Solubility and Electrolytes

• Many barium compounds are poisonous, but barium sulphate is swallowed for clearer X-rays because it's insoluble and not absorbed.
• Silver bromide can saturate a swimming pool to kill microorganisms while remaining harmless to people, despite silver ions being dangerous in large doses.

Solubility

• Solubility relates to how "soluble" a compound is, how "saturated" a solution is, and how solutions can be analyzed, purified, and used to extract metals.
• The focus is on aqueous ion solutions, but may also apply to non-ionic substances.
• Acids, bases, and salts form ionic solutions, but this unit focuses on salts.

Electrolytes

• Electrolytes dissolve to produce an electrically conducting solution containing ions, for example (NH4)3PO4(s)→3 NH+4(aq) + PO3−4(aq) and HCl(g) → H+(aq) + Cl−(aq)

Non-Electrolytes

• Non-electrolytes dissolve to produce a non-conducting solution containing only neutral molecules, for example C2H2(g) → C2H2(aq) and Br2(l) → Br2(aq)

Molecular vs. Ionic solutions

• Molecular solutions contain only neutral molecules.
• Ionic solutions contain ions.

Predicting Ionic/Molecular Solutions

• Whether a compound forms an ionic or molecular solution can often be predicted, though exceptions exist

Rules for Classifying Compounds

• Compounds made of a metal and a non-metal tend to be ionic in solution, for example FeCl3(s) → Fe3+ (aq) + 3 Cl (aq)
• Compounds with polyatomic ions will be ionic in solution, for example (NH4)2Cr2O7(s) → 2 NH+4(aq) + Cr2O2−7(aq)
• Ionic solids are made of one type of positive and one type of negative ion.

Note for Recognizing ions

• Recognizing NH+4 in (NH4)2HPO4 means HPO2−4 is the second ion.
• Recognizing Na+ in NaH3P2O7 means H3P2O−7 is the second ion.

Covalent Compounds

• Covalent compounds (non-metal + non-metal), especially organic compounds (containing carbon), generally form molecular solutions, for example CH3OH (I) → CH3OH(aq) and CIO2(1) → CIO2(aq)

Solubility Defined

• Solubility of a substance is the maximum amount that can dissolve in a solvent at a temperature.
• A saturated solution has dissolved the "maximum amount" of a substance.
• Adding more substance to a saturated solution causes it to remain undissolved.

Supersaturation

• Supersaturation is an unstable condition where a concentrated solution contains more dissolved salt than it normally would when saturated, without solid crystals forming.

Alternative Solubility Definitions

• A saturated solution is formed when the dissolved substance is in equilibrium with some of the undissolved substance.
• An unsaturated solution contains less than the maximum amount of a substance, with no undissolved solid present, and no equilibrium.
• The solubility of a substance is its equilibrium concentration in solution at a given temperature.

Molar Solubility

• Solubility is referred to as the molar solubility when concentration is in "moles/litre".

Saturated Solution Conditions

• Saturation exists when some undissolved material is present.
• Equilibrium exists between the dissolved and undissolved material.

Equilibrium Equation

• Saturated solutions can be represented by an equation showing equilibrium with aqueous products, for example AgBrO3(s) ⇌ Ag+(aq) + BrO3(aq).

Dissolving Reaction

• Solid AgBrO3 dissolves and enters solution: AgBrO3(s) → Ag+(aq) + BrO3(aq)

Crystallization Reaction

• Ag+ and BrO3 ions come together to form AgBrO3: Ag+(aq) + BrO3(aq) → AgBrO3(s).
• Initially, dissolving predominates, then as ions accumulate, the rate of dissolving equals the rate of crystallization.

Calculating Solubility and Ion Concentrations

• After experimentally measuring the mass of a substance in 1 L of solution, it is straightforward to find the solubility.

Molar Solubility Example

• If 1 L of saturated AgBrO3(aq) contains 1.96 g of AgBrO3, the molar solubility calculation is: [AgBrO3] = 1.96 g/L x (1 mol / 235.8 g) = 8.31 x 10^-3 M

Gram Solubility Example

• If the molar solubility of PbI2 is 1.37 x 10^-3 M, expressing it in grams per liter is calculated as: Solubility (g/L) = 1.37 x 10^-3 mol/L x (461.0 g / 1 mol) = 0.632 g/L

Calculator Input

• To enter "1.37 x 10^-3" on a calculator, use the "EXP" or "EE" key for "10 to the power of ".

Molar Solubility Calculation

• If 250 mL of saturated CaCl2 contain 18.6 g of CaCl2 at 20°C, molar solubility is: [CaCl2] = (18.6 g / 0.250L) x (1 mol / 111.1g) = 0.670 M
• Assume all solutions are at 25°C unless indicated.
• Calculation review relating to the solubility of a salt.
• Review of calculations involving the concentrations of ions in solution which involves a salt dissociating into ions when it dissolves
• If 1 mol of Na3PO4 is dissolved and diluted to a total volume of 1L , find: Na3PO4(s) -> 3Na+(aq)+PO43-(aq)

Molar concentrations

• Moles of particles at (in 1L): 1mol -> 3 mol + 1mol
• Molar concentration: 1M -> 3M + 1M
• The final solution has [Na+] = 3M and [PO43-] = 1 m and NO particles of Na3PO4 in the final solution, all of the Na3PO4 present has dissociated into ions

Concentration Calculations

• Based on the amount of Ag2CO3 (s) which has dissolved: Ag2CO3 (s)-> 2Ag+ (aq)+ CO32-(aq)
• It can be seen that if [Ag2CO3]DISSOLVED = 1.2 x 10-4, then [CO32-] = 1.2 x 10-4 M
• AND [Ag+] = 1.2 x 10-4 mol Ag2CO3 x 2 mol Ag+ / 1mol Ag2CO3 = 2.4 x 10-4 M

Dilution Example

• The example calculates the dilution occuring when solutions of ions are mixed with water or each other, as illustrated in the next example
• If 5.0 mL of 0.020 M CI- is added to 15.0 mL of 0.012M Br-, what is the molarity of the CI and Br- ions in the mixture?
• The 5.0 mL of chloride solution is diluted by the 15.0mL of liquid contained in the bromide solution, and vice versa

Perform Dilutions

• [SUBSTANCE]DILUTED equal to [SUBSTANCE]OLD x OLD VOLUME / DILUTED VOLUME
• So that [CI-] DILUTED = 0.020M x 5.0mL / (5.0 + 15.0) mL = 0.0050M
• AND [Br-] DILUTED = 0.012 M x 15.0mL / (5.0 + 15.0) mL = 0.0090M

Predicting Solubility of Salts

• This section examines some of the general rules for predicting the solubility of salts and uses a Solubility Table to predict whether a specific salt is soluble in water
• First, let’s establish what is meant when a salt is said to be "SOLUBLE” or to have a “LOW SOLUBILITY"
• Strictly speaking, nothing is INSOLUBLE in water for example, glass dissolves to an extremely small extent in water
• Normally, the amount of glass which dissolves in water can be NEGLECTED and glass is said to have a NEGLIGIBLE SOLUBILITY in water
• Some compounds may dissolve SLIGHTLY such that the amount which dissolves is extremely small but cannot be neglected
• Such a substance is said to have LOW SOLUBILITY, For example, lead (II) chloride, PbCI2, has a low solubility in water, but the amount which dissolves cannot be neglected.
• For example, the small amount of Pb2+ ion present in saturated PbCl2(aq) is toxic if swallowed
• This is like asking: “What is the least amount of money you need in order to be RICH?” All that can be done is to set an arbitrary value
• Definition: A substance is said to have LOW SOLUBILITY if a saturated solution of the substance is less than 0.1M
• The phrase “having a solubility less than 0.1 M” is often used to indicate low solubility

Solubility Table Usage

• Note that the term ""Alkali ions"" refers to the family of ions: Li, Na, etc; Negative Ions (Anions): positive Ions (Cations) :Solubility of compounds
• Phosphate PO43- or Carbonate Co32- or Sulphite, So32-: alkali ions, H+ , NH4+: Soluble
• Determine whether FeCo3(s) is soluble or has low solubility, find the compounds negative ion CO3 in this case and locate in in the negative ion colum
• Find Fe2+ in the “positive ion” column, Since Fe2 + is not explicitly listed then it must be included in the category “all Others”
• Finally, look in the “”SOLUBILITY OF COMPOUNDS”column (list column) to see that FeCO3(s) has “LOW SOLUBILITY”
• When two ions form a compound having "LOW SOLUBILITY", the mixing of the two ions will cause a PRECIPITATE to form
• Ex: Will a precipitate form when 0.2 M solutions of CaS and Na2SO4 are mixed?
• Look for combinations of ions which might form a precipitate.
• The ions present are: CaS -> Ca2+ + S2-

Precipitation

• Therefore, mixing the solutions produces a precipitate of CaSO4
• Notes: copper forms two different ions: Cu+ and Cu2+
• Cu+ has LOW solubility when combined with CI-, Br- or l- : Cu2+ is soluble when combined with CI-,Br- or 1
• IMPORTANT REMINDER, you either have memorized the formulae and charges of the common ions or will be using the table Names, Formulae, and Charges of Some Common, lons found at the back of this book.
• You MUST be able to write the chemical formula of a compound formed from two ions.

Chemical Generalizations

Compounds containing alkali metal, H+, NH4+ or NO3- will be soluble in water. It is difficult or impossible to precipitate alkali metal ions, H+, NH4+ or NO3- from solution. If you have to write the formula for a soluble compound containing a specific ion, you should include. An alkali metal ion, NH4+ or NO3- to help you ensure the compound will be soluble, note that H+ may be omitted, as it may be described in further detail in Unit IV. If you need to get a particular anion into solution, it is strongly suggested that you choose Na+ to combine with your anion; salts containing sodium are very common. Example: A soluble salt containing CO3^2- would be NA2CO3. If you need to get a particular cation into solution, it is strongly suggested that you choose NO3- to combine with your cation; salts containing nitrate are very common. Example: A soluble salt containing Fe3+ would be Fe(NO3)3.

Writing Formula, Complete, and Net Ionic Equations

• Every reaction that produces a precipitate will be a double replacement reaction.
• Double replacement reaction is written by using a 5 step procedure:
• On the reactant side of an equation, write the two compounds being mixed: Ca(NO3)2 + Na3PO4 ->
• Identify the ions present and their charges for each reactant compound: Ca^2+, No3^-, Na^+ and PO4^3-

Reactions

• Write two product molecules by swapping the negative ions in the reactant molecules. (Allow a small space between the positive and negative ions to leave room for subscripts.)
• Example: Ca(NO3)2 and Na3PO4 -> CaPO4 + NaNO3. Criss cross the charges to find the proper formula for each product molecule AgCl(s) and Ag+(aq) have an equilibrium constant Ksp=[Ag+][Cl−]=1.8×10−10. The values of Ksp and Q for a sparingly soluble salt can be considered in the following way:: Q = is the product of the ion concentrations which actually exist in solution or it is What “we have”.

Equilibrium Concentration Levels

Ksp = to the product of ion concentrations required to establish the solubility equilibrium- we what we need to form a saturated solution 3 Possible outcomes if two solutions are mixed: Case 1: Q < Ksp- so that “what we have" is less than ‘what we need to form a saturated solution”, consequently, there are not enough ions in solution to form a precipitate.

Cases

Case 1 : Precipitate can’t form if Q < Ksp Case 2: . Q=Ksp then the amount “we have" is equal to “what we need to form a saturated solution, resulting in the minimum amount of precipitate forms at his point Case 2: Barely saturated form is formed if Q-> Ksp Case 3 where if Q > Ksp which means “what we have” is greater than “what we need to form a saturated solution”

Precipitates Form

• Precipitates forms when a reduction from Ksp’s value occurs from value of Q has been reduced and will continue from excess ions have been removed from its solutions . Case 3- A Preciptiate forms if Q > Ksp.