Solubility Flashcards

Question

Mass percentage

Answer 1

Ratio of the mass of the solute total mass of solution multiplied by 100 Mass % = ( mass of solute / total mass of solution) x 100%

Answer 2

10^6 multipled by the ratio of the mass of solute to the total mass of the solution Ppm = (mass of solute / total mass of solution) x 10^6 NOT the number of solute molecules per million molecules

Answer 3

Molarity, molality, mole fraction, mass percentage, parts per million, Normality Always given in terms of the form of solute before dissolution E.g. 1 mol of NaCl added to 1 L of water, resulting solution is ~ 1 molar, not 2 molar even though one mole of NaCl dissociates into two moles of ions

Answer 4

Number of equivalents per liter of solution Equivalent: depends on type of reaction taking place in solution - Acid-base reaction: equivalent is defined as mass of acid or base that can donate or accept one mole of protons - 1 molar solution of H2SO4 is called a 2 normal solution because it can donate 2 protons for each H2SO4 molecule

Answer 5

Physical reaction involving three steps: 1. Breaking of intermolecular bonds between solute molecules 2. Breaking of intermolecular bonds between solvent molecules 3. Formation of intermolecular bonds between solvent and solute molecules

Answer 6

Enthalpy change is approximately equal to internal energy change \delta H ~= \delta U Heat of solution (\delta H_sol) = \delta H1 + \delta H2 + \delta H3 First two steps are endothermic and third step is exothermic Hesse’s Law: overall enthalpy change of a reaction can be exothermic even if the reaction contains endothermic steps

Answer 7

More stable if the reaction was exothermic (released heat) new intermolecular bonds are more stable than original ones and new intermolecular attractions within the solution stronger than intermolecular attractions within pure substances

Answer 8

Combining steps 2 and 3 of the formation of a solution, which is the breaking of intermolecular bonds between solvent molecules and the formation of intermolecular bonds between solvent and solute molecules

Answer 9

Usually involves an increase in entropy of system | One exception is when a gas dissolves in a liquid or solid (entropy change will be negative)

Answer 10

Adding heat gives extra energy to break intermolecular bonds between solute-solute and solvent-solvent to form new bonds Without input of heat, \delta G would remain positive and dissolution would not happen at any appreciable level

Answer 11

Gibbs free energy equation is: \delta G = \delta Hsol - T \delta S Entropy is usually positive for dissolution, but \delta Hsol can be positive or negative If \delta Hsol is negative, then will proceed spontaneously If \delta Hsol is positive, then temperature dependent

Answer 12

Certain amounts of liquid molecules escape the liquid intermolecular bonds with enough kinetic energy and go into the gas phase So there are some gas molecules above the liquid As the space above the liquid fills with molecules, some of the molecules crash back into the liquid Equilibrium reached when rate at which molecules are leaving the liquid equal rate at which molecules are re-entering liquid

Answer 13

Partial Pressure of the compound that is required to create equilibrium of liquid molecules going into gas and gas molecules going into liquid - created by molecules in the open space at equilibrium Related to kinetic energy of the molecules, so a function of temperature

Answer 14

Partial pressure of water vapor decreases when the wind blows (velocity increases, so pressure decreases) Pressure of water does not change, since it is fixed at a given temperature Since partial pressure of water vapor now lower than pressure of water (they were in equilibrium), number of water molecules leaving the puddle is greater than number of molecules entering the puddle Water is evaporating from puddle

Answer 15

When vapor pressure of liquid equals local atmospheric pressure, liquid boils A liquid can be brought to a boil by raising the temperature, thereby raising the vapor pressure until it reaches the atmospheric pressure OR by lowering the atmospheric pressure until it equals the vapor pressure

Answer 16

Solids do have a vapor pressure, usually very low Melting point is temperature at which the vapor pressure of solid is equal to vapor pressure of liquid phase of that substance Above the melting point, the vapor pressure of the liquid is _ than that of the solid, so solid moves into liquid

Answer 17

Solute with no vapor pressure If added to a liquid, some solute molecules reach the surface of the solution and reduce the surface area available for liquid molecules Since solute molecules do not break free form solution, but take up surface area, number of molecules breaking free from liquid decreases while surface area of solution and volume of open space above the solution remain the same - Means that vapor pressure of solution is lower than vapor pressure of pure solvent

Answer 18

Pv = \chi_a Pa | Vapor pressure of solution Pv is proportional to vapor pressure of pure liquid Pa and the mole fraction of liquid \chi_a

Answer 19

A solute with a vapor pressure Will also compete for the surface area of a liquid, however some of the molecules of a volatile solute will escape form the solution and contribute to vapor pressure For ideal solution, in which the solute and solvent have similar properties, partial pressures contributed by solvent and solute can be found by applying Raoult’s law separately: Pv = \chi_a Pa + \chi_b Pb

Answer 20

Can predict direction of deviation based on heats of solution If heat of solution is negative, stronger bonds are formed and fewer molecules are able to break free from surface, so negative deviation of vapor pressure from Raoult’s law Opposite for positive heat of solution (endothermic)

Answer 21

Quantifies a solute’s tendency to dissolve in a solvent Maximum number of moles of a solute that can dissolve in a solution (depends on temperature and the ions in solution) Commonly measured in mol/L If a salt, dissolution is reversible and at first the rate of reverse reaction, precipitation, is lower than rate of dissolution As conc. Of dissolved salt increases, rates of dissolution and precipitation equilibrate (solution is said to be saturated)

Answer 22

Ksp where solids and pure liquids are left out of the solubility product expression Computed the same way as other equilibrium constant, K Concentration of products over concentration of reactants, each raised to the power of their coefficients in the balanced equation Fixed for a given temperature, independent of ion concentrations Measure of how inclined a compound is to dissolve

Answer 23

Solubility: # moles of solute / L of solution that can be dissolved in a given solvent For dissolution of BaF2(s) into Ba^2+(aq) + 2F^-(aq), Ksp = [Ba^2+][F^-]^2 At 25 deg C, Ksp = 2.4 x 10^-5 Set x = # moles / L OR solubility 2.4 x 10^-5 = (x)(2x)^2 from Ksp equation Where x ~= 1.8 x 10^-2 mol / L for 1 L of water at 25 deg C

Answer 24

Disturbance of the equilibrium of solubility of an ion that is the same as one that is already in solution E.g. adding NaF to a solution that already contains BaF2 dissolved in equilibrium Pushes equilibrium in direction that will reduce the concentration of that ion Spectator Ion: Na+ would be a spectator, since it does not affect the equilibrium

Answer 25

If Ksp = 2.4 x 10^-5 for 25 deg C adn we are at 25 deg C: 2.4 x 10^-5 = (x)(2x + 1)^2 Can simplify math, because since x will be very small, can say that 2x + 1 ~= 1 2.4 x 10^-5 = (x)(1) -> x = 2.4 x 10^-5 Double check if 2x is much less than 1 4.8 x 10^-5 << 1 A common ion added to a solution that is not saturated does NOT shift the equilibrium, because in unsaturated solution there is no equilibrium to shift

Answer 26

(Cl-, Br-, I-) are soluble | Mercury, lead, and silver compounds are not soluble (Hg2^2+, Pb^2+, Ag^+)

Answer 27

Sulfate: SO4^2-, most are soluble Except for those containing mercury, lead, and heavier alkaline earth metals (Hg2^2+, Pb^2+, Ca^2+, Sr^2+, Ba^2+) -> not soluble

Answer 28

Ca^2+, Sr^2+, Ba^2+ are soluble when paired with sulfides (S^2-) and hydroxides (OH^-)

Answer 29

CO3^2-, PO4^3-, S^2-, OH^- are generally insoluble other than OH- and S^2- with heavier alkaline metals

Answer 30

If solubility less than 0.01 mol / L, than generally said to be insoluble Ions with single charges are generally soluble, and ions with double charges on one ion are less soluble Smaller ions or molecules such as those in Group 1 are soluble Large, heavy cations are less soluble Higher charges means the molecule might have stronger intramolecular ionic bonds, so individual ions are harder to pull apart and dissolve

Answer 31

C = ka1 Pv C is solubility of gas a (mol / L), ka1 is Henry’s law constant (unique to each solute-solvent pair), Pv is vapor partial pressure of gas a Concentration of gas in solution is proportional to vapor partial pressure of gas above solution Pressure and temperature can affect solubility (pressure increases solubility of gas) Law can also be written as: Pv = \chi_a ka2, where \chi_a is mole fraction of a in solution, and ka2 is Henry’s law constant

Answer 32

In the second formulation of Henry’s Law: Pv = \chi_a ka2 And Raoult’s Law: Pv = \chi_a Pa But, ka2 does NOT equal Pa, so they are in conflict Both are approximations and are more accurate in certain contexts Raoult’s law is more accurate when it is applied to vapor partial pressure of a solvent with high concentration Henry’s law is more accurate when applied to vapor partial pressure of a volatile solute where the solute has a low concentration In an ideally dilute solution, the solvent obeys Raoult’s law and the solute obeys Henry’s law When solvent conc is high, each solvent molecule is surrounded by other solvent molecules, so behaves more like pure solvent, more similar to vapor pressure of pure liquid (Raoult’s)

Answer 33

In an exothermic dissolution reaction, heat will be produced Le Chatelier’s principle tells us that more heat should push the reaction to the left, however it does not The dissolution is accompanied by a large increase in entropy which in \delta G = \delta H - T \delta S means that increased temperature can increase spontaneity of dissolution Due to large increase in entropy, water solubility of many solids increases with increasing temperature regardless of the enthalpy change (must be found by experiment to be absolutely certain) Not the same when a gas dissolves in a liquid, bc entropy change is negative for this dissolution. So as temperature increases, gas solubility generally decreases

Answer 34

Double effect 1. Hot water holds less oxygen than cold water (gas solubility decreases with increasing temperature) 2. Hot water floats on the cold water, sealing it off from oxygen in the air above

Answer 35

For example, if we had: Ba(OH)2 (s) -> Ba^2+(aq) + 2OH^- (aq) If we add base to the solution, it will force reaction to the left (Le Chatelier) If we add acid, then we have H^+ (aq) + OH-(aq) -> H2O(l) which removes the hydroxide ions from the solution, and more of the solid compound would be dissolved to maintain equilibrium

Answer 36

Increased temperature of solvent decreases solubility Heavier, larger gases experience greater Van der Waals forces and tend to be more soluble Gases that chemically react with solvent have greater solubility

Answer 37

Electrons are transferred from one atom to another | Atom that loses electrons is oxidized, atom that gains electrons is reduced

Answer 38

Possible charge values that an atom can hold within a molecule In many cases, the charges do not truly exist, and oxidation states simply provide a system for tracking the movement of electrons Oxidation states must add up to total charge on molecule or ion Often the same as the charge of the cation or anion that the element typically forms (difference to Noble gas config), but if atoms had permanent oxidation states, redox reactions could not take place

Answer 39

Rules higher on table have higher priority if contradictions: Oxidation State Atom 0 Atoms in their elemental form -1 Fluorine +1 Hydrogen (except when bonded to a metal, like NaH; then -1) -2 Oxygen (except when in a peroxide like H2O2; then -1)

Answer 40

Rows higher up in table have higher priority during contradictions Oxidation State Group on Periodic Table +1 Group 1 elements (alkali metals) +2. Group 2 elements (alkaline earth metals) +3 Group 15 elements (Nitrogen family) -2 Group 16 elements (Oxygen family) -1 Group 17 elements (halogens)

Answer 41

Reducing agent (AKA reductant) gives electrons to another species, becomes oxidized Oxidizing agent (AKA oxidant) receives electrons from another species, so becomes reduced and oxidizes the other species E.g. CH4 + 2O2 -> CO2 + 2H2O C has gone from -4 to +4, Oxygen goes from 0 to -2 Atom is oxidized or reduced, compound is the oxidizing or reducing agent

Answer 42

Used to find the molarity of a reducing agent (remember, mol / L) Strong oxidizing agent is titrated and the resulting voltage change is measured Solutions must be different, one is standard solution and one has reducing agent As strong oxidizing agent is added to solution with reducing agent, voltage increases first gradually and then quite suddenly Half equivalence point: point near the middle of the gradual increase Equivalence point: voltage suddenly shoots up - Occurs when all moles of reducing agent in solution have been completely oxidized - # moles of oxidizing agent required to reach equivalence point will be either equal to or a multiple of the number of moles of the reducing agent in the solution that is being titrated (one molecule of oxidizing agent may accept a different number of electrons than one molecule of reducing agent gives up)

Answer 43

For example, if you add oxidizing agent Ce^4+, will take electrons from Sn^2+ in following two reactions: Sn^2+ -> Sn^4+ + 2e- Ce^4+ + e- -> Ce^3+ Where two Ce^4+ atoms are required for each Sn^2+ atom If add 2mL of 5 mM Ce^4+ solution to get to equivalence point, molarity of original Sn^2+ solution is determined by: 2 mL Ce^4+ solution x 5 mmol / L x 1L / 1000 mL = 0.01 mmoles of Ce^4+ This means 0.01 mmoles Ce^4+ are required to oxidize 0.005 mmoles Sn^2+ Therefore: (0.005 mmoles Sn^2+ / 100 mL unknown Sn^2+ solution) x (1000 mL / L) = 0.05 mM Sn^2+ solution

Answer 44

Likelihood that the redox reaction will proceed The more positive the potential, the more likely the reaction will proceed Potentials for oxidation and reduction components of reaction can be approximated separately based on a standard hydrogen electrode (SHE) - Each component is a half reaction (can’t occur by itself) - Reverse of a reduction half reaction is an oxidation half reaction Half reaction potentials are usually listed as reduction potentials, but reverse sign for oxidation potential

Answer 45

2H+ + 2e- -> H2 Where E^{o} noted as 0.00 V All electrical potentials for redox reactions are relative to the Standard hydrogen electrode, so positive means greater than SHE and negative means less than SHE

Answer 46

Reduction potentials are intensive properties, so do not multiply by number of ions in each half reaction 2(Au^3+ + 3e- -> Au), E^{o} = 1.50 V 3(Cu -> Cu^2+ + 2e-), E^{o} = -0.34V 1.50V - 0.34V = 1.16V

Answer 47

Balance redox reaction in acidic solution: 1. Divide the reaction into its corresponding half reactions 2. Balance the elements other than H and O 3. Add H2O to one side until the O atoms are balanced 4. Add H+ to one side until the H atoms are balanced 5. Add e- to one side until the charge is balanced 6. Multiply each half reaction by an integer so that an equal number of electrons are transferred into each reaction 7. Add the two half reactions and simplify For redox reactions occurring in basic solution, follow same steps, then neutralize H+ ions by adding the same number of OH- ions to both sides of reaction

Answer 48

Cell in which two distinct electrically conducting chemical phases are placed in contact and a charged species from one phase cannot flow freely to another, creating an electric potential Has an alternative pathway for just the flow of electrons between phases, generating a current from one phase to another in a conversion of chemical energy to electrical energy Made of a multiphase series of components, no components present in more than one phase, and all phases must conduct electricity, but at least one impermeable to e- Ionic conductor: phase impermeable to e- carries the current in the form of ions (electrolyte sol’n in form of salt bridge)

Answer 49

T-E-I-E’-T T: terminals (conductors such as metal wires) E: electrodes (also conductors), anode and cathode I: ionic conductor (often salt bridge) When cell is formed, emf is electric potential difference between T and T’

Answer 50

Anode: marked with a negative sign, where oxidation takes place Cathode: marked with positive sign, where reduction takes place Sometimes electrode also refers to electrolyte solution and strip of metal, or can be referred to wholly as a half cell

Answer 51

symbol E, also called the electromotive force (emf) Potential difference between the terminals when they are not connected Connecting the terminals reduces the potential difference due to internal resistance within the galvanic cell Drop in emf increases as current increases Current from one terminal to other defined as moving in opposite direction to electron flow (e- flow from anode to cathode)

Answer 52

Sum of the standard state potentials of the corresponding half reactions Cell potential for galvanic cell is always positive (a galvanic cell always has chemical energy that can be converted to work) Cell potential depends on: - half reactions - concentrations of reactants and products - temperature Concentrations will always be 1M, so you can use values from table to calculate cell potential

Answer 53

Exactly equal to the reduction potential of the half reaction occurring at the other electrode Many half reaction reduction potentials can be measured using the SHE SHE has no salt bridge, because both electrodes are in contact with the same solution, for example, producing HCl from H2 and AgCl

Answer 54

Type of liquid junction that minimizes potential difference When cell contains two difference solutions, liquid junction is required to separate the solutions, but because ions can move across a liquid junction, liquid junctions create additional small potential difference that affects potential of galvanic cell Salt bridges can be made from an aqueous solution of KCl Allow movement of ions between solutions without creating a strong extra potential, because K+ ions move toward the cathode at about the same rate that the Cl- ions move towards the anode Job is to maintain the neutrality of the half cell

Answer 55

Causes the battery to lose its chemical potential over time, because charge buildup will occur

Answer 56

IUPAC convention to represent a galvanic cell where each phase is listed from left to right, beginning with the terminal attached to the anode and ending with the terminal attached to the cathode Terminals often left out because they are always the same material ( do not take place in reaction) Vertical line is placed between phases, double vertical line indicates a salt bridge, dotted vertical line indicates a boundary between two miscible liquids, and species in the same phase are separated by a comma Pt’(s)|Zn(s)|Zn^2+(aq)||Cu^2+(aq)|Cu(s)|Pt(s)

Answer 57

\delta G = -nFE_max Where n is number of moles of electrons transferred in balanced redox reaction, F is charge on one mole of e- (96,486 C / mol), and E_max is cell potential Says that free energy is the product of total charge nF and voltage E (product of charge (C) and voltage (V) is equal to electrical work) Positive cell potential means spontaneous reaction When all conditions are standard: \delta G^{o} = -nFE^{o}

Answer 58

\delta G = \delta G^{o} + RT ln(Q) and substitute -nFE for \delta G and -nFE^{o} for \delta G^{o}, and then divide by -nF: Nernst Equation: E = E^{o} - (RT / nF) ln(Q) At 298K and base 10 log: E = E^{o} - 0.06/n log (Q) So calculate Q first by plugging in nonstandard concentration to compute the equilibrium constant

Answer 59

Limited form of galvanic cell in which a reduction half reaction takes place in one half cell, while the exact reverse of that half reaction is taking place in the other half cell Cells differ in ion concentrations Potential: E^{o} = 0V If concentrations were equal, concentration cell potential would be zero Nernst equation can be used to calculate potential for concentration cell, but can determine flow of cell if think about entropy wanting to equalize the concentrations on both sides of cell

Answer 60

Remember, that the solution in the reaction is both a reactant and a product, so will be in numerator and denominator for different half cell concentrations: E = E^{o} - 0.06 / 2 log(products/ reactants) with E^{o} = 0V

Answer 61

A type of galvanic cell in which there is a power source connected to wire connecting electrodes which can drive the reaction in the opposite direction Will always have a negative emf Cathode is marked negative and anode is marked positive However the reduction still takes place at the cathode and oxidation takes place at the anode Used in industry for metal plating, purifying metals

Answer 62

E.g. Na+ + e- -> Na, E^{o} = -2.71V 2Cl- -> 2e- + Cl2, E^{o} = -1.36V Reaction will not run in aqueous solution, because water has a less negative reduction potential than sodium, which means sodium will oxidize spontaneously in water

Answer 63

Cell of choice for cars because of high current it can supply Can act as both a galvanic and electrolytic due to rechargeable capabilities Anode made of lead, cathode made of lead dioxide Discharges in the presence of sulfuric acid Oxidation of lead at anode: Pb(s) -> 2e- + Pb^2+(aq) Sulfuric acid reacts with lead ions to precipitate insoluble lead sulfate: Pb^2+(aq) + SO4^2-(aq) -> PbSO4(s) Reduction at the cathode: 2e- + PbO2(s) + 4H+(aq) + SO4^2-(aq) -> PbSO4(s) + 2H2O(l) Both half reactions produce lead sulfate, insoluble toxin

Answer 64

More expensive than lead storage battery Improvement upon lead storage battery due to size (smaller) and because its electrolyte is not consumed during discharge Batteries arranged in cylindrical configurations that reduce internal resistance