1. 16-2. Dilution Calculations and Molar Solubility A. MOLAR SOLUBILITY : Think of molar solubility as the molarity of a saturated solution!! Moles per Litre Example-1: 0.49 g of AgBrO 3 is required to saturate 250 mL of water. What is the molar solubility of AgBrO 3 ? 0.49 g x 1mol AgBrO 3 = 235.8 g 0.002078 mol AgBrO 3 M= mol = 0.002078 mol = L 0.25 L 0.0083 M AgBrO 3

2. Example-2: AlF 3 has a molar solubility of 0.0665 M. How many grams of AlF 3 will dissolve in 1.00 L? mol = M x Litre =0.0665 M x 1.00 L = 0.0665 mol x 84.0 g = 1 mol 5.59 g AlF 3 will dissolve In 1.00 L of solution 0.0665 mol AlF 3

3. B. DILUTION CALCULATIONS : Example-3: Suppose you have 300.0 mL of 1.2 M HCl. How much water do you need to add to dilute it to 0.50 M HCl? C 1 V 1 = C 2 V 2 (1.2M)(300.0mL) = (0.50M)(V 2 ) V 2 = 720mL 420 mL water added Use the formula: C 1 V 1 = C 2 V 2 (initial conc.) x (initial volume) = (final conc.) x (final volume) Or better still: moles before = moles after (and use moles = M x L to substitute in) 720mL – 300.0 mL =

4. Example-4: You have 145 mL of a 8.0 M HNO 3 solution. What is final concentration after adding 650 mL H 2 O? C 1 V 1 = C 2 V 2 (8.0M)(145mL) = (C2)(145mL + 650mL) C 2 = 1.46 M =1.5 M HNO 3 final concentration

5. C. Concentration of Individual Ions Use the stoichiometry (mole ratio) to calculate ion concentrations MgBr 2(s) sssd Mg +2 (aq) + 2 Br - (aq) H2OH2O 1 M sssd 1 M + 2 M 1 mol sssd 1 mol + 2 mol 0.3 M sssd 0.3 M + 0.6 M

6. [Fe 3+ ] = 1.5 M Fe 2 (SO 4 ) 3 x 2 mol Fe 3+ 1 mole Fe 2 (SO 4 ) 3 Example-5: What are the concentrations of ions in a 1.5M solution of Fe 2 (SO 4 ) 3 ? = 3.0 M Fe 3+ Fe 2 (SO 4 ) 3(aq) sssd 2Fe 3+ (aq) + 3 SO 4 2- (aq) [SO 4 2- ] = 1.5 M Fe 2 (SO 4 ) 3 x 3 mol SO 4 2- 1 mole Fe 2 (SO 4 ) 3 = 4.5 M SO 4 2-

7. Example-6 : 8.5g of MgCl 2 is dissolved in a solution and diluted up to 2.0L. What is the [Mg 2+ ] and [Cl - ]? [MgCl 2 ] = 8.5 g MgCl 2 x 1mole MgCl 2 2.0 L95.1 g 0.045 M Mg 2+ ; 0.089 M Cl - = 0.04469 M MgCl 2 MgCl 2(aq) sssd Mg +2 (aq) + 2 Cl - (aq) 0.04469 M 0.08937 M

8. Example-7: 500.0 mL of 0.80 M Li 2 CO 3 is mixed with 500.0 mL of 0.50 M BeF 2. and no reaction occurs. Find the concentration of all ions! [Li 2 CO 3 ] = 0.80M Li 2 CO 3 x 0.5000L = 0.40 M Li 2 CO 3 1.000L Dilute compounds first and then find the ions (final volume = 500 mL + 500mL = 1000mL), a 1:2 dilution [BeF 2 ] = 0.50M BeF 2 x 0.5000L = 0.25 M BeF 2 1.000L [Li + ] = 0.80 M [CO 3 2- ] = 0.40 M [Be 2+ ] = 0.25 M [F - ] = 0.50M

9. Example-8: 300.0 mL of 0.60 M AlCl 3 is mixed with 200.0 mL of 0.50 M BeCl 2. and no reaction occurs. Find concentration of all ions! [AlCl 3 ] = 0.60M AlCl 3 x 0.3000L = 0.360 M AlCl 3 0.5000L Dilute compounds first and then find the ions (final volume = 300.0 mL + 200.0mL = 500.0mL), [BeCl 2 ] = 0.50M BeCl 2 x 0.2000L = 0.20 M BeCl 2 0.5000L [Al 3+ ] = 0.36 M [Cl - ] = 1.1 M [Be 2+ ] = 0.20 M [Cl - ] = 0.40 M [Cl - ] = 1.1 M +0.40 M 1.5 M