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Science 20. Solutions. Solutions. What are solutions? a homogeneous mixture of two or more substances; the different substances cannot be mechanically separated or seen. A solution may be a solid (stainless steel), a liquid (iced tea) or a gas (the air we breathe).
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Science 20 Solutions
Solutions What are solutions? • a homogeneous mixture of two or more substances; the different substances cannot be mechanically separated or seen. • A solution may be a solid (stainless steel), a liquid (iced tea) or a gas (the air we breathe). • Solvent: the substance that is present in the largest amount - usually determines the phase of the solution • Solute: the substance present in the smallest amount – what is dissolved in the solvent. • We will deal mainly with solutions where water is the solvent (aqueous solutions) • Why is water considered to be the universal solvent? • Water is able to dissolve many types of solutes.
Water is a polar molecule • The negative oxygen attracts positive ions, and they hydrogen ends attracts the negative ions of ionic compounds • What about molecular compounds like sugar (C12H22O11)? http://programs.northlandcollege.edu/biology/Biology1111/animations/dissolve.html http://www.mhhe.com/physsci/chemistry/essentialchemistry/flash/molvie1.swf
Electrolyte: a substance that conducts electricity when dissolved in water. • Acids, bases and soluble ionic solutions are electrolytes because they dissolve into ions. Insoluble ionic compounds are very weak electrolytes • Non-electrolyte: a substance that does not conduct electricity when dissolved in water. • Molecular compounds are non-electrolytes because they do not dissolve into ions
1. Neutral Molecular • Solutions containing molecular compounds (non metals sharing electrons) • No ions form in solution, therefore do NOT conduct electricity (non-electrolytes) • No change in color with litmus paper • Form colorless solutions • Examples: 2. Neutral Ionic • Dissociate in water to form positive and negative ions • Conduct electricity (electrolytes) • No change in color with litmus paper • Form colorless and colorful solutions • Examples:
3. Acidic Solutions • Form when an acid is dissolved in water • acid dissociates into positive hydrogen ions and negative ions • The concentration of H+ ions in solution indicate the pH (strength of acid) • Electrolytes (conduct electricity) • Turn blue litmus paper red 4. Basic Solutions • Bases - presence hydroxide ions (OH-) • Electrolytes • Turn red litmus paper blue • Conc. of OH- ions indicates pH level • Examples
Acids and Bases Defined Acids are empirically defined as substances that: • Taste sour • Conduct electricity (they are electrolytes) • React with metals to produce hydrogen gas • Turn blue litmus paper red • Their pH ranges between 0 and 7 Bases are empirically defined as substances that: • Taste bitter • Conduct electricity (they are electrolytes) • Feel soapy or slippery • Turn red litmus paper blue • Their pH ranges between 7 and 14
Testing Acids, Bases and Ionic Substances • To distinguish between acids, bases and ionic substances a few simple diagnostic tests can be used. • The litmus test - used primarily to distinguish acids from bases. • The conductivity test – used to determine if a solution is ionic, acidic, basic, or if the solution is molecular.
A - calcium carbonate B - sulfurous acid C - potassium hydroxide D - sucrose
NaOH 5. CH3OH • HCl 6. NaCl • Ag2O 7. CH3COOH • CH4
Substances Present in Water • Solutions made by dissolving solutes in water are called aqueous solutions. • If the temperature of the solution rises as a solute is dissolved, the reaction is said to be exothermic; energy is released to the solvent * energy is a product in the chemical reaction. • If the temperature of the solution decreases as a solute is dissolved, the reaction is said to be endothermic; energy is absorbed in the solvent. * energy is a reactant in the chemical reaction.
Dissociation, Ionization & Separation • We use our solubility table to determine if something dissociates or not • Dissociation: the separation of ions from an ionic compound when it dissolves in water. • Ex. If magnesium chloride is dissolved in water, it will form magnesium ions and chloride ions. Mg(Cl)2(s) Mg2+(aq) + 2Cl-(aq) • Please note that the equation must be balanced the same as any other equation. Also note that the charges on both sides of the equation are balanced. • Ex. K2SO4(s) 2K+(aq) + SO42-(aq) Ca(NO3)2(s) Ca2+(aq) + 2NO3-(aq)
If the ionic compound does not dissolve in water, it will remain as an ionic compound in water. • Ex. AgBr(s) AgBr(s) • Molecular compounds do not dissociate into respective ions in water. They dissolve in solution. • Ex. C6H12O6(s) C6H12O6(aq) • The formation of charged ions from neutral atoms by the action of the solvent is called ionization. • Acids undergo ionization in aqueous solution. • When acids are not dissolved in water, they are molecular compounds (two nonmetals). • When acids are dissolved in water they conduct electricity. Therefore, we assume that they ionize in solution; they form ions once they have dissolved in water. • Ex. HCl(g) H+(aq) + Cl-(aq)
Qualitative Analysis of Ions in Solutions • Flame Test: • Some ions impart specific colours when subjected to flame • Solution Colour: • Some ions create specific colours when in solution. • Precipitates: • By understanding the solubility table, we can determine what ions are in solution by whether or not a precipitate is formed.
Concentration There are many ways to measure concentration. 3 common ways: • Percentage of solute compared to solvent by volume (%v/v) • Moles of solute per unit volume of solution (mol/L). The units for concentration are usually written asmol/L. • You will also see ppm – parts per million (mg/L).
Ways to Write Concentration Using Percents • We can use two methods of representing concentrations based on percent • % by mass (%m/v) or % by volume (%v/v) • Simply put the solute is put in a ratio with 100 ml (because percent is out of 100) of solution • We will focus on (%v/v)
(%v/v) Examples • 45 ml of methyl alcohol is mixed with 55 ml of water. What is the % v/v of the solution • 45 is the solute, (45 + 55) is the total solution • 45 ml / 100 ml is 45% • how many grams of NaCl must be added to 25.0 ml of water to prepare a 7.50% solution • 7.5 / 100 = X / 25 =1.875 g
PPM • one way of representing solution concentration is ppm (parts per million) • simply put its mg / L • 8 mg of solute in 1 L of solution is 8 ppm • Example: some water has 6.3 X 10-3 g of lead per 375 mL of solution. What is the ppm concentration? • 6.3 X 10-3 g / 375 mL convert g to mg and mL to L so 6.3 mg / .375 L = X / 1L =16.8 ppm
Concentration in Solution • Several ways to indicate the strength or concentration of a solution. • Generally expressed as the number of moles of solute dissolved in one liter of solution. • Concentration is often referred to as molarity (M). • The symbol for concentration is C. • To calculate the concentration of a solution, use the following formula: C = n V C = concentration (mol/L) n = number of moles (mol) V = volume (L)
Molar Concentration Examples • What is the concentration made from dissolving 8.0 moles of CuSO4 in 2.2 litres of water • Step 1 • Write down what you know and don’t know • n = 8.0 mols • M = 159.62 g/mol • v = 2.2 L
Step 2 • Write down and rearrange your formula • c = n / v • Step 3 • Plug in your number and include your units • C = (8.0 mols) / 2.2 L = 3.636363636 mol/L • SIG DIGS = 3.6 mol/L
More Examples: • Example 2: What is the concentration when 5.2 moles of hydrosulfuric acid are dissolved in 500 mL of water? Step one: Convert volume to liters, mass to moles. 500 mL = 0.500 L Step two: Calculate concentration. C = 5.2 mol / 0.500 L = 10 mol/L
Concentration and Mass Calculations a) To calculate concentration, given the number of moles and the volume, use factor-analysis or the equation C = n/V. b) To calculate the mass of solute required to make a solution, given the concentration and volume, follow these steps: 1) Calculate the number of moles required n = C·V 2) Calculate the molar mass of the solute 3) Calculate the mass required m = n·M
you may have to use both formulas we know to solve questions: m=nM and c=n/v • Example: What is the mass of solute in 4.50L of a 2.4 mol/L solution of HCl ?
Solution Preparation When a chemist prepares astandard solutionfrom a solid reagent, the following steps should be taken: • Calculate the required mass of solute. • Obtain the required mass of solute in a clean, dry beaker. • Dissolve the solute in distilled water using about one half of the final solution volume. • Transfer the solution and all the water used to rinse the equipment into a clean volumetric flask. • Add distilled water until the desired volume is reached. • Stopper the flask and mix the solution by inverting the flask several times.
Dilution Calculations • The process of decreasing the concentration of a solution usually by adding more solvent. • During a dilution procedure the addition of water changes the volume of the original solution as well as the concentration. HOWEVER the original number of moles remains the same. ni(before diluting)= nf(after diluting) therefore, Ci · Vi = Cf · Vf Ci = initial concentration Cf = final concentration Vi = initial volume Vf = final volume
Dilution Problems • Eg. Water is added to 200mL of a 2.40mol/L of NH3 cleaning solution until the final volume is 1.00L. Find the molar concentration of the final diluted solution. • 0.200L x 2.40mol/L = 1.00L x c = 0.480mol/L • Change a solution of 30 mL of 0.430 mol/L of KIO solution to the solutions of the following volumes: 40mL, 25mL, 180mL
More Examples • Example 3: What is the concentration of a solution if 30.0 mL of 4.00 mol/L NaBr(aq) was diluted to a new volume of 1.50 L? • Example 4: What is the original concentration of a stock solution if 40 mL of the stock solution was used to make 2.00 L of a 0.0100 mol/L solution?
Diluting the Solution Once a standard solution has been prepared, a chemist may wish to dilute the solution further. To do this, please follow these steps: • Calculate the volume of standard solution required to make the diluted solution. • Add about one half of the final volume of distilled water to the volumetric flask. • Measure the required volume of stock solution using a pipet (see page 532 of your textbook). • Transfer the stock solution to the volumetric flask. • Add distilled water to the bottom of the calibration line. • Stopper and invert the flask several times to mix the solution.
Neutralization Reactions • A neutralization reaction is a double replacement reaction where an acid reacts with a base to produce water and a neutral substance. • Following are a few examples of neutralization reactions: H2CO3(aq) + 2NaOH(aq) Na2CO3(aq) + 2HOH(l) acid base neutral water 2HCl(aq) + Mg(OH)2(aq) MgCl2(aq) + 2HOH(l) acid base neutral water
Precipitates • Sometimes when two soluble ionic compounds are mixed, an insoluble product is formed. • This insoluble product forms a solid, which is known as a precipitate. • The reaction to form a precipitate is a double replacement reaction.