 # COMMUNICATING CONCENTRATION: Section 5.3.  Most solutions are similar in that they are colorless and aqueous, so there is no way of knowing, by looking.

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COMMUNICATING CONCENTRATION: Section 5.3

 Most solutions are similar in that they are colorless and aqueous, so there is no way of knowing, by looking at them, how much of the solute is present in the solution.  To do quantitative analysis (how much is present), numerical representation is needed  Concentration is a ratio comparing the quantity of solute to the quantity of solution; in general, the concentration, c, of any solution is expressed as:  Concentration – general formula  C = quantity of solute quantity of solvent --

 “Dilute” – relatively small quantity of solute per volume of solution  “Concentrated” – relatively large quantity of solute per volume of solution

Communicating Concentration  There are many different ways to express concentration:  - percent concentration  - parts per million  - amount concentration  We will focus on these 3 in Chem 20!

Bicarbonates: 320 PPM Calcium Ion: 150 PPM Fluoride: 0.12 PPM Magnesium: 4.2 PPM Potassium: 1.2 PPM Sulfates: 43 PPM Sodium: 11 PPM

 Amount Concentration (aka Molar Concentration) (mol/L) – is the chemcial amount of solute dissloved in 1L  c = n solute (mol) V solution (L)

 Practice (Show all of your work)  A solution contains 140 mL of pure acetic acid in a 500 mL bottle of solution. What is the percentage by volume concentration of acetic acid?  How much hydrogen peroxide (in grams) is in 200mL of commercial 3.0% hydrogen peroxide topical solutions? --

 A gold ring has a mass of 12.0g and contains 11.1 g of pure gold. What is the percentage weight by weight concentration of gold in the metal?  A chemist added 0.2 mg of salt to 100 mL of water. What is the concentration of salt in parts per million? (remember 1 ppm = 1 mg/L)

 A scientist has found 1.2 x10 -3 g of lead in a 400 g children’s toy. What is the concentration of lead in parts per million?  A laboratory technician made a solution using 0.255 mol of sodium hydroxide in 1.50 L of solution. Calculate the amount (molar) concentration of sodium hydroxide. -

Stop and Practice  Complete Pg. 205-206 #1-6

Calculations involving concentration  Calculations involving concentration always involve: quantity of solute, volume of solution and concentration of solution. If any two of these values are known, the third can be calculated using unit analysis.  Start with the non-ratio unit (has only one unit) whenever possible and then use the concentration ratio as the conversion factor.

 1) What volume of 30.0% W/V hydrogen peroxide solution can be made from 125 g of pure hydrogen peroxide?  30.0% W/V = 30.0g or 100 mL 100 mL 30.0 g Which conversion factor will you use? V H 2 O 2 = 125 g x 100 mL = 417 mL 30.0 g

Homework  Complete pg. 205-206 #1-6  P. 208 # 7- 12

Mass, Volume and Concentration Calculations  5) A chemical analysis requires 2.00 L of 0.150 mol/L AgNO 3(aq). What mass of solid silver nitrate is required to prepare this solution? (see pg. 209)  Na 2 CO 3 = 169.88g2.00 L x 0.150 mol x 169.88 g = 51.0 g 1 mol 1 L 1 mol  6) A student dissolves 5.00 g of solid sodium carbonate to make 125 mL of a solution. What is the concentration of this solution?  Na 2 CO 3 = 105.99g5.00 g x 1 mol x 1. = 0.377 mol/L mol 105.99g 0.125 L  If you prefer to use mathematical formulas: c (concentration mol/L) = n (moles mol) V (volume L)

Concentration of Ions Calculations  The dissociation or ionization equations for ionic compounds or strong acids allow you to determine the molar concentration of either the ions or the compounds in solution.  The ion concentration is always equal to the whole number multiple of the compound concentration. Square brackets will be used around formulas to indicate amount conc.

 If [CaCl 2 (aq) ] = 1 mol/L then [Ca 2+ (aq) ]= 1 mol/L and [Cl – (aq) ]= 2 mol/L because the dissociation equation shows that 2 mol of chloride ions are produced from 1 mol of calcium chloride – a 2:1 mole ratio.

 Al 2 (SO 4 ) 3 (aq)  2 Al 3+ (aq) + 3 SO 4 2- (aq)  c = 0.30 mol/L  [Al 3+ (aq) ] = 0.30 mol/L x (2) = 0.60 mol/L 1  [SO 4 2- (aq) ] = 0.30 mol/L x (3) = 0.90 mol/L 1

 2) Determine the amount concentration of barium and hydroxide ions in a solution made by dissolving 7.34 g of barium hydroxide to make a volume of 150 mL.  Ba(OH) 2 (aq)  Ba 2+ (aq) + 2 OH - (aq) c = ?  7.34 g x 1 mol x 1 = 0.28557 mol/L 171.35 g 0.150L  [Ba 2+ (aq) ] = 0.28557 mol/L x (1) = 0.286 mol/L 1  [OH - (aq) ] = 0.28557 mol/L x (2) = 0.571 mol/L 1

 Complete pg. 210 #13-16  Complete pg. 212 # 17-19  Extra Practice:  pg. 214# 4-12 (evens only)  Concentration quiz on Thursday !

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