1. Solutions: Concentration Chapter 14

2. Solution  Homogenous mixture of 2 or more substances in single phase  = 1 layer  Component present in largest amt = solvent  Other component(s) = solute  Alloys, air are all solutions (solns)

3. Colligative properties  = change of mp, bp, osmotic pressure of soln, change in vapor pressure  Irrespective of solute identity  Dependent on concentration

4. Concentration  Molarity (M) = moles solute/L of solution = mol/L  We’re all familiar with this  Drawback  won’t give you proper amount of solvent used to make soln  Molality (m) better

5. Problem  Give the concentration (in M) of 0.0012 grams of NaCl in 545 mL of water  MW of NaCl = 58.442 g/mol

6. Molality  Molality (m) = moles solute/kg of solvent  Let’s look to the right   molality  molarity

7. Problem  Give the concentration (in m) of 0.0012 grams of NaCl in 545 mL of water  Density of water @ 25°C = 0.9970 g/mL  MW of NaCl = 58.442 g/mol

8. Mole fraction  Mole fraction of A (X A ) = n A /n tot  Amt of component A/total components  Soln contains 1.00 mol ethanol and 9.00 mol water

9. Weight percent  Ex: 46.1 g ethanol & 162 g water  Commonly used in household products like vinegar & bleach

10. Problem Concentrated sulfuric acid has a density of 1.84 g/cm 3 and is 95.0% (w/w) H 2 SO 4. MW H 2 SO 4 = 98.079 g/mol.   Calculate the molarity and the molality of this solution.

11. Solution: molarity

12. Solution: molality

13. Problem A 10.7 molal solution of NaOH has a density of 1.33 g/cm 3 at 20°C. MW NaOH = 39.996 g/mol & MW H 2 O = 18.0153 g/mol.   Calculate the mole fraction of NaOH, the weight percentage of NaOH and the molarity of the solution.

14. Solution

15. More practice  An aqueous soln of NaCl is created using 133 g of NaCl diluted to a total soln volume of 1.00 L.  Calculate the molarity, molality, and mass percent of the soln, given a density of 1.08 g/mL and MW of NaCl = 58.442 g/mol.

16. Solution

17. Part per million  = PPM (in grams)  Ex: 1.0 ppm = 1.0 g of substance in system w/ 1.0 million g total mass  @ STP water density  1.0 g/mL  So, mg/L and ppm are   Used predominately by environmental and analytical chemists

18. Solution process  One can add only so much solute to solvent  Since no more dissolves  soln said to be saturated  NaCl = 35.9 g/100 mL water (25°C)  Albeit, nothing changes visually, soln is constantly dissolving and re-solidifying ions

19. Solution process  Essentially, solubility = solute concentration in equilibrium w/undissolved solute in saturated soln  Unsaturated soln = soln w/less than saturated amt of solute  NaCl < 35.9 g/100 mL water (25°C)  Supersaturated soln = soln w/more solute than sat. soln  NaCl > 35.9 g/100 mL water (25°C)

20. Making supersaturated solutions  Pour in excessive amount of solute  Heat up the soln  Stir until all solute dissolves  Cool it slowly  No shaking, no jarring of soln  Gives lower freezing point  Once disturbed (energy in), causes crystallization to occur  excess crystallized out of soln  Exothermic  Heat packs of sodium acetate (can reach 50°C!)  http://www.npr.org/programs/wesun/features/2001/dec/heatpack/0 11229.heatpack.html http://www.npr.org/programs/wesun/features/2001/dec/heatpack/0 11229.heatpack.html http://www.npr.org/programs/wesun/features/2001/dec/heatpack/0 11229.heatpack.html  Your second lab deals with this

21. Liquids as solutes  Miscible = mixable  Immiscible = unmixable  Used in language too:  Mestizo, mischling  Ability to dissolve based on similar polarities (or lack thereof) of solute/solvent  Like dissolves in like

22. Let’s try these   Considering intermolecular forces, give reasons for the following observations:   a) Octane, C 8 H 18, is very miscible with CCl 4.   b) Methanol, CH 3 OH, mixes in all proportions with water.   c) Sodium bromide is not very soluble in diethyl ether (CH 3 CH 2 —O—CH 2 CH 3 ).   d) Octanol, C 8 H 17 OH, is not very soluble in water.