solubility (midterm 2)

Question 1

what is solubility equilibria?

Answer 1

In this section of the course we will consider equilibria arising from sparingly soluble solids dissolving to form aqueous solutions of ions.

Question 2

what eventually happens to the equilibria?

Answer 2

Eventually, a dynamic equilibrium is reached:

Ca3(PO4)2(s) 3 Ca2+(aq) + 2 PO43-(aq)

At this point, no more solid will dissolve
(Rate of Dissolution = Rate of Precipitation)

Conversely, when Ca2+ and PO43- are mixed, this equilibrium will be established in reverse.

Increasing the concentration of Ca2+ or PO43- will favour precipitation while lowering the concentration of either or both of the dissolved ions by adding pure water will cause more of the solid to dissolve.

Question 3

what is solubility product?

Answer 3

Solubility Product (Ksp ) is an equilibrium constant, for a given ionic compound at a given temperature.

Question 4

what is solubility?

Answer 4

Solubility, in contrast, is the extent to which the compound dissolves in solution expressed usually in g/L or moles/L of saturated solution.

Question 5

what happens when a common ion is present?

Answer 5

If a common ion is present, however, the solubility will vary with the concentration of the common ion (i.e. represses dissolution).

Question 6

what happens to salts that belong to the same formula class?

Answer 6

(i.e. those salts that produce the same number of ions in the solubility reaction), the solubility decreases with decreasing Ksp.

Question 7

what happens to ionic compounds that belong to different formula classes?

Answer 7

formula classes (e.g. MgF2 vs. Ba3(PO4)2), the relative solubilities cannot be assessed simply by comparing their Ksp values because they produce different numbers of ions when they dissolve.

In these instances, their solubilities must be derived from their Ksp values.

Question 8

how is Ksp determined from solubility measurements?

Answer 8

The method by which Ksp values are obtained is the opposite of the solubility calculations procedure we have just considered.

Question 9

what is common ion effect on the solubility of precipitates?

Answer 9

what happens when the water contains one of the ions present in the dissolving salt (common ion).

Le Chatelier’s principle, would predict that the solubility of a solid in a solution containing an ion in common with the solid is less than its solubility in pure water. In other words, the solubility of the salt is lowered if the dissolving solution already contains one or both of the ions that make up the salt.

Question 10

what is precipitation?

Answer 10

Precipitation is the reverse of dissolution.

Question 11

how do we determine if a precipitate forms?

Answer 11

To determine if a precipitate forms when two solutions of soluble ionic salts are mixed, we calculate the inverse ion product (Qppt) at the instant of mixing and compare it to the Kppt.

The Qppt differs from the Kppt in that the concentrations used in the Qppt calculation are the initial values and not necessarily those at equilibria.

Question 12

when is a solution saturated?

Answer 12

Qppt = Kppt, the system is at equilibrium (i.e. the solution is saturated).

Question 13

when is the solution unsaturated?

Answer 13

If the Qppt > Kppt, the system is not at equilibrium, no precipitate will form (i.e. the solution is unsaturated).

Question 14

when is the solution super saturated?

Answer 14

If the Qppt < Kppt, the system is not at equilibrium, a precipitate will form immediately after the instant of mixing (i.e. the solution is supersaturated).

Question 15

what is a complex ion equilibria?

Answer 15

A complex ion is a stable ion composed of many atoms that has been formed by coordination (attachment) of anions or neutral molecules to a central metal atom or ion.

The anions or neutral molecules bound to the metal centre are called ligands.

Each ligand donates at least one pair of electrons to the central metal atom.

Question 16

describe a lewis acid

Answer 16

Species that is the acceptor of a lone pair of electrons in the formation of a coordinate covalent bond

Question 17

describe a lewis base

Answer 17

Species that is the donor of a lone pair of electrons in the formation of a coordinate covalent bond.

Question 18

what is the BN bond?

Answer 18

The B-N bond is called a coordinate covalent or dative bond.

Question 19

formation of a transition model can be viewed as:

Answer 19

Similarly, the formation of a transition-metal complex can be viewed as a Lewis acid-Lewis base reaction.

Question 20

what are monodentate ligands?

Answer 20

The formation of complexes involving monodentate ligands (groups attached via one atom) is a stepwise process, with each ligand added successively to the central metal ion.

Question 21

steps for formation of complexes involving monodentate ligands

Answer 21

step one find K1
step 2 find K 2
The product K1 x K2 is the overall stability or formation constant for the generation of the bis amine adduct
Kf

Question 22

what does a larger Kf value indicate?

Answer 22

The large value for the Kf indicates that the complex ion product is strongly favoured.
Most formation constants for complex ions are quite large. This suggests that these species are very stable compared to the free ligand and uncomplexed metal ion.

Question 23

what does complex ion formation result in?

Answer 23

Complex-ion formation results an increase in solubility of precipitates.

Question 24

why do we write in 0 (sat) but exclude it for k equation?

Answer 24

sat solutions have an activity of 1 due to micro crystals

Question 25

why do solids generally form more ions (more soluble) in hot water?

Answer 25

dissolving of most solids is an endothermic reaction

Question 26

what ph change is considered large?

Answer 26

+ or - 1

Question 27

How to calculate the pH of a STRONG ACID ANALYTE (e.g., HCl, HNO3) reacted with STRONG BASE TITRANT (e.g., NaOH, KOH):

Answer 27

(A) Before addition of titrant: pH is calculated as minus common logarithm of original concentration of H+ in aqueous acid.

(B) Before the equivalence point: pH is calculated as minus common logarithm of excess concentration of H+ in mixture.

(C) At the equivalence point: pH is calculated as minus common logarithm of concentration of H+ in water, that is, pH = -log (1.00E-7).

(D) After the equivalence point: pOH is calculated as minus common logarithm of excess concentration of OH- in mixture. Then, pH = 14.00 - pOH.

Question 28

How to calculate the pH of a STRONG BASE ANALYTE (e.g., NaOH, KOH) reacted with STRONG ACID TITRANT (e.g., HCl, HNO3):

Answer 28

(A) Before addition of titrant: pOH is calculated as minus common logarithm of original concentration of OH- in aqueous base. Then, pH = 14.00 - pOH.

(B) Before the equivalence point: pOH is calculated as minus common logarithm of excess concentration of OH- in mixture. Then, pH = 14.00 - pOH.

(C) At the equivalence point: pH is calculated as minus common logarithm of concentration of H+ in water, that is, pH = -log (1.00E-7).

(D) After the equivalence point: pH is calculated as minus common logarithm of excess concentration of H+ in mixture.

Question 29

How to calculate the pH of a WEAK ACID ANALYTE (e.g., CH3-COOH, HF, HNO2, HClO; and acidic salts such as NH4Cl, alkylammonium halide) reacted with STRONG BASE TITRANT (e.g., NaOH, KOH)

Answer 29

(A) Before addition of titrant: pH is calculated as minus common logarithm of concentration of H+ in aqueous acid, which results from the ionization equilibrium of weak acid: HA = A- + H+.

(B) Before the equivalence point: The mixture is a buffer solution. Then, pH = pKa(HA) + log( [A-] / [HA] ).

(C) At the equivalence point: pOH is calculated as minus common logarithm of concentration of OH-, which results from the ionization equilibrium of conjugate base, that is, A- + H2O = HA + OH-. Then, pH = 14.00 - pOH.

(D) After the equivalence point: pOH is calculated as minus common logarithm of excess concentration of OH- in mixture. Then, pH = 14.00 - pOH.

Question 30

How to calculate the pH of a WEAK BASE ANALYTE (e.g., NH3, amine; and basic salts such as NaCH3COO, NaF, Na2CO3, NaClO) reacted with STRONG ACID TITRANT (e.g., HCl, HNO3):

Answer 30

(A) Before addition of titrant: pOH is calculated as minus common logarithm of concentration of OH- in aqueous base, which results from the ionization equilibrium of weak base: B + H2O = BH+ + OH-. Then, pH = 14.00 - pOH.

(B) Before the equivalence point: The mixture is a buffer solution. Then, pH = pKa(BH+) + log( [B] / [BH+] ).

(C) At the equivalence point: pH is calculated as minus common logarithm of concentration of H+, which results from the ionization equilibrium of conjugate acid, that is, BH+ = B + H+.

(D) After the equivalence point: pH is calculated as minus common logarithm of excess concentration of H+ in mixture.

Question 31

review these + graphs drawn in notes

Answer 31

To determine the point with the specific data given in the exercise, follow these steps:

1. • Identify which substance acts as the analyte (i.e., a sample of a fixed amount of substance) and which acts as the titrant (i.e., the substance that is dispensed gradually).
2. • Calculate the equivalent amount of each reactant as follows: divide the amount in mol of reactant (n = cV) by the reactant’s stoichiometric coefficient in the balanced chemical equation.
3. • Compare the equivalent amounts of analyte and titrant in order to identify the point of the titration curve that the reacting mixture is at:

(A) Before addition of titrant: Original amount of analyte and 0 mol of titrant added.

(B) Before the equivalence point: The equivalent amount of analyte is larger than that of titrant added.

(C) At the equivalence point: The equivalent amount of analyte is equal to that of titrant added.

(D) After the equivalence point: The equivalent amount of analyte is less than that of titrant added.