is needed to oxidize 25.00 mL of a FeSO 4 solution in an acidic medium. What is the concentration of the FeSO 4 solution in molarity? The net ionic equation is need to find given want to calculate

25. Redox Titration Solution Solution The number of moles of KMnO 4 (in 16.42 mL) = M x L From the net ionic equation we see that 5 mol Fe 2+  1 mol MnO 4 - M = moles of solute liters of solution = 1.090 x 10 -2 mol 0.025 L = 0.436 M FeSO 4 (0.1327 M KMnO 4 ) x (0.01642 L) = 2.179 x 10 -3 mol

26. 26 Titration Bell Ringer 50 mL of Ba(OH) 2 is measured out into an Erlenmeyer flask. The concentration is unknown. It takes 8.5 mL of 2.5 M H 3 PO 4 standard to reach the equivalence point and neutralize the unknown base. Determine the concentration of Ba(OH) 2 solution.

27. 27 Chemistry in Action: Metals from the Sea CaCO 3 (s) CaO (s) + CO 2 (g) Mg(OH) 2 (s) + 2HCl (aq) MgCl 2 (aq) + 2H 2 O (l) CaO (s) + H 2 O (l) Ca 2+ (aq) + 2OH - (aq) Mg 2+ (aq) + 2OH - (aq) Mg(OH) 2 (s) Mg 2+ + 2e - Mg 2Cl - Cl 2 + 2e - MgCl 2 (aq) Mg (s) + Cl 2 (g) 1.3 g of Magnesium/ Kg seawater Many metals are found in the earth’s crust, but it is cheaper to “mine” from the sea Precipitation Slightly soluble Precipitation Electrolysis of MgCl 2 (redox)

28. 28 Gravimetric Analysis  Analytical technique based on the measurement of mass Precipitation: the analyte is precipitated out of solution by adding another reagent to make it insoluble. Then filtered and weighed. Volatilization: the analyte is converted to a gas and removed. The loss of mass from the starting material indicates the mass of gas.

29. 29 Gravimetric Analysis 1.Dissolve unknown substance in water (if not already dissolved) 2.React unknown with precipitating reagent to form a solid precipitate Reagent: chemical added to another substance to bring about a change 3.Filter, dry, and weigh precipitate. 4.Use chemical formula and mass of precipitate to determine amount of unknown analyte (chemical of interest in experiment)

30. A sample of an unknown soluble compound contains Ba +2 and is dissolved in water and treated with excess sodium phosphate. If 0.411 g of Barium phosphate precipitates out of solution, what mass of Barium in found in the unknown compound? 3 x 137.3 g Ba +2 601.9 g Ba 3 (PO 4 ) 2 x 100% = 68.4% Ba in Ba 3 (PO 4 ) 2 0.684 x 0.411 g = 0.253 g Ba +2 Gravimetric Analysis Problem #1 We need to find 1 st find the Mass % of the analyte in the precipitated compound using their respective molar masses *It is not mandatory to convert to percentage form. The mass fraction can be used directly. 411.9 g Ba +2 601.9 g Ba 3 (PO 4 ) 2 x 0.411g Ba 3 (PO 4 ) 2 = 0.253 g Ba +2

31. We have 250 mL of a Copper (Cu +1 ) solution. We add excess Na 2 CO 3 to precipitate out 3.8 g of Cu 2 CO 3. What is the [Cu +1 ] Molarity of the original solution? Gravimetric Analysis Problem #2 1 mol Cu 2 CO 3 = 0.041 mole Cu +1 M = 0.250 L = 0.16 M Cu +1 Since we need to find moles of Cu +1, it will be quicker to use train-tracks instead of % composition (either would work). 3.8 g Cu 2 CO 3 187.0 g Cu 2 CO 3 2 mol Cu +1 1 mol Cu 2 CO 3 mol Cu + 0.041

32. A 0.5662-g sample of an ionic compound containing chloride ions and an unknown metal is dissolved in water and treated with an excess of AgNO 3. If 1.0882 g of AgCl precipitate forms, what is the percent by mass of Cl in the original compound? Gravimetric Analysis Problem #3 35.45 g Cl - 143.4 g AgCl x 1.0882 g AgCl = 0.269 grams Cl

33. More Gravimetric Review Problems 1) An unknown ionic compound contains Carbonate (CO 3 -2 ). To precipitate the carbonate, we add excess CaCl 2 and collect 25.3 grams of CaCO 3 precipitate. Calculate mass of Carbonate present in the original compound. 2) 50 mL of a solution contains an unknown amount of Ni + ions. We add excess Na 3 PO 4 to precipitate out 5.67 grams of Ni 3 PO 4. What is the Molarity of Ni + ? 3) 340 mL of an unknown solution contains the Silver ion (Ag + ). When excess Na 2 S is added, 15.8 grams of precipitated Ag 2 S are formed. What is the Molarity of Ag + in the solution?