Concentration of Solutions

Review: Solutions are made up of 1)Solute - substance dissolved or present in lesser proportion 2) Solvent - substance that is the dissolving medium - what the solute is dissolved in - many times this is water

Solutions can be classified as 1)Electrolyte solutions - solutions that conduct electricity (solute is either an ionic compound or forms ions when it dissolves) 2)Nonelectrolyte solutions - solutions that do not conduct electricity (solute is a molecular compound that does not form ions as it dissolves)

Components that make up mixtures can be 1)miscible - will form solutions in most any proportions (only liquids or gases) 2)immiscible substances - will not form solutions - example: oil and water

There is a limit to the amount of solute that can be dissolved in a given amount of solvent - this is the "solubility" of the solute. Solubility of a solute can be expressed in the following terms: 1)soluble – The substance mostly dissolves 2)insoluble – Very little of the substance dissolves 3)slightly soluble – In between: some dissolves but it may not be enough to affect the properties of the solution.

Concentration The concentration of a solution refers to the amount of solute dissolved in the solvent Qualitative terms used are dilute (not much solute) and concentrated (alot of solute) - concentrated does not mean pure.

Molarity Molarity is one way to measure the concentration of a solution. moles of solute volume of solution in liters Molarity (M) = Molarity is the most used - other units include: molality, normality, formality, mole fraction, % weight, % volume (proof)

Making a Solution…

Sample Problem # g of NaCl is dissolved in enough water to make 81.0 mL of sol’n. Calculate the molarity of the solution g NaCl 1 mol NaCl X =.812M NaCl.0810 L sol’n 58.5 g NaCl

Sample Problem #2 How many grams of NaCl are required to make 450 mL of.500 M soln?.500 moles NaCl 58.5 g NaCl.45 L soln X X = 13 g NaCl 1 L soln 1 mole NaCl

Sample Problem #3 How many mL of a.250 M NaCl soln can be prepared using 7.51 g NaCl? 1 mole NaCl 1 L soln 1000 mL 7.51 g NaCl X X X = 514mL NaCl soln 58.5 g NaCl.250 mole NaCl 1 L

Titration

Sample Problem # mL of.325 molar hydrochloric acid (HCl) completely neutralizes 35.0 mL of a calcium hydroxide solution. What is the molarity of the calcium hydroxide solution? 25.0 mL35.0 mL.325 M ? M 2 HCl + Ca(OH) 2  2 H 2 O + CaCl L HCl.325 mole HCl 1 mole Ca(OH) X X =.116 M Ca(OH) L Ca(OH) 2 1 L HCl 2 mole HCl

Sample Problem #5 How many mL of.525 M nitric acid (HNO 3 ) solution would completely neutralize 22.5 mL of.275 M Ca(OH) 2 base solution?.525 M.275 M ? mL 22.5 mL 2 HNO 3 + Ca(OH) > 2 H 2 O + Ca(NO 3 ) mole Ca(OH) 2 2 mole HNO 3 1 L HNO L Ca(OH) 2 X X X L Ca(OH) 2 1 mole Ca(OH) moles HNO 3 =.0236 L or 23.6 mL HNO 3

Dilutions

Sample Problem # mL of.250 M HCl solution is added to 30.0 mL of.150 M HCl solution. What is the concentration of the resulting solution?.0200 L X.250 moles/L = moles HCl.0300 L X.150 moles/L = moles HCl L moles HCl total total moles total molarity = total Liters moles HCl M = =.190 M HCl.0500 L soln.

Sample Problem # mL of.375 M HCl solution is diluted by adding water to a new volume of 50.0 mL of solution. What is the concentration of the resulting solution? total moles total molarity = total Liters.0100 L X.375 moles/L = moles HCl moles HCl M = =.0750 M HCl.0500 L soln.

Sample Problem # mL of.375 M HCl solution is diluted by adding water to a new volume of 50.0 mL of solution. What is the concentration of the resulting solution? M 1 V 1 = M 2 V 2 (.375M)(10.0mL) = M(50.0mL) M =.0750 M

Sample Problem #8 Indicate the concentration of each ion present in the solution formed by mixing 44.0 mL of M Na 2 SO 4 and 25.0 mL of M KCl. Na 2 SO 4 + KCl  X (No reaction!) Total Volume = 44.0mL mL = 69.0mL=.0690L

Sample Problem # g of aluminum reacts with 75.0 mL of M ZnI 2 solution. How many grams of zinc are produced? M 1.00 g 75.0 mL x g 2 Al (s) + 3 ZnI 2 (aq)  2 AlI 3 (aq) + 3 Zn (s) mole ZnI 2 2 mole Al 27.0 g Al.0750 L soln x x x = g Al 1 L soln 3 mole ZnI 2 1 mole Al Al is excess or ZnI 2 is limiting mole ZnI 2 3 mole Zn 65.4 g Zn.0750 L soln x x x = 1.47 g Zn 1 L soln 3 mole ZnI 2 1 mole Zn

Sample Problem # mL of M ZnI 2 soln is added to 125 mL of M AgNO 3 soln. How many grams of the precipitate AgI are produced? M M 75.0 mL 125 mL x g ZnI 2 (aq) + 2 AgNO 3 (aq)  2 AgI (s) + Zn(NO 3 ) 2 (aq) mole ZnI 2 2 mole AgNO 3 1 L soln L soln x x x = L AgNO 3 1 L soln 1 mole ZnI mole AgNO 3 excess AgNO 3 or ZnI 2 is limiting mole ZnI 2 2 mole AgI 235 g AgI L soln x x x = 10.6 g AgI 1 L soln 1 mole ZnI 2 1 mole AgI