Aqueous Solutions

4.1 The Concept of Solubility:

Electrical Conductivity:

• an electrolyte solution conducts electricity

• Electrical conductivity requires the presence of ions in the solution. The more ions present, the greater electrical conductivity.

• Ionic compounds made up of metal and non-metals.

  • Salts are iconic compounds

  • Salts differ in their abilities to dissolve in water.

  • All salts dissolve in water to produce electrolyte solutions.

  • Salts that dissolve well, produce solutions with high ion concentrations, so these salts are strong electrolytes.

• Molecular compounds made up of non-metal atoms

  • Molecular solutes do not form ions in water.

  • When molecular compounds dissolve, its molecules enter the solution intact.

  • Some compounds, even though they appear molecular form ions!

  • Recall: solutions of acids are ionic to varying degrees

    • Strong acids completely ionize in solutions

    • Weak acids are able to form ions but to an extent

      • High concentrations of weak acids have varying low ion concentrations and so, weak acids are weak electrolytes.

Concentrations of Ions in Solutions:

• When an ionic compound dissolves in water, the water molecules attract and separate the anions and cations in the solute.

• The process of separating positive and negative ions in solution is called dissociation.

Calculating the Molarity of a Solution:

• When solute dissolves in water, the concentrations of ions is proportional to the concentrations of the solution formed.

Saturated Solutions and Solubility:

• To make a solution we dissolve solute in solvent.

  • If keep adding more solute, it will get to the point where no more solute will be dissolved.

  • At this point it is saturated.

    • In a saturated solution, there must be undissolved solid, a precipitate.

• Solubility of a substance is the amount of solute required to make a certain volume of saturated solution.

  • It’s the max amount of solute that can be dissolved in a particular volume of solution.

• Recall: solubility with most salts increase with temperature.

  • Solubilities are determined at different temps.

  • It also depends on other ions.

Equilibrium in Solutions:

• A saturated solution must contain some undissolved solute.

• An equilibrium exists between the undissolved solid and dissolved ions.

4.2 Qualitative Analysis-Identifying Unknown Ions

Precipitate Formation:

• Salts have different abilities to dissolve in water.

  • When salt is added to water, it will dissolve if the attraction of the cation and anion to each other is overcome or replaced by and attraction of the ions to a polar water molecule.

    • Example: NaCl is added to water.

    • The attraction between the Na cations to the Cl anions is replaced.

    • Instead the cations is attracted to the negative dipole on the oxygen atom of the water molecule.

    • And anion is attracted to the positive dipole on the hydrogen atom of the water molecule.

  • Because the ions are attracted to the solvent, dissociation occurs.

How to Use Solubility Table:

• Soluble: more than 0.1 mol of salt will dissolve in 1.0L of solution.

• Low Solubility: less than 0.1 mol of salt will dissolve in 1.0L of solution.

  • does NOT mean the salt will not dissolve.

  • Solubility of salt changes with the temperature. The terms soluble and low solubility refers to the amounts of solute that dissolve in room temp.

• Salts with Alkali ions or ammonium ions are soluble

• Acids are soluble

• Salts with nitrate ions are soluble.

Equations Representing Precipitate Reactions:

• There are 3 types of equations used to represent double replacement/metathesis reactions:

  • Formula Equation: shows chemical formulas of the compounds and their states.

    • Use solubility rules to find the product with low solubility

    • Ex) 2KI (aq) + Pb(NO3)2 (aq) = 2KNO3 (aq) + PbL2 (s)

  • Complete Ionic Equation: soluble salts are represented in their dissociation form. Because the precipitate has a low solubility, it does not show its dissociation form.

    • Ex) 2K+ (aq) + 2I- (aq) + Pb2+ (aq) + 2NO3- (aq) = 2K+ (aq) + 2NO3- (aq) + PbI2 (s)

      • Make sure to show charges and states!

  • Net Ionic Equation: Only the ions that take part of the reactions are shown. Ions that are seen both on the reactant and product sides are called spectator ions. They only watch, they don’t take part.

    • For example in the Complete Ionic Equation above, Potassium (K+) and Nitrate (NO3) are spectator ions.

    • Ex) Pb 2+ (aq) + 2I- (aq) = PbI2 (s)

Qualitative Analysis:

• Qualitative Analysis refers to a procedure that can be used to help identify unknown ions in a solution!

• There are many different methods of finding unknown ions

  • Looking at a coloured solution can help.

    • Ex) Copper (II) ions are blue.

    • Ex) Chromates can be yellow.

    • Ex) Dichromates are orange.

    • Ex) Permanganate is a deep purple.

    • Nickel (II) is light green.

  • Flame method. Where you put the solutions into a flame.

    • Copper (II) ions turn flame blue-green

    • Lithium ions turn flame red

    • Barium Ions turn flame green.

      • This method coincides with “Atomic absorption spectroscopy.” Where the samples are vaporized and passed through a flame as an instrument record quantity of energy or wavelength that is absorbed by the atoms to give off a particular spectra.

        • Think! The experiment we did in the costume lab where all the lights are turned off and we looked through a machine that emits a different colour when we looked at a certain solution.

  • Another method is adding a reagent to selectively precipitate certain ions.

    • The colour often indicates of a present ion.

    • Amount of ppt formed or the speed at which it formed can also help identify.

    • You can use solubility rules to do this!

Identifying Precipitates:

• The formations of a ppt can identify and ion, but sometimes the ppt may have more than one possible identity.

• We can add a reagent to dissolve precipitates.

Hard Water:

• Hard water: the presence of Ca 2+, Mg2+, and/or Fe2+ ions in water at relatively high concentrations.

• Soft Water: Without the presence of the ions mentioned above.

• Limestone (CaCO3) and Dolomites (CaMg(CO3)2) are in soil all over the world, when the CO2 from atmosphere dissolves in water from streams or rain, this solution dissolves limestone.

• The presence of Ca 2+ poses hazards for municipalities water systems.

  • As water is heated CO2 id driven off the gas, reforming a “scale” of CaCO3.

    • This can lower efficiency of heating systems and can potentially block water pipes.

• Soaps also react with Ca 2+ and Mg2+ to form soap scum.

  • That’s why some laundry detergents have water softeners to avoid getting this scum on clothes

  • Shampoos have lathering agents, so when you’re washing your hair in harder water, it may be difficult to lather your hair.

• Water treatment for hard water often includes the addition of CaO or Na2CO3 to underground water reservoirs to precipitate out Ca2+ and Mg2+ ions.

4.3 The Solubility Product Constant Ksp

The Solubility Product Constant:

• Recall: Some chemical systems establish equilibrium.

  • For these systems we learned how to find the Keq.

• In a saturated solution, and equilibrium can be established between the dissolving and recrystallization of a salt.

  • We can represent equilibrium using the following dissociation equation:

  • Ex) PbCl (s) = Pb2+ (aq) + 2Cl- (aq)

    • Recall! Solids do not show up on the Keq, so for this equilibrium the Keq is shown as:

    • Keq= [Pb2+][Cl-]²

      • This is a special type of equilibrium, so its given its own equilibrium constant type: Ksp

      • Ksp: The solubility product constant

• Solubility product is the value obtained when the maximum concentrations of ions are multiplied together.

  • As solubility depends on temperature, so does Ksp.

    • Ksp can change with temperature!

• Molar Solubility of a substance is the molar concentration of solute in a saturated solution.

• Solubility Product Constant (Ksp) is the product of the ion concentrations in a saturated solution raised to the power of the coefficients in the equilibrium.

  • As the solubility of a substance increases so does the concentration of the ions. So the value of Ksp also increases.

Calculating the Ksp From Solubility:

• Solubility of a substance and solubility product constant are two different things but are related.

  • We can calculate one from the other.

  • When solving any Ksp problem, start by writing the equilibrium equation for the saturated solution AND THEN the Ksp expression.

  • Example from the textbook:

Calculating the Solubility from Ksp:

• The Ksp value is NOT a concentration.

  • It’s the product of the ion concentrations in a saturated solution raised to the power of their coefficients from the balanced dissociation equation.

  • Example from the textbook:

Types of Salts:

• We introduced 2 types of salts:

  • AB (example: AgCl)

  • AB2 (example: PbCl2)

    • For AB salts: Ksp = (solubility)²

    • For A2B or AB2 salts: Ksp = 4(solubility)³

4.4 Precipitation Formation and the Solubility Product Ksp

Precipitates That Form When Solutions are Mixed Together:

• Concentrations of the ions are directly related to one another.

  • Example) Attempt to dissolve AgCl in water. The [Ag+] equals the [Cl-].

• In this lesson, we will be looking at situations where ions forming the precipitate DO NOT come from the same solute.

  • The concentrations of the ions in solution are NOT related to each other.

  • Instead, they depend on the concentrations and volumes of the solutions being combined

Predicting Whether a Precipitate Will Form:

• When two solutions are mixed, we can predict whether a precipitate will form.

• If the product of the ion concentrations exceeds the Ksp value, ions will not remain dissolved and will form a precipitate.

• If the product of the ion concentrations are less than the Ksp value, ions will remain dissolved and a precipitate will not form.

• Note: Ksp expression is the equilibrium concentrations of ions in a saturated solution.

  • If an equilibrium is not present in a solution we calculate the trial ion product (TIP)

    • It’s also called the trial Ksp of reaction quotient (Q)

• Qsp>Ksp precipitate forms

• Qsp<Ksp no precipitate forms

• Qsp=Ksp the solution is saturated

Using Ksp to Calculate the Concentration of Ions in Solution:

• Similar to what we studied in the previous section, however these ions that form a precipitate are from different stock reagent sources.

  • First determine concentrations of any ion present in the solutions being combined

  • Once another solution is added, consider potential precipitates that can form

  • Then write an equation for the equilibrium present in a saturated solution (dissociation of the ppt)

  • Then write Ksp expression.

    • Example: A solution of 0.025 M Pb(NO3)2.

      • write dissociation for Pb(NO3)2, to calculate the concentration of each ion.

      • Pb(NO3)2 (aq) = Pb2+ (aq) + 2NO3- (aq)

        • [Pb(NO3)2]=0.025M [Pb2+]=0.025M [NO3-]=0.050M

      • If NaCl is added, the ppt that may form is PbCl2(s). If a ppt forms, then saturated solution of PbCl2 is present and is governed by its Ksp.

        • PbCl2 (s) = Pb2+ (aq) + 2Cl- (aq)

        • Ksp= [Pb2+][Cl-]²

      • NOTE: The original [Pb2+] and [Cl-] are not related because they come from different sources!

        • The max concentration of Cl- can exist in this solution can be calculated from the Ksp and the concentration of Pb2+ provided by the Pb(NO3)2 source.

          • 1.2×10^-5 = (0.025)[Cl-]

          • [Cl-]= 0.022M = [NaCl]

The Common Ion Effect:

• Defined solubility as the maximum amount of solute that will dissolve in a given volume of solvent at a specific temperature.

  • Solubility depends on temperature.

  • Also depends on the solvent’s identity.

    • So far in this unit we have only considered pure water as the solvent.

    • The presence of other ions in the solvent has an effect on the solubility of the solute.

• Solubility of a substance depends on the presence of other ions in solution and the temperature.

  • The Ksp of a substance depends on temperature only.

• Example: AgsCO3 (s) = 2Ag+ (aq) + CO3 2- (aq)

  • The amount of Ag+ and CO32- in the solution is governed by the Ksp.

    • However if the solution already contained Ag+ ions, [Ag+] is increased.

    • To maintain the Ksp value, the [CO3 2-] must decrease.

    • And increase in silver causes the eq’m to shift left. This will cause the decrease of carbonate and the increase of solid silver carbonate to increase (Ag2CO3).

      • The presence of the silver ions in the solvent decrease the solubility of Ag2CO3. This is the Common Ion Effect.

        • The solubility of Ag2CO3 is decreased due to the common ion, Ag+ in the solvent.

          • If CO3 2- was present in the solvent it would cause a similar effect resulting in the decrease of solubility.

• The solubility of a solute is decreased by the presence of a second solute in a solvent containing a common ion.

• How can we fix it?

  • According to Le Châteliers principle, the solubility of Ag2CO3 can be increased by removing the Ag+ or CO3 2- ions.

    • This would cause a shift right.

    • Example: The presence of HCl in the solvent. Carbonates dissolve in acid solutions.

      • The presence of H3O+ from the acid causes the [CO3 2-] to decrease.

      • That will then cause more Ag2CO3 (s) to dissolve.

        • Additionally, the chloride ion from the acid precipitates the silver ions from the solution.

          • The removal of the silver ion causes an even further shift to the right. It then increases the solubility of the silver carbonate even further.

Calculating Solubility With a Common Ion Present:

• The presence of a common ion in the solvent decreases solubility of a solute.

Extra Information to Know:

• Solutions that have a transition metal as the ion are usually a coloured solution.

• Salts consist of positive ions from a metal with negative ions from a non-metal.

• To find which has the greater conductivity, is to count the ions.

  • Example: Which of the following aq solutions would have the greatest conductivity?

    • A) 0.2 M NaOH

    • B) 0.2 M RbCl

    • C) 0.2 M K3PO4

    • D) 0.2 M H2SO3

      • Explanation: The strong electrolyte with the greatest ions is the best conductor.

      • D is a weak electrolyte.

      • A had 2 ions

      • B had 2 ions

      • C had 4 ions

        • Since A and B have the same number of ions they cannot be right, C would be the answer.