Solutions

= = Key Terms Solution: The process by which a gas, liquid, or solid is dispersed homogeneously in a gas, liquid, or solid without chemical change. Solute: The substance dissolved in a given solution. Solvent: A substance that dissolves another to form a solution. Saturated: Containing the maximum amount of solute capable of being dissolved under given conditions. Unsaturated: Not saturated; having the power to dissolve still more of a substance. Supersaturated: To increase the concentration of (a solution) beyond saturation; saturate abnormally. Concentration: A measure of the amount of dissolved substance contained per unit of volume. Molarity: The number of moles of solute per liter of solution. Molality: The number of moles of solute per kilogram of solvent. Dilution: The act of making a liquid thinner or weaker by the addition =  = =Colligative Properties= The Physical properties of solutions, dependent on the amounts of solute and solvent in the solution, not their chemical identities.

Henry's Law
Henry's Law deals with the effect pressure has on the solubility of gasses. It states that the solubility of a liquid is directly proportional to the partial pressure of the gas on the solution. An example of this is a carbonated drink; the CO2 in the drink dissolves, because it is bottled under pressure. When it is opened, the CO2 fizzes out, and the drink returns back to atmospheric pressure. If left out, the drink will go flat from absence of CO2. Henry's Law is expressed in the following ways:

Cgas = (Kg)(Pg) where Cgas is the concentration of the gas, Kg is the given constant for the gas, and Pg is the partial pressure of the gas above the solution. However, a more useful version of Henry's Law is:

C1/P1 = C2/P2 where the constant Kg is left out, since it is the same for both. Using this equation, one can find any of these variables, if given initial amounts.


 * Example:** At about 20 degrees C the solubility of LiNo3 in water is .149 g/Liter when the partial pressure of Lithium Nitrate is 460 torr. What will the solubility of LiNo3 in water at 20 degrees C be when the partial pressure is 590 torr?

.000324 * 590 = **.19116 g/L = C2** **The solubility of LiNO3 under higher pressure is .19116g/Liter**
 * Given:** C1 = .149g/L P1 = 460 torr P2 = 590 torr
 * Find:** C2
 * Solution:** .149 (C1) / 460 (P1) = C2 / 590 (P2)

**Raoult's Law**
Raoult's Law deals with the lowering of a solution's vapor pressure with a nonvolatile solution. Since all liquid solutions have a lower vapor presure than their pure solvents, how much the vapor pressure is lowerd relies on whether the solute is non-ionizable or ionic. For non-ionizable solutions, they will dissolve in water, but won't ionize since they have very strong covalent bonds. The equation **Psolution = Xsolvent * Pinitial solvent** can be be used to compare the vapor pressure to the mole fraction of the solvent. For volatile solutions (when two or more components can evaporate) we can use Dalton's Law and create two equations: **PA = (XA)(PinitialA)** and **PB = (XA)(PinitialB)** which, when the partial pressure values PA and PB are found can be put into the equation: **PA + PB = Ptotal**


 * Example 1:** If 0.280 mol of a nonvolatile nonelectrolyte are dissolved in 3.30 mol of water, what is the vapor pressure of the resulting solution? The vapor pressure of pure water is 23.8 torr at 25 C.

.922 * 23.8 = **21.9 torr The vapor pressure will be less than the initial amount.
 * Given:** .280 mol nonvolatile nonelecytolyte, 3.3 mol H2O, Pinitial solvent = 23.8 torr
 * Find:** Xsolvent (mol fraction),
 * Solution:** Xsolvent = Moles A / Total Moles = Xsolvent = 3.3 / (3.3 + .28) **Xsolvent = .922**

Example 2:** A solution is composed of 1.20 mol cyclohexane PA=97.6 torr and 2.50 mol acetone PB=229.5 torr. What is the total vapor pressure P total above this solution?

(PA) + (PB)
 * Given:** PinitialA = 97.6 torr, PinitialB = 229.5 torr, 1.2 mol cyclohexane, 2.5 mol acetone
 * Find:** XA, XB, P total
 * Solution:** P total = (PinitialA * XA) + (PinitialB * XB)

XA = Moles A / Total moles = (1.20) / (1.20 + 2.50) = **0.324** **mol** XB= 1- Moles A = 1 - 0.324 = **0.676** **mol**

P total = (97.6 torr)(0.324) + (229.5 torr)(0.676) = **187 torr**

Boiling Point Elevation
The boiling point of a substance increases when the only volatile component of the solution is the solvent. Since the only volatile component is the solvent, the vapor pressure of the solution is less than the the solvent's vapor pressure at any temperature. When the solution is raised to the regular boiling point of the pure solvent, it does not boil, because the vapor pressure of the solution is not equal to the atmospheric pressure. Therefore, when a nonvolatile solute is placed in a solvent, the boiling point is elevated. An equation can be used in order to calculate boiling point elevation. **The change in temperature (number of degrees the boiling point is raised) is directly proportional to its molal concentration.** This is modeled in the equation: **Change in T = (Kb)(m)** where Kb is a given boiling point constant and m is the molality (mol/kg).


 * Example:** At what temperature will a .25 m solution of sugar in water boil when the pressure is 1 atm? Kb for water is .51.


 * Given:** m = .25, Kb = .51, Normal Boiling Point of H2O = 100 degrees C
 * Find:** Change in T, New Boiling Point
 * Solution:** Change in T = .51 * .25 = **.1275 degrees C** 100 degrees C + .1275 degrees C = **100.1275 degrees C New Boiling Point for .25m solution**

Freezing Point Depression
Solutions will freeze at lower points than their sure solvents because when a pure solvent freezes, pure crystals start to form. As less and less energy is put into the system, the crystals will form solids with other crystals. However, in solutions, the solvent crystals get in the way of the solute crystals, forcing the system to compensate by requiring it less kinetic energy in order to freeze. This is known as freezing point depression. **The number of degrees a solution's freezing point is depressed is proportional to the molality.** Knowing this, this equation can be formed: **Change in T (# of degrees the freezing point is depressed by) = (Kf)(m)** where Kf is a freezing point depression constant, different for every solvent, and m is the molality.

5.45 - 2.535 = 2.915 = the new freezing point of Benzene**
 * Example:** At what temperature will .5m solution of sugar in Benzene freeze?
 * Given:** Kf = 5.07 (given), m = .5, Initial freezing point of Benzene = 5.45 (given)
 * Find:** Change in T
 * Solution:** 5.07 * .5 = **2.535 degrees C** **The freezing point of benzene (initially 5.45) is lowered by 2.535.

=Solubility = = =  Solubility is the property of a solid, liquid, or gaseous chemical substance called solute to dissolve in a liquid solvent to form a homogeneous solution. The solubility of a substance strongly depends on the used solvent as well as on temperature and pressure. The extent of the solubility of a substance in a specific solvent is measured as the saturation concentration where adding more solute does not increase the concentration of the solution. The solvent is generally a liquid, which can be a pure substance or a mixture. The extent of solubility ranges widely, from infinitely soluble or fully miscible such as ethanol in water, to poorly soluble, such as silver chloride in water. The term insoluble is often applied to poorly or very poorly soluble compounds. = Factors Affecting Solubility =

   

Temperature
Generally in many cases solubility increases with the rise in temperature and decreases with the fall of temperature but it is not necessary in all cases. However we must follow two behaviors: In endothermic process solubility increases with the increase in temperature and vice versa. For example: solubility of potassium nitrate increases with the increase in temperature. In exothermic processes solubility decreases with the increase of temperature. For example: solubility of calcium oxide decreases with the increase in temperature. Gases are more soluble in cold solvent than in hot solvent.

**Pressure **
The effect of pressure is observed only in the case of gases. An increase in pressure increases the solubility of a gas in a liquid. For example carbon dioxide is filled in cold drink bottles (such as Coca Cola, Pepsi, 7up etc.) under pressure.

The Nature of the Solute and the Solvent
Solubility of a solute in a solvent purely depends on the nature of both solute and solvent. A polar solute dissolved in polar solvent. Solubility of a non-polar solute in a solvent is large. A polar solute has low solubility or insoluble in a non-polar solvent.

=<span style="font-family: Arial,Helvetica,sans-serif;"> Solubility Rules =

<span style="font-family: Arial,Helvetica,sans-serif;"> The Solubility Rules are a set of rules that define which salts are soluble in water and which are not. =<span style="font-family: Arial,Helvetica,sans-serif;"> = =<span style="font-family: Arial,Helvetica,sans-serif;"> = > Thus, AgCl, PbBr2, and Hg2Cl2 are all insoluble. <span style="font-family: Arial,Helvetica,sans-serif;"> =<span style="font-family: Arial,Helvetica,sans-serif;">Polarity =
 * 1) <span style="font-family: Arial,Helvetica,sans-serif;"><span style="font-family: Arial,Helvetica,sans-serif;"> Salts containing Group I elements are soluble (Li+, Na+, K+, Cs+, Rb+). Exceptions to this rule are rare. Salts containing the ammonium<span style="font-family: Arial,Helvetica,sans-serif;"> ion (NH4+) are also soluble.
 * 2) <span style="font-family: Arial,Helvetica,sans-serif;">Salts containing nitrate ion (NO3-) are generally soluble.
 * 3) <span style="font-family: Arial,Helvetica,sans-serif;">Salts containing Cl -, Br -, I - are generally soluble. Important exceptions to this rule are halide salts of Ag+, Pb2+, and (Hg2)2+.
 * 1) <span style="font-family: Arial,Helvetica,sans-serif;">Most silver salts are insoluble. AgNO3 and Ag(C2H3O2) are common soluble salts of silver; virtually anything else is insoluble.
 * 2) <span style="font-family: Arial,Helvetica,sans-serif;">Most sulfate salts are soluble. Important exceptions to this rule include BaSO4, PbSO4, Ag2SO4 and SrSO4.
 * 3) <span style="font-family: Arial,Helvetica,sans-serif;">Most hydroxide salts are only slightly soluble. Hydroxide salts of Group I elements are soluble. Hydroxide salts of Group II elements (Ca, Sr, and Ba) are slightly soluble. Hydroxide salts of transition metals and Al3+ are insoluble. Thus, Fe(OH)3, Al(OH)3, Co(OH)2 are not soluble.
 * 4) <span style="font-family: Arial,Helvetica,sans-serif;">Most sulfides of transition metals are highly insoluble. Thus, CdS, FeS, ZnS, Ag2S are all insoluble. Arsenic, antimony, bismuth, and lead sulfides are also insoluble.
 * 5) <span style="font-family: Arial,Helvetica,sans-serif;">Carbonates are frequently insoluble. Group II carbonates (Ca, Sr, and Ba) are insoluble. Some other insoluble carbonates include FeCO3 and PbCO3.
 * 6) <span style="font-family: Arial,Helvetica,sans-serif;">Chromates are frequently insoluble. Examples: PbCrO4, BaCrO4
 * 7) <span style="font-family: Arial,Helvetica,sans-serif;">Phosphates are frequently insoluble. Examples: Ca3(PO4)2, Ag3PO4
 * 8) <span style="font-family: Arial,Helvetica,sans-serif;">Fluorides are frequently insoluble. Examples: BaF2, MgF2 PbF2.

=<span style="font-family: Arial,Helvetica,sans-serif;"> =

<span style="font-family: Arial,Helvetica,sans-serif;">
<span style="font-family: Arial,Helvetica,sans-serif;">Molecules do not always share electrons equally in covalent bonding. The shapes of molecules differ depending on the electronegativity of each atom. If Polar molecules can bond together due to dipole–dipole intermolecular forces between one molecule (or part of a large molecule) with asymmetrical charge distribution and another molecule also with asymmetrical charge distribution. Molecular polarity is dependent on the difference in electronegativity between atoms in a compound and the asymmetry of the compound's structure. For example, a molecule of water is polar because of the unequal sharing of its electrons in a "bent" structure, whereas methane is considered non-polar because the carbon shares the electrons with the hydrogen atoms uniformly. Polarity underlies a number of physical properties including surface tension, solubility, and melting- and boiling-points. Electrons are not always shared equally between two bonding atoms: one atom might exert more of a force on the electron cloud than the other. This "pull" is termed electronegativity and measures the attraction for electrons a particular atom has. The unequal sharing of electrons within a bond leads to the formation of an electric dipole: a separation of positive and negative electric charge. Atoms with high electronegativities — such as fluorine, oxygen, and nitrogen — exert a greater pull on electrons than atoms with lower electronegativities. In a bonding situation this can lead to unequal sharing of electrons between atoms, as electrons will spend more time closer to the atom with the higher electronegativity. Bonds can fall between one of two extremes — being completely non-polar or completely polar. A completely non-polar bond occurs when the electronegativities are identical and therefore possess a difference of zero. A completely polar bond is more correctly termed ionic bonding and occurs when the difference between electronegativities is large enough that one atom takes an electron from the other. The terms "polar" and "non-polar" bonds usually refer to covalent bonds. III.Dilution of Solutions Calculations (M1V1=M2V2) Key Concepts The concentration of a solution is usually given in moles per liter (mol L-1 OR mol/L). This is also known as molarity. Concentration, or Molarity, is given the symbol M. A short way to write that the concentration of a solution of hydrochloric acid is 0.01 mol/L is to write [HCl]=0.01M the square brackets around the substance indicate concentration. The solute is the substance which dissolves. The solvent is the liquid which does the dissolving. A solution is prepared by dissolving a solute in a solvent. When a solution is diluted, more solvent is added to it. Since M = n ÷ V, and n (the moles of solute) is the same for the original solution and the new diluted solution, it follows that M1V1 = M2V2 where M1=original concentration of solution V1=original volume of solution M2=new concentration of solution after dilution V2=new volume of solution after dilution To calculate the new concentration (M2) of a solution given its new volume (V2) and its original concentration (M1) and original volume (V1): M2 = (M1 x V1) ÷ V2 To calculate the new volume (V2) of a solution given its new concentration (M2) and its original concentration (M1) and original volume (V1): V2 = (M1 x V1) ÷ M2 Examples 1. M2=(M1V1) ÷ V2 Calculate the new concentration (molarity) if enough water is added to 100mL of 0.25M sodium chloride to make up 1.5L. M2=(M1V1) ÷ V2 M1 = 0.25M V1 = 100mL = 100 ÷ 1000 = 0.100L (volume must be in liters) V2 = 1.5L [NaCl(aq)]new = M2 = (0.25 x 0.100) ÷ 1.5 = 0.017M (or 0.0.017 mol/L or 0.0.17mol L-1) 2. V2=(M1V1) ÷ M2 Calculate the volume to which 500mL of 0.02M copper sulfate solution must be diluted to make a new concentration of 0.001M. V2=(M1V1) ÷ M2 M1 = 0.02M V1 = 500mL = 500 ÷ 1000 = 500 x 10-3L = 0.500L (since there are 1000mL in 1L) M2 = 0.001M V(CuSO4)new = V2 = (0.02 x 0.500) ÷ 0.001 = 10.00L Solution Problems Using Stoichiometry

Stoichiometry tables can be used in analysis of disassociation reactions, in order to determine the various amounts of material in both the reactants and the products. Here are some Stoichiometric calculations in regard to solutions.

1. How many grams of S-2 ion are present in a .72 molal (NH4)2S solution if there is 663ml of H20?

(NH4)2S → 2NH4+1 + S-2

step 1: determine number of moles (NH4)2S Known: molality: .72 *molality = moles solute/kg solvent also known: mass solvent = 663g -- .663 kg H2O need to find: moles (NH4)2S

.72 = moles (NH4)2S / .663 moles (NH4)2S = .47736 moles

step 2: use stoichiometry table to determine mass


 * || (NH4)2S || → || 2NH4+1 || S-2 ||
 * mass ||  ||   ||   || 15.307g--answer ||
 * molar mass ||  ||   ||   || 32.066g/mol ||
 * moles || .47736mol ||  ||   || .47736 mol ||

2. An unknown amount of CaCl2 is dissolved in water. After the compound has disassociated, 11.8g of the Cl- around found to be dissolved in the solution. What was the original mass of the CaCl2 added to the water?

CaCl2 → Ca+2 + 2Cl-

step 1: determine the number of moles of Cl-. To do this, we'll use the equation moles = mass / molar mass 11.8g ÷ 35.4527g/mol = .332838 moles

step 2: determine mole ratio between the Cl- and the CaCl2 -- it's 2:1 step 3: use a stoichiometry table
 * || CaCl2 || → || Ca+2 || 2Cl- ||
 * mass || 18.4697g--answer ||  ||   || 11.8g ||
 * molar mass || 110.9834g/mol ||  ||   || 35.4527g/mol ||
 * moles || .166419mol ||  ||   || .332838 mol ||

3. In 567 ml .42 M Al2(SO4)3 solution, what is the mass of the SO4-2 dissolved? the mass of the Al+3 dissolved?

Al2(SO4)3 → 2 Al+3 + 3(SO4)-2

step 1: determine number of moles Al2(SO4)3 molarity = moles/L .42 = moles Al2(SO4)3 / .567L moles Al2(SO4)3 = .23814

step 2: step up a stoichiometry table and use it mass
 * || Al2(SO4)3 || → || 2 Al+3 || 3(SO4)-2 ||
 * ||  || 12.85g--answer || 68.63147g --answer ||
 * molar mass ||  ||   || 26.98g/mol || 96.066g/mol ||
 * moles || .23814mol ||  || .47628mol || .71442mol ||

4. 10g of an unknown solute disassociate in 270 ml of water. The molar concentration of the free polyatomic ion produced in the reaction is .6 M. Find the molar mass of the original compound.

A2(BC)3 → 2A+3 + 3BC-2

step 1 : find number of moles of BC-2 molarity = moles solute/ L solution .6 = moles BC-2 / .270 moles BC-2 = .162 step 2: use a stoichiometry table
 * || A2(BC)3 || → || 2A+3 || 3BC-2 ||
 * mass || 10g ||  ||   ||   ||
 * molar mass || 185.18519g/mol --answer ||  ||   ||   ||
 * moles || .054 mol ||  ||   || .162 mol ||

Problems Dealing With the Titration of Solutions A titration is the neutralization of an acid or base by the addition of a solution of vast different pH. For example, an HCl solution, a heavily acidic solution, is titrated, or neutralized, by the addition of an NaOH solution (a basic solution). 1. A 100ml solution of Potassium Hydroxide is titrated with 1.2 M 175ml solution of Sulfuric acid. What is the molar concentration of the Potassium Hydroxide solution?
 * these are very similar to the solutions stoichiometry problems

2KOH + H2SO4 → K2SO4 + 2H20

step 1: calculate number of moles H2SO4 1.2 = moles H2SO4 / .175

moles H2SO4 = .21 step 2: we can use a small stoichiometry table to figure the rest... .42 moles / .1 L = 4.2 M--answer
 * || 2KOH || H2SO4 || → ||
 * mass ||  ||   ||   ||
 * molar mass ||  ||   ||   ||
 * moles || .42mol || .21mol ||  ||

2. An unknown amount of Sodium Hydroxide is titrated with 53ml of 4.5 M Phosphoric acid solution. How many grams of Sodium Hydroxide were used originally?

3NaOH + H3PO4 → Na3PO4 + 3H20

step 1: find the number of moles of H3PO4 4.5 = moles H3PO4 / .053L moles H3PO4 = .2385 step 2: we can use a small stoichiometry table to figure the rest...
 * || 3NaOH || H3PO4 || → ||
 * mass || 28.612679g-answer ||  ||   ||
 * molar mass || 39.99 ||  ||   ||
 * moles || .7155 || .2385 ||  ||