TOPIC 7 SOLUTIONS By: Chemistry Lecturer School of Allied Health Sciences City University College of Science and Technology

Solute and Solvents  A solution consists of a liquid (the solvent) with a substance (the solute) dissolved in it.  Solubility - The amount of solute which can be dissolved in a solvent.

Saturation  the amount of solute which can be dissolved in a solvent.  Once a solution has reached the limit of the solute's and solvent's solubility, the solution is said to be saturated. Meaning that it can hold no more solute.

Supersaturation  Whenever a solution contains more solute than it can hold, it is said to be supersaturated  If more of the solute is added, the extra solute will settle out of the supersaturated solution.

Unsaturation  Solution that is capable of dissolving more solute than it already contains at the same temperature.

Solubility curve  Trace the solubility of a substance with increasing temp.  Each curve represents saturation  Above the curve the solution is supersaturated  Below the curve the solution is unsaturated

Supersaturated Temperature  ↑ Grams solute/100g water Saturated Unsaturated

SolutionSaturated or Unsaturated? A solution that contains 70g of NaNO 3 at 30°C A solution that contains 50g of NH 4 Cl at 50°C A solution that contains 20g of KClO 3 at 50°C A solution that contains 70g of KI at 0°C

What mass of solute will dissolve in 100mL of water at the following temperatures? 1)KNO 3 at 70°C = ____________ 2) NaCl at 100°C= ____________ 3)NH 4 Cl at 90°C= ____________

TYPES OF MIXTURE  A homogeneous mixture is defined as the mixture which has uniform composition through out its mass  Example: Air, sugar solution, salt solution, alloys, soft drinks (Pepsi, Coca-Cola etc.)  Mixture that do not have uniform composition through out their mass are known as heterogeneous mixtures  Example: Soil, rocks etc. HOMOGENEOUSHETEROGENEOUS

CONCENTRATION The concentration of a solution measured in moles of solute per liter of solution. mol = M L

DILUTION  Dilution consists of adding more solvent to a solution so that the concentration of the solute becomes lower.

Calculating dilution  Dilution calculations are simplified by using the following equation:  M 1 V 1 = M 2 V 2  Where:  M 1 = concentration of the first solution V 1 = volume of the first solution M 2 = concentration of the second solution V 2 = volume of the second solution

LET’S TRY!!  You have a 5 M aqueous solution of table sugar. You want to dilute the solution to 0.05 M with 250mL. What do you do? Solution:  To solve the problem, you simply plug in the three numbers you know:  M 1 V 1 = M 2 V 2  (5 M) (V 1 ) = (0.05 M) (250mL)  2.5 L = V 1

You have 4 L of an 90% aqueous solution of hydrochloric acid. You want 1 L of a 25% solution. How much of the stock solution do you need to use? Solution  M 1 = 90% V 1 = ? M 2 = 25% V 2 = 1 L  M 1 V 1 = M 2 V 2  (90%) (V 1 ) = (25%) (1L)  0.28L = V 1

DILUTION PROCESS

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