Solution Chemistry Vocabulary Refresher A solution can be

Solution Chemistry Vocabulary Refresher: A solution can be defined as a homogeneous mixture of two or more substances in a single phase. Solutions consist of a solute(s) and a solvent. The solvent is the dissolving medium in a solution. It causes the solute to dissolve. It is what does the dissolving. The solute is the dissolved substance in a solution. It’s what gets dissolved. If something is capable of being dissolved, it is said to be soluble.

If two liquids dissolve together, like alcohol and water, they are said to be miscible. If not, then they are characterized as immiscible. i. e. oil and water.

We are going to focus on aqueous solutions. Solutes can be classified in many ways; one way is according to their ability to conduct electricity when in solution. Electrical current is the flow of electrons; thus electricity is moving charges. Solutions of ionic compounds can conduct electricity, and are classified as electrolytes. The ions that are dissolved can move and flow. An electrolyte is a substance that dissolves in water to give a solution that conducts an electric current.

Solutions of strong electrolytes conduct electricity well. The solutes of these solutions typically have broken down into their ions, and thus there are many ions to conduct the electricity. Solutions of weak electrolytes will conduct electricity slightly. The solutes of these solutions typically have only partly broken down into their ions; having few ions to conduct the current. A non-electrolyte is a substance that dissolves in water to give a solution that does not conduct an electric current. A solution of a non-electrolyte will not conduct electricity. These solutes remain in their molecular form with no ions present in the solution.

How does a solution conduct electricity? First you need two electrodes and a source of electricity. Electrodes are electrical conductors (copper, zinc, carbon) partially immersed in a solution and connected to a source of electricity. The positive electrode is called the anode, while the negative electrode is the cathode. The ions within the solution serve as carriers of electricity. An anion is a negatively charged ion, whereas a cation is a positively charged ion.

How do I make a solution? First of all that depends on what the solute and solvent are. Will the solute dissolve in your solvent? If so, how much of it will dissolve? What affects the rate of dissolution (rate of dissolving)? • shaken or stirred • particles are made smaller • temperature is increased Why?

In order for a solvent to dissolve a solute, the solvent molecules must come in contact with the solute. Stirring moves “fresh” solvent molecules next to the solute. Solvent particles are brought into contact more with solute particles. Fresh molecules? ? ? Here is more theory to explain what is going on at the molecular level. Let’s use H 2 O for example. Water is known as the universal solvent, because many things will dissolve in it. Why?

Well, it’s polar. It has an unequal sharing of electrons in its bonds. This coupled with its bent shape create “poles” within the molecule. The oxygen atom takes on a partial negative charge, whereas the hydrogen atoms have a partial positive charge. d- This creates a polar molecule with dipole forces. H–O–H O H d+

The forces causing an ionic solid to dissolve in water are called ion -dipole forces; the attraction of water dipoles for cations and anions. The attractions of water dipoles for ions pull the ions out of the crystalline lattice and into the aqueous solution. The extent to which an ionic solid dissolves in water is determined largely by the competition between the inter-ionic attractions that hold ions in a crystal and the ion-dipole attractions that pull them into solution. The breaking part of compounds into the ions that created them is called dissociation.

The breaking part of compounds into the ions that created them is called dissociation. Na. Cl(s) dissociates into 1 Na+1(aq) and 1 Cl-(aq) Na. Cl(s) 1+ Na (aq) + Cl (aq) Ca. Cl 2(s) dissociates into 1 Ca 2+(aq) and 2 Cl-(aq) Ca. Cl 2(s) Ca 2+(aq) + 2 Cl-(aq) The process of dissolving a solute with a solvent is called solvation. The process of dissolving a solute with water as the solvent is called hydration.

Formula unit H bonds

So, agitating a solution will increase the rate of dissolution. Increasing the surface area of the solute will also rapidly increase the rate of dissolution. More surface area = more points of contact between the solute and solvent. Lastly, heating always increases the rate of dissolution of solids in liquids. Higher temperatures make the molecules of the solvent move around faster and contact the solute harder and more often.

How much can I dissolve? ? ? Increased temperature increases the rate at which a solid dissolves. The rate at which a substance dissolves does not alter the substances solubility. We can’t predict whether it will increase the amount of solid that dissolves. We must read it from a graph of experimental data. The solubility of a substance is the amount of that substance required to form a saturated solution with a specific amount of solvent at a specified temperature.

For solids in liquids, as the temperature rises, solubilities also tend to increase. About 95% of all ionic compounds have aqueous solubilities that increase significantly with increasing temperature. Most of the remainder have solubilities that change little with temperature. A very few have solubilities that decrease with increasing temperature For gases in a liquid, as the temperature rises, the solubilities decrease. Gas molecules can escape from the solution. For the gases in a liquid – as the pressure goes up, the solubility goes up.

Pressure effects? ? ? Changing the pressure doesn’t affect the amount of solid or liquid that dissolves. They are incompressible. Gases on the other hand, can be affected by pressure. The dissolved gas reaches an equilibrium with the gas above the liquid

Henry’s Law The effect of pressure on the solubility of a gas is known as Henry’s law. The gas is at equilibrium with the dissolved gas in this solution. The equilibrium is dynamic. If you increase the pressure the gas molecules dissolve faster. The equilibrium is disturbed. The system reaches a new equilibrium with more gas dissolved. This is an illustration of Henry’s Law.

Effervescence is the rapid escape of a gas from a liquid in which it is dissolved.

What does it mean if a solution is saturated? A saturated solution is one that contains the maximum amount of dissolved solute. It will not dissolve any more solute under the existing conditions.

An unsaturated solution is a solution that contains less solute than a saturated solution under the existing conditions. It will dissolve more solute (until the saturation point is reached) under the existing conditions.

A supersaturated solution in one that temporarily contains more dissolved solute than a saturated solution contains under the same conditions. A supersaturated solution is created when a warm, saturated solution is allowed to cool without the precipitation of the excess solute. The addition of more solute will disrupt the solution and cause it to crystallize.

Molarity Most chemical reactions take place in solution. Substances enter into chemical reactions according to certain molar ratios. We already know how to go from mass to moles, but volumes of solutions are more convenient to measure than their masses. So we need the concentration of the solutions. Concentration is a measure of the amount of solute dissolved in a certain amount of solvent.

There are several methods of expressing the concentration of a solution. Qualitatively solutions are termed either: concentrated or dilute A concentrated solution contains a relatively large amount of solute as compared to the solvent. A dilute solution contains a relatively small concentration of solute as compared to the solvent. These expressions are not very precise and can have many numerical equivalencies. To be more specific, chemists express concentrations quantitatively.

These expressions are not very precise and can have many numerical equivalencies. To be more specific, chemists express concentrations quantitatively. Molarity is the amount of solute (in moles) divided by the volume of solution (in liters). Molality is the amount of solute (in moles) divided by the mass of of solvent(in kilograms). Percent by mass is the amount of solute (in grams) multiplied by 100% and divided by the amount of solution (in grams) Percent by volume is the volume of solute (in liters) multiplied by 100% and divided by the volume of solution (in liters). Mass/Volume percent is the amount of solute (in grams) multiplied by 100% and divided by the volume of solution (in liters).

The most often used method in chemistry is MOLARITY (M). Molarity is defined as the moles of dissolved substance (solute) per volume of solution expressed in cubic decimeters (dm 3) and liters (L). M = Moles of solute Volume of solution in either dm 3 or L

As chemists we express concentration in terms of molarity because we measure solutions by volume. Yet, we need to count the number of particles of the solution. A 0. 372 M solution of Ba(NO 3)2 contains 0. 372 moles of Ba(NO 3)2 in 1 dm 3 of solution. One mole of sodium chloride, is 58. 44 grams. If 58. 44 grams of Na. Cl are dissolved in enough water (water is typically the solvent) to make 1 cubic decimeter of solution, the solution is a 1 M solution of Na. Cl. Similarly, if 2 moles of Na. Cl (116. 88 g) are dissolved in enough water to make 1 dm 3 of solution, the solution is a 2 M solution.

Fifty (50) m. L and 500 m. L of a 2 M solution will have the SAME CONCENTRATION. The total number of particles changes, but the concentration does not change. Recall: 1 m. L = 1 cm 3 1000 m. L = 1 L 1 dm 3 = 1000 cm 3

FIND THE MOLARITY (M): UNITS: moles/L or moles/dm 3 Given amount of solute moles of solute = Given amount of solution L or dm 3 of solution EXAMPLE: Given: 2. 50 x 102 cm 3 solution and 9. 46 g cesium bromide. Wanted: What is the molarity (M)? 9. 46 g Cs. Br 2. 50 x 102 cm 3 1 moles Cs. Br 1000 cm 3 solution 212. 81 g Cs. Br 1 dm 3 solution =

Example: Calculate the Molarity of a solution with 34. 6 g of Na. Cl dissolved in 125 m. L of solution. 34. 6 g Na. Cl 125 m. L 1 mole Na. Cl 1000 m. L solution 58. 44 g Na. Cl 1 L solution = 4. 74 mole/L Na. Cl = 4. 74 M Na. Cl =

FIND THE MASS OF SUBSTANCE NEEDED: The volume of solution is given, and the key is to use the MOLARITY as a conversion factor!!!! 6 M solution of Na. Cl means: 6 moles of Na. Cl in every 1 L of solution 6 moles of Na. Cl = 1 L of solution EXAMPLE: Make a 0. 133 M manganese (II) selenate solution with a final volume of 500. cm 3. What the. Molarity formula as foramanganese selenate? Useisthe Conversion(II) Factor!!!! Mn. Se. O 4

FIND THE MASS OF SUBSTANCE NEEDED: EXAMPLE: Make a 0. 133 M manganese (II) selenate solution with a final volume of 500. cm 3. Start the problem with the given (not the Molarity) 500. cm 3 solution = We want to What use are the Molarity units as fora. Molarity? conversion factor. 0. 133 M Mn. Se. O 4 = 0. 133 moles of Mn. Se. O 4 1 L of solution What should we convert the given into in order to use Molarity as a conversion factor?

FIND THE MASS OF SUBSTANCE NEEDED: EXAMPLE: 500. cm 3 solution Make a 0. 133 M manganese (II) selenate solution with a final volume of 500. cm 3. 1 dm 3 solution 0. 133 moles Mn. Se. O 4 1000 cm 3 solution 1 dm 3 solution 197. 90 g Mn. Se. O 4 1 mole Mn. Se. O 4 =

Example: How many grams of Ca. Cl 2 are needed to make 625 m. L of a 2. 0 M solution? 625 m. L solution 1 L solution 2. 0 moles Ca. Cl 2 110. 98 g Ca. Cl 2 1000 m. L solution 1 mole Ca. Cl 2 We will dissolve 140 g of calcium chloride in enough water to make 625 m. L of solution. =

Now let’s c if u can do it… Example: 10. 3 g of KCl are dissolved in a small amount of water then diluted to 250 m. L. What is the molarity? 10. 3 g KCl 1 mole KCl 1000 m. L solution 250 m. L 74. 55 g KCl 1 L solution = = 0. 55 M KCl Example: How many grams of glucose are needed to make 125 m. L of a 0. 50 M C 6 H 12 O 6 solution? 125 m. L solution 1 L solution . 50 moles C 6 H 12 O 6 180. 18 g C 6 H 12 O 6 1000 m. L solution 1 mole C 6 H 12 O 6 = = 11 g C 6 H 12 O 6

FIND THE VOLUME OF SOLUTION NEEDED: The mass of solute is given, and the key is to use the MOLARITY as a conversion factor!!!! EXAMPLE: To what volume should 5. 0 g of KCl be diluted in order to prepare a 0. 25 M solution? 5. 0 g KCl 1 mole KCl 1 L solution 74. 55 g KCl 0. 25 moles KCl = What should we convert the given into in order to use Molarity as a conversion factor?

Calculating Ion Concentrations In Solution Bracket symbols, [ ], are used to represent molar concentrations of ions or molecules in solution. Given 0. 010 M Na 2 SO 4 : • 2 Na 1+ ions are formed in solution; therefore the concentration of Na 1+ is 0. 020 M. • 1 SO 42 - ion is formed in solution; therefore the concentration of SO 42 - is 0. 010 M.

Example: What would the concentration be if you used 27 g of Ca. Cl 2 to make 500. m. L of solution? 27 g Ca. Cl 2 500. m. L 1 mole Ca. Cl 2 1000 m. L solution 110. 98 g Ca. Cl 2 1 L solution = = 0. 4865741575 M Ca. Cl 2 = 0. 49 M Ca. Cl 2 What is the concentration of each ion? 27 g Ca. Cl 2 1 mole Ca+2 1000 m. L 500. m. L 110. 98 g Ca. Cl 2 1 mole Ca. Cl 2 1 L = = 0. 49 M [Ca 2+] 27 g Ca. Cl 2 1 mole Ca. Cl 2 2 mole Cl-1 1000 m. L 500. m. L 110. 98 g Ca. Cl 2 1 mole Ca. Cl 2 1 L = = 0. 97 M [Cl 1 -]

Example: Calculate the concentration of a solution made by dissolving 45. 6 g of Fe 2(SO 4)3 to 475 m. L. What is the concentration of each ion? 45. 6 g Fe 2(SO 4)3 475 m. L 1 mole Fe 2(SO 4)3 1000 m. L solution 399. 88 g Fe 2(SO 4)3 1 L solution = 0. 240 M Fe 2(SO 4)3 45. 6 g Fe 2(SO 4)3 475 m. L 1 mole Fe 2(SO 4)3 2 mole Fe+3 399. 88 g Fe 2(SO 4)3 1 mole Fe 2(SO 4)3 1000 m. L 1 L = 0. 480 M [Fe+3] 3 mole Fe+3 399. 88 g Fe 2(SO 4)3 1 mole Fe 2(SO 4)3 1000 m. L 1 L = 0. 720 M [SO 4 -2]

Molarity by Dilution Acids are usually acquired from the chemical suppliers in a highly concentrated solution. A solution for which a precise concentration is known is called a standard solution. The following are standard solutions of concentrated acids. HNO 3 H 2 SO 4 15. 8 M 18 M

A lab that we are going to do requires a more dilute (less concentrated) solution. How can we make a solution with a lower Molarity? How much water? M 1 V 1 = M 2 V 2 M 1= initial Molarity V 1= initial volume M 2= final Molarity V 2= final volume

EXAMPLE: Our next lab requires 1. 0 M HCl. We need 3. 0 L of HCl for the entire class. How much concentrated HCl is needed to prepare the solution? (Assume HCl is 12 M concentrated. ) M 1 V 1 = M 2 V 2 M 1= 12 M HCl V 1= ? ? ? M 2= 1. 0 M HCl V 2= 3. 0 L M 1 V 1 = M 2 V 1 = 1. 0 M HCl 12 M HCl V 1 = M 2 V 2 M 1 3. 0 L

EXAMPLE: How much H 2 O should 250. m. L of 12. 0 M HCl be added in order to produce a 1. 00 M solution? M 1 V 1 = M 2 V 2 M 1= 12. 0 M HCl V 1= 250. m. L M 1 V 1 = M 2 V 2 12. 0 M HCl V 2 = M 2= 1. 00 M HCl V 2= ? ? ? V 2 = M 1 V 1 M 2 250. m. L 1. 00 M HCl 300 NOT

EXAMPLE: How much H 2 O should 250. m. L of 12. 0 M HCl be added in order to produce a 1. 0 M solution? NOT 300 What does V 2 represent? Are we looking for V 2? How do we find the 300 amount of water that was added? VH = O 2 2750 m. L H 2 O

Arrhenius Concept of Acids and Bases Arrhenius acid – a substance that, when dissolved in water, increases the concentration of hydronium ion, H 3 O+(aq). • Chemists often use the notation H+(aq) for the H 3 O+(aq) ion, and call it the hydrogen ion. • Remember, however, that the aqueous hydrogen ion is actually chemically bonded to water, that is, H 3 O+. Arrhenius base – a substance that, when dissolved in water, increases the concentration of hydroxide ion, OH-(aq). – The Arrhenius concept limits bases to compounds that contain a hydroxide ion. Copyright © Cengage Learning. All rights reserved. 15 | 50

Solutions of Strong Acid or Base By dissolving substances in water, you can alter the concentrations of H+(aq) and OH-(aq). – In a neutral solution, as in pure water, [H+] = [OH-]. – In an acidic solution, [H+] > [OH-]. – In a basic solution, [H+] < [OH-]. Copyright © Cengage Learning. All rights reserved. 15 | 51

At 25°C, you observe the following conditions. Solutions can be characterized as: Acidic: Neutral: Basic: [H 3 O+] > 1. 0 × 10 -7 M [H 3 O+] = 1. 0 × 10 -7 M [H 3 O+] < 1. 0 × 10 -7 M Copyright © Cengage Learning. All rights reserved. 15 | 52

A has 5 H 3 O+ and 5 OH-. It is neutral. B has 7 H 3 O+ and 3 OH-. It is acidic. C has 3 H 3 O+ and 7 OH-. It is basic. Listed from most acidic to most basic: B, A, C. Copyright © Cengage Learning. All rights reserved. 15 | 53

Although you can quantitatively describe the acidity of a solution by its [H+], it is often more convenient to give acidity in terms of p. H. The p. H of a Solution ―the negative logarithm of the molar hydrogen-ion concentration. p. H = –log[H 3 O+] ―For a log, only the decimal part of the number has significant figures. The whole number part, called the characteristic, is not significant. Copyright © Cengage Learning. All rights reserved. 15 | 54

Calculate the p. H of typical adult blood, which has a hydronium ion concentration of 4. 0 × 10 -8 M. [H 3 O+] = 4. 0 × 10 -8 M p. H = –log [H 3 O+] p. H = – log (4. 0 × 10 -8) = – (– 7. 40) p. H = 7. 40 The two significant figures are the two decimal places. Copyright © Cengage Learning. All rights reserved. 15 | 55

The p. H of a Solution In a neutral solution, whose hydrogen-ion concentration is 1. 0 x 10 -7, the p. H = 7. 00. For acidic solutions, the hydrogen-ion concentration is greater than 1. 0 x 10 -7, so the p. H is less than 7. 00. Similarly, a basic solution has a p. H greater than 7. 00. Copyright © Cengage Learning. All rights reserved. 15 | 56

The p. H Scale Copyright © Cengage Learning. All rights reserved. 15 | 57

A Problem to Consider A sample of orange juice has a hydrogen-ion concentration of

A Problem to Consider The p. H of human arterial blood is 7. 40.

The p. H of a Solution A measurement of the hydroxide ion concentration, similar to p. H, is the p. OH. The p. OH of a solution is defined as the negative logarithm of the molar hydroxide-ion concentration. Then because Kw = [H+][OH-] = 1. 0 x 10 -14 at 25°C, you can show that Copyright © Cengage Learning. All rights reserved. 15 | 60

A Problem to Consider An ammonia solution has a hydroxide-ion concentration of 1. 9 x 10 -3 M. What is the p. H of the solution? You first calculate the p. OH: Then the p. H is: Copyright © Cengage Learning. All rights reserved. 15 | 61

[H 3 O+] = 10 -p. H [OH-] = 10 -p. OH The p. H of natural rain in 5. 60. What is its hydronium ion concentration? p. H = 5. 60 [H 3 O+] = 10 -p. H = 10 -5. 60 [H 3 O+] = 2. 5 × 10 -6 M Because the p. H has two decimal places, the concentration can have only two significant figures. Copyright © Cengage Learning. All rights reserved. 15 | 62

[H 3 + O] p. H Copyright © Cengage Learning. All rights reserved. [OH

The p. H of a Solution The p. H of a solution can accurately be measured using a p. H meter. • Although less precise, acid-base indicators are often used to measure p. H because they usually change color within a narrow p. H range. Copyright © Cengage Learning. All rights reserved. 15 | 64