October 31, 2011
CH-111
Chapter 4 - Overview
General Properties of Aqueous Solutions
Solution: a homogenous mixture of two or more substances.
Solvent: substance present in the gretest quantity.
Solutes: substance that is dissolved in the solvent.
Ions present in solution carry electrical charge. Conductivity of solutions indicates the presence of ions. Absence of conductivity indicates lack of ions.
Electrolyte: a substance whose aqueous solutions contain ions.
Nonelectrolyte: a substance that does not form ions in solution.
When an ionic solid is dissolved in water, each ion separates from the solid structure and disperses throughout the solution. The ionic solid dissociates into its component ions as it dissolves. Water is an effective solvent for ionic compounds because the O atom is partially negative and the H atom is partially positive. Cations are attracted to the partially negative end of H2O and anions by the positive end.
As they are dissolved, ions become surrounded by H2O molecules and are said to be "solvated". This is denoted by the (aq) after the atom and charge: Na+(aq) and Cl-(aq).
Solvation: helps stabilize ions insolution and prevents cations and anions from recombining. Ions become dispersed uniformly throughout the solution.
Molecular Compounds in Water:
A molecular compound dissolved in water usually consists of intact molecules dispersed throughout the solution. Most molecular compounds are nonelectrolytes. The exceptions are acids, ex; HCl, which ionizes in solution to H+(aq) and Cl-(aq).
Strong and Weak Electrolytes:
Strong Electrolyte: solutes that exist in solution completely or nearly completely as ions (dissociate completely or nearly completely).
Weak Electrolyte: solutes that exist in solution mostly in the form of neutral molecules with only a small fraction in the form of ions.
Solubility is not not direct*ly proportionate to electrolytic strength. Ca(OH)2 is only partially soluble in water, but the portion that does dissolve will dissociate completely. Conversely, CH3COOH is extremely soluble in water but is a weak electrolyte. The two properties are not interdependent.
At any given time, a reaction can be significant in both directions. Molecules are dissociating and recombining at once. The balance between these opposing processes determines the relative numbers of ions and neutral molecules, producing a state of chemical equilibrium.
Chemical Equilibrium: state in which the relative numbers of each type of ion or molecule in the reaction are constant over time.
Precipitation Reactions:
Reactions that result in the formation of an insoluble product are called precipitation reactions. Precipitation reactions occur when pairs of oppositely charged ions attract each other so strongly that they form an insoluble ionic solid.
Precipitate: an insoluble solid formed by a reaction in solution.
Solubility Guidelines for Ionic Compounds:
Solubility: the amount of the substance that can be dissolved in a given quantity of solvent at the given temperature.
Any substance with a solubility less than 0.01 mol/L will be referred to as insoluble. In these situations, the attraction between teh oppositely charged ions in the solid is too great for the water molecules to separate the ions to any significant extent and the substance remains largely undissolved.
Soluble Ionic Compounds:
NO3-
CH3COO-
Cl- with the exceptions of compounds containing Ag+, Hg2^2+, and Pb^2+
Br- with the exceptions of compounds containing Ag+, Hg2^2+, and Pb^2+
I- with the exceptions of compounds containing Ag+, Hg2^2+, and Pb^2+
SO4^2- with the exceptions of compounds containing Sr^2+, Ba^2+, Hg2^2+ and Pb^2+
Insoluble Ionic Compounds:
S^2- with the exception of compounds containing NH4+, alkali metal cations, Ca^2+, Sr^2+, Ba^2+
CO3^2- with the exception of compounds containing NH4+, alkali metal cations
PO4^3- with the exception of compounds containing NH4+, alkali metal cations
OH- with the exception of compounds containing NH4+, alkali metal cations, Ca^2+, Sr^2+, Ba^2+
To predict whether a precipitate forms when we mix aqueous solutions of two strong electrolytes, we must:
1. note the ions present in the reactants
2. consider the possible cation-anion combinations
3. use the table above to determine if any of these combinations is insoluble
Example: Mixing solutions of Mg(NO3)2 and NaOH. Both are soluble ionic compounds and strong electrolytes.
1. Note the ions present in the reactants:
Mg^2+, NO3-, Na+, OH-
2. Consider the possible cation-anion combinations:
We have Mg(NO3)2 + NaOH, which based on the chart above are both soluble combinations. The only other set of possible combinations are to switch cation-anions: MgOH + Na(NO3)
Balance the equation:
Mg(NO3)2 + 2NaOH ----------- Mg(OH)2 + 2NaNO3
Make sure to cross charges to ensure equation is balanced properly, as in MgOH, Mg is 2+ and OH is -1, cross charges to get Mg(OH)2.
We can look at this new equation and the table to see that OH- is typically insoluble, and that Mg is not a noted exception to that rule. Therefore Mg(OH)2 will form the precipitate. The rest of the substances are soluble and remain in solution. We can now denote this in the equation:
Mg(NO3)2 (aq) + 2NaOH (aq) -----------------> Mg(OH)2 (s) + 2NaNO3 (aq)
Metathesis Reactions (Exchange Reactions):
Reactions in which cations and anions appear to exchange partners and conform to the general equation AX + BY = AY + BX. The chemical equation we did above to determine the precipitate in Mg(NO3)2 reaction is an example of a metathesis reaction. To complete and balance the equation for a metathesis reaction, we can use the following steps:
1. use the chemical formulas of the reactants to determine which ions are present.
2. write the chemical formulas of the products by combining the cation from one reactant with the anion of the other, using the ionic charges to determine the subscripts in the chemical formulas.
3. check the water solubilities of the products. For a precipitation reaction to occur, at least one product must be insoluble in water.
4. Balance the equation.
We're going to take a quick stop here to do a metathesis reaction problem:
a) Predict the identity of the precipitate that forms when aqueous solutions of BaCl2 and K2SO4 are mixed
b) Write the balanced chemical equation for the reaction
1. Ba^2+, Cl2^-, K2+, SO4^2-
2. BaSO4 + KCl
3. BaCl2 + K2SO4 --------> BaSO4 + 2KCl
a)We can see by the balanced equation that Ba^2+ is an exception to the solubility of SO4^2- and thus BaSO4 will form the solid precipitate in the reaction. Hence:
b) BaCl2(aq) + K2SO4(aq) ---------------> BaSO4(s) + 2KCl(aq)
Ionic Equations:
It is useful to indicate whether dissolved substanes are present predominantly as ions or as molecules chemical equations.
Molecular Equation: an equation showing the complete chemical formulas of reactants and products that shows the chemical formulas without indicating ionic character, ex:
Pb(NO3)2(aq) + 2KI(aq) -----------> PbI2(s) + 2KNO3(aq)
We can break this equation down to indicate which species exist as ions in the solution:
Pb^2+ (aq) + 2NO3-(aq) + 2K+(aq) + 2I-(aq) ------> PbI2(s) + 2K+(aq) + 2NO3-(aq)
This form of equation, with all soluble strong electrolytes shown as ions, is called a complete ionic equation.
Spectator ions: Ions that appear in identical forms on both sides of a complete ionic equation which play no direct role in the reaction.
In the equation above, the spectator ions would be 2No3- and 2K+ because both are present on the reactant side and the product side. We can cancel them out much like an algebraic equation to get the
Net Ionic Equation: the equation that includes only the ions and molecules directly involved in the reaction.
In this case it would be:
Pb^2+(aq) + 2I-(aq) ----------------> PbI2(s)
Note that charge, like mass, is conserved in reactions, therefore the sum of the ionic charges must be the same on both sides of a balanced net ionic equation. In the above equation, 1 atom of Pb with a 2+ charge combines with 2 I atoms, each with a -1 charge, totalling the charge on each side at 0, or neutral. It is the same on both product and reactant sides.
The following steps summarize the procedure for writing net ionic equations:
1. Write a balanced molecular equation for the reaction.
2. Rewrite the equation to show the ions that form in solution when each soluble strong electrolyte dissociates into its ions. Only strong electrolytes dissolved in aqueous solution are written in ionic form.
3. Identify and cancel spectator ions.
Let's do one... for FUN!
Q: Write the net ionic equation for the precipitation reaction that occurs when aqueous solutions of calcium chloride and sodium carbonate are mixed.
A: For one, let's write out the compounds so we can better see what we're working with.
Calcium chloride is CaCl2 (because Ca is +2 and Cl is -1, reverse the charges)
Sodium carbonate is Na2CO3 (because Na is +1 and CO3 is -2, reverse)
So from here we can write the first part of our equation, which we know is a precipitation reaction and thus will comply with the rules of metathesis reactions:
CaCl2 + Na2CO3 -------> ???? + ????
Switch the cation-anion combinations to determine the products of the reaction:
CaCl2 + Na2CO3 --------> CaCO3 + NaCl
Balance the molecular equation:
CaCl2 + Na2CO3 ---------> CaCO3 + 2NaCl
Now we can look at the equation and determine what our precipitate is based on our solubility guidelines. CO3^2- is not soluble and Ca^2+ is not an exception in this case, so CaCO3 is our precipitate. We can rewrite the equation now to include solubility:
CaCl2(aq) + Na2CO3(aq) ---------> CaCO3(s) + 2NaCl(aq)
Now separate the ionic components according to solubility:
Ca^2+(aq) + 2Cl-(aq) + 2Na+(aq) + CO3^2- ---------> CaCO3(s) + 2Na+(aq) + 2Cl-(aq)
Looking at the above equation with the ionic component separated, we can determine that Na and Cl are spectator ions in this equation and thus cancel them out:
Ca^2(aq) + CO3^2-(aq) -----------> CaCO3(s)
The result is our net ionic equation and both sides are equal in charge so we are finished!
But oh no, there's much more.
Acids, Bases, and Neutralization Reactions:
Acids: substances that ionize in aqueous solution to form hydrogen ions H+(aq). Because a hydrogen atom consists of a proton and an electron H+ is merely a proton. For this reason, acids are often called "proton donors". Molecules of different acids ionize to form different numbers of H+ ions.
Monoprotic acids will ionize 1 H+ ion, where diprotic acids will ionize 2 H+ ions per molecule of acid. Ionization of diprotic acids occurs in two steps, watch the difference between monoprotic and diprotic:
Monoprotic: HCl(aq) ---------> H+(aq) + Cl-(aq)
Diprotic: H2SO4(aq) ----------> H+(aq) + HSO4-(aq) and then HSO4-(aq) <=====> H+(aq) + SO4^2-(aq)
Note the <===> which denotes a reaction is occurring in both directions.
Acids containing COOH groups, as in CH3COOH (acetic acid), can only give up 1 H+ per COOH group when dissolved in water, the remaining H+ ions do not break their bonds with CH. Acetic acid, for example, only has one COOH group and therefore ionizes 1 H+ in water. Citric acid, on the other hand, has 3 COOH groups and can therefore give up 3 H+ ions.
Bases: substances that accept (react with) H+ ions. Bases produce hydroxide (OH-) compounds. Compounds do not necessarily have to contain OH- ions to be bases. Ammonia (NH3) for example, is a common base. The following represents the reaction that occurs when ammonia is added to water:
NH3(aq) + H2O(l) <====> NH4+(aq) + OH-(aq)
Strong acids and strong bases: completely ionized in solution, also strong electrolytes.
Weak acids and weak bases: partially ionized in solution, also weak electrolytes.
Strong Acids to commit to memory:
Hydrochloric - HCl
Hydrobromic - HBr
Hydroiodic - HI
Chloric - HClO3
Perchloric - HClO4
Nitric - HNO3
Sulfuric - H2SO4
Strong Bases to commit to memory:
Group 1A metal hydroxides: LiOH, NaOH, KOH, RbOH, CsOH
Heavy group 2A metal hydroxides: Ca(OH)2, Sr(OH)2, Ba(OH)2
Neutralization Reactions and Salts:
When a solution of an acid and a solution of a base are mixed, a neutralization reaction occurs. The products of the reaction have neither acidic or basic characteristic properties. Ex:
HCl(aq) + NaOH (aq) ----------> NaCl(aq) + H2O(l)
Acid + Base ---------------> Salt + Water
Salt: any ionic compound whose cation comes from a base and whose anion comes from an acid.
In general, a neutralization reaction between an acid and a metal hydroxide produces water and a salt.
The above reaction can be further broken down into its water soluble strong electrolyte components:
H+(aq) + Cl-(aq) + Na+(aq) + OH-(aq) -----> Na+(aq) + Cl-(aq) + H2O(l)
We can see the spectator ions are Cl- and Na+ and therefore net ionic equation is:
H+(aq) + OH-(aq) ----> H2O(l)
Neutralization reactions between acids and metal hydroxides are metathesis reactions:
Mg(OH)2(s) + 2HCl(aq) ---------> MgCl2 + 2H2O
Mg(OH)2(s) + 2H+(aq) + 2Cl-(aq) -----> Mg^2+(aq) + 2Cl-(aq) + 2H2O
We can remove the spectator ions, in this case just Cl-:
Mg(OH)2(s) + 2H+(aq) -------->Mg^2+(aq) + 2H2O
By this, we see the neutralization of the metal hydroxide Mg(OH)2 by HCl yeilds 2 water molecules and a magnesium chloride salt. We can also see the metathesis reaction where Mg switches anions (OH for Cl) and OH switches cations (Mg for H, making H2O). They're like chemical swingers, heh. heh.
...
Neutralization Reactions with Gas Formation:
So as we saw above, some acid/base neutralization reactions produce water and a salt. Many bases other than OH- react with H+ ions to form molecular compounds. These reactions form gases that have low solubility in water:
2HCl(aq) +Na2S(aq) ------> H2S(g) + 2NaCl
I pre-balanced the equation, but the given species were simply HCl, Na2S, and H2S. This is another metathesis reaction, as cations and anions are switching partners (insert dumb joke). The H2S product is a gas.
Similarly, Carbonates and bicarbonates will react with acids to form CO2 gas. Reaction of CO3^2- first gives carbonic acid (H2CO3), which is unstable and will decompose to H2O and CO2 if concentration if present in sufficient concentrations in the solution:
1.) HCl(aq) + NaHCO3(aq) ---------> NaCl(aq) +H2CO3(aq) *****yeild of carbonic acid
2.) H2CO3(aq) ----------> H2O(l) + CO2(g) *****carbonic acid decomposes to H2O and CO2
3.) H+(aq) + HCO3-(aq) ----------> H2O(l) + CO2(g) ******net ionic equation
Note again in #1 that we're using metathesis reactions again, the cations and anions switch-a-roo.
Let's do a problem to solidify the concept:
Q: By analogy to examples given in the text, predict what gas forms when Na2SO3(s) reacts with HCl(aq)
A: We start by looking at what we're working with, we have
Na2SO3(s) + HCl(aq) ----------> ???? + ????
Remembering the metathesis reaction rules, we switch the anion-cation combinations, giving us the products:
Na2SO3(s) + 2HCl(aq) -----------> H2SO3(aq) + 2NaCl(aq)
We have come to mid-point in the reaction, we have yeilded a diprotic acid (sulfurous acid), which must be further decomposed:
H2SO3(aq) -----------> H2O(l) + SO2(g)
So Na2So3 reacting with HCl will produce sulfur dioxide gas.
Oxidation-Reduction Reactions:
A third type of reaction in which electrons are transferred from one reactant to another are called oxidation-reduction reactions or redox reactions. For the purposes of this section, the focus is put on redox reactions where one of the reactants is a metal in its elemental form.
Corrosion is a familiar concept to us all. Copper turns green, iron rusts; these physical changes we observe represent the chemical reaction taking place that we can't see. Corrosion is the conversion of a metal into a metal compound by a reaction between the metal and some substance in its environment. When a metal corrodes each metal atom loses electrons and forms a cation, which can combine with an anion to form an ionic compound.
When an atom, ion, or molecule becomes more positively charged (loss of electrons), we say it has been oxidized.
Oxidation: loss of electrons by a substance.
When an atom, ion, or molecule becomes more negatively charged (gains electrons), we say it has been reduced.
Reduction: gain of electrons by a substance.
Important! When one substance is oxidized, another substance must be reduced! Oxidation ALWAYS results in reduction and vice versa!
Think about it like this, if you're taking electrons from one atom, you MUST be putting them somewhere. You can't take them and do nothing with them, therefore in reactions, we have one substance being oxidized and another that is correspondingly being reduced. One loses, the other gains. This always happens in redox reactions.
To keep track of our electrons gained and lost, we use oxidation numbers as a sort of bookkeeping system. Oxidation numbers are assigned to each atom in a neutral substance or ion. For monatomic ions the oxidation number is equal to the charge. For neutral molecules and polyatomic ions, the oxidation number of a given atom is a hypothetical charge. This charge is assigned by artificially dividing up the electrons among the atoms in the molecule or ion. The following are the rules for assigning oxidation numbers:
1. For an atom in its elemental form, the oxidation number is always zero. Each H in an H2 atom has an oxidation number of ZERO.
2. For any monatomic ion the oxidation number equals the ionic charge, thus K+ has an oxidation number of +1, S^2- has an oxidation number of -2, etc. Alkali metals always have +1 charge (group 1A), alkaline earth metals (group 2A) always have a +2 oxidation number (corresponding to their charge), and Al is always 3+ (group 3A), notice a pattern here? Groups 1, 2, and 3 have charges corresponding to their group number. Group 1A = +1, Group 2A = +2, and Group 3A = +3.
3. Nonmetals usually have negative odixation numbers, although these can sometimes be positive (Group 4A +/-4 charge).
a) O is usually -2 in both ionic and molecular compounds, with the major exception being peroxides, which contain the O2^2- ion, giving each oxygen an oxidation number of -1 (-1 for each O atom). Makes sense because Oxygen is in group 6A. Group 6A atoms tend to have a -2 charge. Group 5A = -3, Group 7A = -1 (halogens, read on...)
b) The oxidation number of hydrogen is usually +1 when bonded to nonmetals and -1 when bonded to metals.
c) The oxidation number of flourine is -1 in all compounds. The other halogens (see above) have an oxidation number of -1 in most binary compounds, however, when combined with oxygen, as in oxyanions, they have positive oxidation states.
4. The sum of the oxidation numbers of all atoms in a neutral compound is zero. The sum of the oxidation numbers in a plyatomic ion equals the charge of the ion.
Let's work a problem, seeing as how I only have 30 minutes left until I launch out of this box like a hydrogen powered rocket ship of enthusiastic joy for leaving work. Let us rollllll.........
Q: Determine the oxidation number of sulfur in a) H2S, b) S8, c) SCl2
A: We'll rules the previously written rules to determine these oxidation numbers, starting with a).
a) H2S, Sulfur is a nonmetal bonded to hydrogen, so we can refer to rule 3, seciton b to see that hydrogen when bonded to a nonmetal has an oxidation number of +1. We can also see that H2S is neutral, as no charge is indicated for the compound, which means all charges must equal zero. so we can make it into a sort of algebraic equation where our 2 H atoms at +1 charge + x charge will equal zero:
2(+1) + x = 0
+2 + x = 0
x= -2
So we determine that Sulfur in the H2S molecule will have a charge of -2.
b) Looking, we can see that S8 refers back to rule 1, where an atom is in its elemental form. Hence, the oxidation number must be zero.
c) SCl2 is a binary compound therefore the Cl atom must have an oxidation number of -1, and because it is a neutral compound, all charges must add up to zero. So we can make an equation plugging in the values and using x as our variable for the charge of S:
2(-1) + x = 0
-2 + x = 0
x= +2
So we see that S in the compound SCl2 has a +2 oxidaton number.
Let's try a more difficult one: determine the oxidation number of S in Na2SO3
It's a neutral compound so we know all charges will add up to zero. We know that alkali metals always have a charge of +1 in compounds (rule 2), and oxygen always has -2 in compounds, we use the variable x for Sulfur again and form an equation:
2(+1) + 3(-2) + x = 0
+2 -6 + x = 0
-4 + x = 0
x= +4
Sulfur, in the compound Na2SO3, has an oxidation number of +4.
Continuing with sections 4.5 and 4.6 in just a bit............
And we're back!
Oxidation of Metals by Acids and Salts:
The reaction between a metal and either an acid or a metal salt conforms to the general pattern:
A+BX--------->AX + B
Note which species are switching.
EX: Zn(s) + 2HBr(aq) ---------------> ZnBr2 + H2
Displacement Reaction: reactions in which the ion in solution is displaced (replaced) through oxidation of an element.
Many of these displacement reactions occurring between metals and acids produce salt and hydrogen gas. Let's take a closer look at exactly what is happening during this reaction so that we can better understand it:
Mg(s) + 2HCl(aq) --------------> MgCl2(aq) + H2(g)
Here we see a Mg(s) with no charge, which forms the product MgCl2, knowing that Cl is -1 we can see that Mg went from a zero charge on the reactant side to a +2 charge on the product side, this means that it lost 2 electrons. We already know that when a species loses electrons it has been oxidized, so we can say that magnesium in this equation has been oxidized.
Now let's take a look at the rest of the reaction. We have magnesium being oxidized, so some other element must be reduced. Let's lay out the formula with all charges to make this easier to see:
Mg(s) + 2HCl(aq) --------------> MgCl2(aq) + H2(g)
0 +1 -1 ---------------> +2 -1 0
So magnesium went from zero to +2, meaning it was oxidized. Given the equation and charges above, we can see that hydrogen, in the HCl molecule, went from being +1 on the reactant side to zero on the product side, meaning it gained electrons. Because the charge is +1 and there are 2 hydrogen atoms on either side, we can say that it gained 2 electrons to make the charge zero. So Mg loses two electrons and H gains two; Mg is oxidized and H is reduced.
We can further break this chemical reaction down into its net ionic equation by identifying and cancelling out the spectator ions:
Mg(s) + 2H+(aq) + 2Cl-(aq) ----------> Mg^2+(aq) + 2Cl-(aq) + H2(g)
Cancel out the spectator ion Cl-:
Mg(s) + 2H+(aq) ------------> Mg^2+(aq) + H2(g)
Let's do another problem....
Q: Write the balanced molecular and net ionic equations for the reaction of aluminum with hydrobromic acid.
A: To solve, let's first write out the factors we're working with:
Al(s) + HBr(aq) ------------------> ???? + ????
We know that when a metal (such as Al) reacts with an acid (such as HBr), they produce a salt and hydrogen gas, so we find the products:
Al(s) + HBr(aq) --------------> AlBr + H2
Balance the equation out using charges and atomic values
2Al(s) + 6HBr(aq) ----------------> 2AlBr3(aq) + 3H2(g)
Now we have to break up the equation into its ionic components in order to identify and cancel out spectator ions:
2Al(s) + 6H+(aq) + 6Br-(aq) ----------> 2Al^3+(aq) + 6Br-(aq) + 3H2(g)
We identify Br- as the spectator ions and cancel them out, giving us the net ionic equation:
2Al(s) + 6H+(aq) ---------> 2Al^3+(aq) + 3H2(g)
We can look at this equation and see that Al went from a zero charge to plus three, thus Al was oxidized (lost 6 electrons), and H went from a charge of +6 to a zero charge on the product side, thus H was reduced (gained 6 electrons). The number of charges is equal on both sides. It's PERFECT!
And that's all I'm covering out of chapter 4, the rest is solution stoichiometry which was covered extensively in the labs. Two's.