 # Solutions Solubility -the amount of solute that can be dissolved to form a solution. Solvent – the substance in a solution present in the greatest amount.

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Solutions Solubility -the amount of solute that can be dissolved to form a solution. Solvent – the substance in a solution present in the greatest amount. Solute – the substance in a solution present in the least amount. Saturate – a solution that has come to equilibrium. The rate of dissolving is equal to the rate of re-crystalizing. High solubility – more than.10 mol/L will dissolve (AgNO 3 ) Low solubility – less than.10 mol/L will dissolve (AgCl) Remember that a solution is a homogeneous mixture

Classification of solutions Every combination of solids, liquids, and gases is possible. This means that there are nine possible kinds of solutions. 1.Liquid 2.Liquid 3.Liquid 4.Gas 5.Gas 6.Gas 7.Solid 8.Solid 9.Solid Solute Solvent Solid Liquid Gas Solid Liquid Gas Solid Liquid Gas

Concentration Concentration is one way to express the relationship between the solute and solvent in any solution. Since the mixture is homogeneous, the relative number of particles is the same in a drop as in a liter. One way to show this relationship is by expressing the number of moles of solute dissolved in one liter of solvent. That is: C = n / V M = n / V Molarity

Examples 1) What is the concentration of a solution made by dissolving 2.00 moles of NaCl (s) in a volume of 4.00 L of H 2 O (l) ? Sheet 20/5—1 #1-12 Multiple choice #1-10 2) How many moles of HCl (aq) are present in 5.00 L of a 0.500 mol/L solution of hydrochloric acid? 3) What volume of 1.00 mol/L NaOH (aq) can be made from dissolving 0.300 mol of NaOH (s) in H 2 O (l) ?

Classifying solutions - Qualitative Solutions may be – acidic - basic - neutral How would you test this? pH paper, indicators, litmus, other? Solutions may be – electrolytes - non electrolytes How would you test this? conductivity meter, conductivity apparatus, other?

Dilution Problems In dilution problems, only the amount of solvent changes. Therefore, the number of moles of solute in the final solution is the same as the number of moles of solute in the original solution. Examples 1) What volume of 12.0 mol/L HCl (aq) is needed to produce 2.00 L of 0.250 mol/L HCl (aq) ?

2) What volume of 4.00 mol/L NaOH (aq) is needed to produce 1.00 L of 0.750 mol/L NaOH (aq) ? 3) What is the maximum volume of 0.300 mol/L HNO 3(aq) that could be prepared from 45.0 mL of 8.00 mol/L HNO 3(aq) ? 4) What is the final concentration of a solution where 25.0 mL of 18.0 mol/L H 2 SO 4(aq) is added to 125.0 mL of H 2 O (l) ? Do Sheet 20/5 - -5 #5 – 9, then do 1 - 4

Addition Problems We can use the same method to solve addition problems. In this type of problem, one solution is diluted with a lower concentration of the same kind of solution. Examples: 1) 2.00 L of a 6.00 mol/L HCl (aq) solution is mixed with 0.500 L of a 5.00 mol/L HCl (aq) solution. What is the [HCl] in the final mixture? 2)300 mL of 5.00 mol/L NaOH (aq) is combined with 200 mL of an unknown concentration of NaOH (aq). The concentration of the final mixture is 6.00 mol/L. What is the concentration of the unknown solution?

Sheet 20/5--6 #5 - 10

How to make a standard solution This is a solution of a known and exact concentration Procedure 1.Calculate the mass of solute required and weigh out this amount on a scale. 2.Dissolve the solute in a small amount of water. 3.Top up the water to the final desired volume. 4.Stopper and mix well. Example: Mix up 2.00 L of a 0.120 mol/L NaOH (aq) solution.

Acids and Bases Svante Arrhenius was the first to do extensive study with acids and bases. ACIDS HCl (aq) → H + (aq) + Cl – (aq) HNO 3(aq) → H + (aq) + NO 3 - (aq) HI (aq) → H + (aq) + I - (aq) Acids – substances that give up a hydrogen ion in solution.

Bases NaOH (s) → Na + (aq) + OH – (aq) Ba(OH) 2(s) → Ba 2+ (aq) + 2 OH - (aq) Al(OH) 3(s) → Al 3+ (aq) + 3 OH - (aq) Bases – substances that give up hydroxide ions in solution.

In a solution: If the [H + ] > [OH - ] the solution is acidic. If the [OH - ] > [H + ] the solution is basic. If the [H + ] = [OH - ] the solution is neutral.

pH pH – method of expressing the [H + ] in solution – power of Hydrogen (exponential power) - Scale normally runs from 0 → 14 pH = -log [H + ] 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14

Converting [H + ] to pH Calculate the pH of a solution that has a [H + ] of 1.00 x 10 -2 mol/L. Answer pH = 2.000 This is acidic. If the [H + ] is 1.00 x 10 – 9, the pH is 9.000 This is basic. If the [H + ] is 1.00 x 10 -7. the pH is 7.000 This represents neutral conditions.

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 neutral acidicbasic

Note that these are log numbers. Each change of 1 pH unit is a 10 fold change in [H+]. Example: A solution with a pH of 4.00 has 1000 times more H + than a solution with a pH of 7.00

Neutralization When an acid and a base are reacted, the products will be an ionic salt and water. The salt present depends on which acid and which base were used in the neutralization. Example: combine HCl (aq) with NaOH (aq) HCl (aq) + NaOH (aq) → H 2 O (l) + NaCl (aq) Acid base water salt (sodium chloride)

Write net neutralization reactions for: KOH (aq) + HI (aq) →

Dissociation Reactions Dissociation - means to separate Ionic compounds and acids will dissociate into ions when they are dissolved in water. These reactions must be balanced. Example: Ca 3 (PO 4 ) 2(s) → 3 Ca 2+ (aq) + 2 PO 4 3- (aq) Complete Sheet 20/5 - 14

Volumetric Analysis

Titration is a physical process that is used to find the concentration of an unknown solution. Titration involves adding one solution (titrant) from a buret to another solution (sample) in an Erlenmeyer flask. (titration flask)

(titrant) known volume indicator is added (sample)

Titration procedure (quantitative) 1)Measured volume of unknown acid [ ] is added to a flask. 2)An appropriate indicator is added. 3)Measured amount of base of known [ ] is added using a buret. 4)Continue until solution changes color. This is called the end point.

24.19 mL

Procedure Fill the left side buret with acid. (HCl (aq) ) Fill the right side buret with base.(NaOH (aq) ) Record the initial levels on each buret. Add about 15-20 mL of acid to an erlenmeyer flask. Add 1 drop of indicator. Add base from the second buret until the expected color change occurs. Record the final buret levels on each buret.

Notes Always keep the acid on the left and the base on the right. Stop at the first permanent pink. Don’t forget to add the indicator. Back titration is o.k. You may wash the sides of the flask down with distilled water. Make sure that all glassware is very clean. Remember that you will be marked on accuracy as well. [NaOH (aq) ] = 0.120 mol/L

Dissociation (ionization) Reactions When ionic compounds dissolve in water, they separate into ions – one positive and one negative. Examples: Ca(NO 3 ) 2(s) → Ca 2+ (aq) + 2 NO 3 – (aq) Ca 3 (PO 4 ) 2 → 3 Ca 2+ (aq) + 2 PO 4 3- (aq) Write dissociation reactions for each of the following: BaSO 4(s) → Al(OH) 3(s) → PbSO 4(s) → NH 4 NO 3(s) → Ba 2+ (aq) + SO 4 2- (aq) Al 3+ (aq) + 3 OH – (aq) Pb 2+ (aq) + SO 4 2- (aq) NH 4 + (aq) + NO 3 – (aq)

Net Ionic Equations Combine solutions of NaCl (aq) and AgNO 3(aq)

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