Properties of Solutions

Water Molecule

• A highly polar molecule
• The electronegativity of the oxygen isn't equal with the hydrogens.
• The hydrogens are positive while the oxygen is negative.

Solutions: Solutions contain a solvent and a solute. The solvent is in the greater amount than the solute. The solute is the particles being dissolved. Another type of solution are aqueous solutions. These solutions contain water as the solvent. Other types of solutions are saturated, unsaturated and supersaturated solutions. Saturated has the maximum solute dissolved. Unsaturated means it can dissolve more. Supersaturated is when the solution can dissolve more solute than it could.

Factors that Affect Solubility

• As temperature increases the solubility also increases only for solids and liquids.
• An increase in pressure increases solubility for all states of matter.
• Stirring increases solubility for solids and liquids.
• Grinding also increases solubility as the surface area is increased. (solids)

Aqueous Solutions

• In aqueous solutions the water only dissolves other particles that have the same polarity. It also dissolves ionic compounds.
• Water is polar so it dissolves polar particles. Nonpolar dissolves nonpolar particles

Aqueous Solutions

• For example urea CO(NH2)2 is soluble in water but not in hexane, a non polar compound. This means that urea is polar.

Concentration and Dilution

• Concentration is the measure of solute in a solution.
• It can be expressed in Molarity (M) and Percent solutions.
• Molarity (M) = mol/L, the volume has to be in liters (L). 1000 mL = 1 L. Moles = molar mass of compound.

Concentration and Dilution Cont.

• Dilution is using more solvent in the solution, it reduces the moles of solute. The equation is M1xV1=M2xV2 (V= volume, M= molarity)

Concentration and Dilution Examples

• Find the concentration of a solution with 0.25 mol dissolved in 125 mL of water.
• M= mol/L, so 1000 mL = 1 L, 125 mL / 1000 mL = 0.125 L
• M= 0.25/0.125
• M= 2, but look at significant figures. 0.25 is 2 sig. figs. and 0.125 is 3 sig. figs. so we take the lower sig. fig. which is 2.
• M=2.0

Concentration and Dilution Examples Cont.

• How many milliliters of a solution of 2.00M MgSO4 would you need for 100.0 mL of 0.400M MgSO4.
• First list what is given. M1=2.00M, V1=?, M2=0.400M, V2= 100.0mL.
• (M1)(V1)=(M2)(V2) so (2.00M)(?)=(0.400M)(100.0mL).

Concentration and Dilution Examples Cont.

• Isolate what you need to find. (2.00M)(V1)/2.00M=(0.400M)(100.0mL)/2.00M
• V1=(0.400M)(100.0mL)/2.00M Now you can calculate the V1. V1=40mL/2.00
• V1=20 ,but 3 significant figures so V1=20.0mL