Notes

Solubility Rules 1.Most nitrate (NO 3 1- ) salts are soluble. 2.Most salts of Na +, K +, and NH 4 + are soluble. 3.Most chloride salts are soluble. Notable exceptions are AgCl, PbCl 2, and Hg 2 Cl 2. 4.Most sulfate salts are soluble. Notable exceptions are BaSO 4, PbSO 4, and CaSO 4. 5.Most hydroxide compounds are only slightly soluble.* The important exceptions are NaOH and KOH, Ba(OH) 2 and Ca(OH) 2 are only moderately soluble. 6.Most sulfide (S 2- ), carbonate (CO 3 2- ), and phos- phate (PO 4 3- ) salts are only slightly soluble. *The terms insoluble and slightly soluble really mean the same thing. Such a tiny amount dissolves that it is not possible to detect it with the naked eye. General Rules for Solubility of Ionic Compounds (Salts) in Water at 25 o C

Solubility of Common Compounds NO 3 - salts Na +, K +, NH 4 + salts Soluble compounds Cl -, Br -, I - salts S 2-, CO 3 2-, PO 4 3- salts OH 1- salts SO 4 2- salts Except for those containing Ag +, Hg 2 2+, Pb 2+ Except for those containing Ba 2+, Pb 2+, Ca 2+ Insoluble compounds Except for those containing Na +, K +, Ca 2+

Dilutions Concentration…a measure of solute-to-solvent ratio concentrated vs. dilute “lots of solute” “not much solute” “watery” Add water to dilute a solution; boil water off to concentrate it.

Dilution  Preparation of a desired solution by adding water to a concentrate.  Moles of solute remain the same.

In an experiment, a student needs mL of a M CuCl 2 solution. A stock solution of 2.00 M CuCl 2 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 CuCl 2 )(V c ) = (0.100 M CuCl 2 )( L) V c = L or 12.5 mL To make the solution: 1)Pipet 12.5 mL of stock solution into a mL volumetric flask. 2) Carefully dilute to the calibration mark. M c × V c = M d × V d

How would you prepare 60.0 mL of 0.2 M HNO 3 from a stock solution of 4.00 M HNO 3 ? M i V i = M f V f M i = 4.00 M f = 0.200V f = 0.06 L V i = ? L V i = MfVfMfVf MiMi = x = L = 3 mL 3 mL of acid + 57 mL of water= 60 mL of solution

Serial Dilutions  A serial dilution is any dilution where the concentration decreases by the same quantity in each successive step.  Serial dilutions are multiplicative.  If a solution has a 1/10 dilution the number represents 1 part of the solute added to 9 parts of diluent(solvent).  So the volumes used would be 10-1= 9.  This represents 1 part solute added to 9 parts of diluent(solvent).

Dilutions  If a 1/8 dilution of the stock solution is made followed by a 1/6 dilution what is the final dilution.  The final dilution is: 1/8 x 1/6 = 1/48  These type of dilutions are trickier and not used very frequently in the clinical lab.

Doubling Dilutions  “Doubling dilutions” are very popular.  This is a series of ½ dilutions. Each successive tube will ½ the amount of the original concentrated solution.  If this is done 6 times this is what you would end up with:

Doubling Dilution 6 Times  1st dilution = 1 /2  2nd dilution = 1 /2 x 1 /2 = 1/4  3rd dilution = 1/4 x 1 /2 = 1/8  4th dilution = 1/8 x 1 /2 = 1/16  5th dilution = 1/16 x 1 /2 - 1/32  6th dilution = 1/32 x 1 /2 = 1/64  This results in a series of dilutions, each a doubling dilution of the previous one

…normal boiling point (NBP)…higher BP FREEZING PT. DEPRESSION BOILING PT. ELEVATION Colligative Properties  depend on concentration of a solution Compared to solvent’s… a solution w/that solvent has a… …normal freezing point (NFP)…lower FP

1. salting roads in winter FP BP water0 o C (NFP)100 o C (NBP) 2. antifreeze (AF) /coolant FP BP water0 o C (NFP)100 o C (NBP) water + a little AF–10 o C110 o C 50% water + 50% AF–35 o C130 o C water + a little salt water + more salt –11 o C 103 o C –18 o C 105 o C Applications (NOTE: Data are fictitious.)

3. law enforcement white powder starts melting at… finishes melting at… penalty, if convicted A109 o C175 o Ccomm. service B150 o C180 o C2 years C194 o C196 o C20 years

Calculations for Colligative Properties The change in FP or BP is found using…  T x = change in T o (below NFP or above NBP) K x = constant depending on… (A) solvent (B) freezing or boiling m = molality of solute = mol solute / kg solvent i = integer that accounts for any solute dissociation any sugar (all nonelectrolytes)……………...i = 1 table salt, NaCl  Na 1+ + Cl 1– ………………i = 2 barium bromide, BaBr 2  Ba Br 1– ……i = 3  T x = K x m i

Freezing Point Depression  T f = K f m i Then use these in conjunction with the NFP and NBP to find the FP and BP of the mixture. (K b = ebullioscopic constant, which is 0.51 K kg/mol for the boiling point of water) (K f = cryoscopic constant, which is 1.86 K kg/mol for the freezing point of water) Boiling Point Elevation  T b = K b m i

(NONELECTROLYTE) 168 g glucose (C 6 H 12 O 6 ) are mixed w /2.50 kg H 2 O. Find BP and FP of mixture. For H 2 O, K b = 0.512, K f = –1.86. i = 1  T b = K b m i = (0.373) (1) = 0.19 o C BP = ( ) o C = o C  T f = K f m i = –1.86 (0.373) (1) = –0.69 o C FP = (0 + –0.69) o C = –0.69 o C

168 g cesium bromide are mixed w /2.50 kg H 2 O. Find BP and FP of mixture. For H 2 O, K b = 0.512, K f = –1.86.  T b = K b m i = (0.316) (2) = 0.32 o C BP = ( ) o C = o C  T f = K f m i = –1.86 (0.316) (2) = –1.18 o C FP = (0 + –1.18) o C = –1.18 o C Cs 1+ Br 1– i = 2 CsBr  Cs 1+ + Br 1–

Homework Assignment  Calculate the boiling point of a solution made from 227 g of MgCl 2 dissolved in 700. g of water. What is the boiling point of the solution? Kb = 0.512° C/m.  If 90.0 g of nonionizing C6H12O6, are dissolved in 255 g of H2O, what will be the boiling point of the resulting solution?  Which salt, NaCl or CaCl 2, has a greater effect on freezing point? Explain  Imagine you are a famous chef cooking spaghetti for your favorite chemistry teacher – he is very interested in your technique and would like to know more about the chemistry behind your amazing cooking! How much would the boiling point of 1L of your pasta water change if you added 20g of NaCl to it?  If 45 mL of water are added to 250 mL of a 0.75 M K2SO4 solution, what will the molarity of the diluted solution be?  If water is added to 175 mL of a 0.45 M KOH solution until the volume is 250 mL, what will the molarity of the diluted solution be?  How much water would need to be added to 750 mL of a 2.8 M HCl solution to make a 1.0 M solution?