Objectives: Reactions in solution, electrolytes, reaction equations, composition of solutions
Chemical reactions in aqueous solutions
The water molecule has a dipole moment, is a polar molecule, so it attracts charged ions.
Figure 2 (figures are at the end)
The bond angle is 105 degrees.
As long as solubility is concered, the rule of thumb is "like likes like" which means that the polar water dissolves best charged ions and polar molecules:
H2O + - NH4NO3(s) NH4 (aq) + NO3 (aq) aq stands for the ion being in aqueous solution and being hydrated. hydration: the charges on the ions orient the water dipoles around it, so that the + side of the dipole is close to an anion, a negative ion, and that the - side of the dipole is close to a cation, a positive ion.
The solubility depends on if the attraction between the ions in the solid is smaller or larger than the ion - water attraction.
Nature of aqueous solutions
Figure 3 (figures are at the end)
Many ionic solids are strong electrolytes, some are weak electrolytes and no ionic solid is a non-electrolyte (like sugar, which is a polar molecule).
Strong electrolytes: completely ionized (dissociated into ions) in solution
H2O example: NaCl(s) Na+(aq) + Cl-(aq) reaction to 100% complete no NaCl molecules in the solution!
1. Soluble salts: NaCl, BaCl2 and many others
2. Strong acids: An acid is a substance that produces H+(aq) ions in solution, like e.g. HCl, HNO3, H2SO4, HClO4
H2O HCl H+(aq) + Cl-(aq)
H2O + - H2SO4 H (aq) + HSO4 (aq)
The hydrogen sulfate dissociates further:
H2O - + 2- HSO4 H (aq) + SO4 (aq)
3. strong bases: A substance that produces OH-(aq) ions in solution e.g. NaOH, KOH, Ba(OH)2
4. Also some hardly soluble electrolytes like BaSO4 are strong electrolytes:
Only very little is dissolved, but also only ions in solution, no molecules Weak electrolytes: Do not ionize completely when dissolved
Example: acetic acid: CH3 - COOH:
Figure 4 (figures are at the end)
It is a weak electrolyte and thus we have a dissociation equilibrium where there are also molecules in the solution.
Because it is a weak electrolyte it is also a weak acid.
Other ways of writing:
H2O - + C2H3O2H(aq) <=> C2H3O2 (aq) + H (aq)
H2O HAc(aq) <=> H+(aq) + Ac-(aq) (<=> denotes the double arrow of equilibrium)
In an 0.1 M (mol solute per L solution) solution the degree of dissociation is only 1%
That means, in equilibrium there are 99 % HAc molecules in the solution but only 1% of the acetic acid is dissociated into ions. So for each 100 HAc molecules dissolved, there are 99 undissociated molecules and only 1 dissociated ion pair.
Another weak electrolyte is ammonia, NH3, which therefore is a weak base:
+ - NH3 + H2O <=> NH4 (aq) + OH (aq)
Non-electrolytes: only molecules in the solution, no ions:
H2O sugar: C12H22O11 C12H22O11(aq)
H2O ethanol: C2H5OH C2H5OH(aq) Composition of solutions
Type and concentrations (molarity, M, for example) of solutions? number of moles of solute (mol) molarity = volume of the solution in liter (L = dm3 )
n mol m _M = ( ) n = V L MM
m : solute mass , MM : molar mass of the solute
Examples:
Calculate the molarity, M, of a solution prepared by dissolving 15.0 g of
NaOH in water to give 905 mL solution.
MM = (22.99 + 16.00 + 1.008) g/mol = 40.00 g/mol 15.0 g/(40.0 g/mol) mol M = = 0.414 = 0.414 M 0.905 L L
Give the concentrations of each ion in the following solutions:
(a) 0.1 M ZnCl2 (b) 1 M Fe2(SO4)3
H2O 2+ - (a) ZnCl2 Zn (aq) + 2 Cl (aq)
2+ 2+ 1 mol Zn 2+ [ Zn ] = 0.1 M ZnCl 2 = 0.1 M Zn 1 mol ZnCl 2
- - 2 mol Cl - [Cl ] = 0.1 M ZnCl 2 = 0.2 M Cl 1 mol ZnCl 2
H2O 3+ 2- (b) Fe2(SO4)3 2 Fe (aq) + 3 SO4 (aq) (6 positive and 6 negative charges)
3+ 2- it follows: in 1 M Fe2(SO4)3 solution we have 2M Fe and 3M SO4 ions
- Calculate the number of moles of NO3 (nitrate) ions in 420 mL of a 0.010 M solution of Co(NO3)3.
M = n/V and thus n = MV, where n is the number of moles of the solute
-2 -3 n = 1.0 x 10 mol/L x 0.420 L = 4.2 x 10 mol Co(NO3)3 are in the 420 mL of
3+ - solution. Since 1 mol of Co(NO3)3 dissociates into 1 mol of Co and 3 mol of NO3 ions, the number of moles of nitrate ions in the solution is 3n: n(nitrate) = 3 n(cobalt nitrate) = 3 x 4.2 x 10-3 mol = 12.6 x 10-3 mol
= 1.3 x 10-1 mol nitrate (2 significant figures)
Preparation of a solution of given molarity by dilution
500 mL of a 0.100 M NaOH solution contains n = MV = 0.100 mol/L x 0.500 L = 0.0500 mol NaOH corresponding to 0.0500 mol x 40.00 g/mol (MM) = 2.000 g NaOH
Dilution: Adding of more solvent to a solution, making the molarity smaller.
Example: How many mL of a 0.250 M solution of NaOH must be used to prepare 500 mL of a 0.100 M solution of NaOH?
Figure 5 (figures are at the end)
final (index 2) molarity and volume are 0.100 M and 500 mL, initial (1) molarity is 0.250 M. Since M1V1 = M2V2 the volume V1 of the 0.250 M solution we need to dilute to 500 mL is
M 2V 2 0.100 M 500 mL V 1 = = = 200 mL M 1 0.250 M
So we must add 300 mL of water to the 200 mL 0.250 M solution to obtain 500 mL
0.100 M solution (300 mL water is only correct if the densities of the solutions and the water are all the same). Objectives: Reaction equations in solutions, types of reactions in solutions, precipitation reactions
Reaction equations in solutions
When we combine a solution of potassium chromate with a solution of barium nitrate a yellow solid separates out (precipitation).
For ionic reactions like that there are three types of equations possible.
Molecular equations
Write the ionic compounds as if they are molecules, but put an (s) for the solid
(insoluble) precipitate (the ionic compounds must always be written as charge neutral) and we put (aq) at each electrolyte that is in solution:
K2CrO4(aq) + Ba(NO3)2(aq) BaCrO4(s) + 2 KNO3(aq)
The yellow precipitation is the barium chromate which is insoluble.
Complete ionic equations
Write each hydrated ion extra from the molecular equation and keep the insoluble solid together:
+ 2- 2+ - + - 2K (aq) + CrO4 (aq) + Ba (aq) + 2NO3 (aq) BaCrO4(s) + 2 K (aq) + 2 NO3 (aq)
In this complete ionic equation ions are included which do not take part in the reaction.
These ions are called "spectator ions", because they just swim around and do nothing.
Since they occur unchanged on the left and on the right hand side, they can be subtracted und thus cancelled from the equation which gives the last kind of equation. Net ionic equation
Here you take the complete ionic equation and cancel out all spectator ions that appear unchanged on both sides:
2+ 2- Ba (aq) + CrO4 (aq) BaCrO4(s)
Write net ionic equations for the reactions that occur when you combine the following solutions
(a) silver nitrate (AgNO3) solution with barium chloride (BaCl2) solution
(b) potassium nitrate (KNO3) solution with barium chloride (BaCl2) solution
(c) lead nitrate (Pb(NO3)2) solution with sodium sulfate (Na2SO4) solution
(d) potassium hydroxide (KOH) solution with iron(III) nitrate (Fe(NO3)3) solution
(a) molecular: 2 AgNO3(aq) + BaCl2(aq) 2 AgCl(s) + Ba(NO3)2(aq)
+ - 2+ - complete ionic: 2 Ag (aq) + 2 NO3 (aq) + Ba (aq) + 2 Cl (aq) 2 AgCl(s) +
2+ - + Ba (aq) + 2 NO3 (aq) net ionic: 2 Ag+(aq) + 2 Cl-(aq) 2 AgCl(s) divide by 2 to get the final result:
Ag+(aq) + Cl-(aq) AgCl(s)
(b) since KCl and Ba(NO3)2 are completely soluble, nothing happens
2+ 2- (c) Pb (aq) + SO4 (aq) PbSO4(s)
3+ - (d) Fe (aq) + 3 OH (aq) Fe(OH)3(s)
Types of chemical reactions in solutions
Among the many different kinds of possible reactions in solutions, 3 groups are the most important ones:
1. Precipitation reactions: an insoluble salt precipitates as solid out of a solution 2. Acid - base reactions: neutralization of an H+(aq) ion from the acid with an OH-
+ - (aq) ion from the base: H (aq) + OH (aq) H2O(l): neutralization
3. Redox (reduction-oxidation) reactions
Precipitation reactions
Appearence of an insoluble salt that falls as a solid out of the solution
When you mix a potassium chromate solution with barium nitrate solution, then you know already that potassium chromate and barium nitrate are soluble and cannot precipitate out.
So you only have to check if the remaining possible combinations of ions in the combined solution is soluble or not.
Here the two other combinations are potassium nitrate (which is soluble) and barium chromate, BaCrO4, (which is insoluble) and precipitates as a yellow solid.
K2CrO4(aq) + Ba(NO3)2(aq) BaCrO4(s) + 2 KNO3(aq)
+ 2- 2+ - So the possible ions to combine are K , CrO4 , Ba , and NO3
K2CrO4 and Ba(NO3)2 cannot precipitate, because we combine their solutions.
Further possible are KNO3, which is soluble and does not precipitate, and BaCrO4
(which is insoluble) and thus does precipitate.
Other example:
AgNO3(aq) + KCl(aq) AgCl(s) + KNO3(aq)
The precipitation is the white, insoluble solid AgCl. To answer such questions, there are the solubility rules, which unfortunately must be memorized:
- 1. NO3 : all nitrates are soluble
- - - 2. Cl , Br , I : all chlorides are soluble, except AgCl(s), Hg2Cl2(s) and PbCl2(s) which
are insoluble
2- 3. SO4 : all sulfates are soluble, except SrSO4(s), BaSO4(s) and PbSO4(s), which are insoluble
2- 2- 3- 4. CrO4 , CO3 , PO4 : all carbonates and phosphates are insoluble, except those of + element group I and of NH4 (ammonium) ions which are soluble
5. OH-: all hydroxides are insoluble, except those of element group I and those of the
+ 2+ 2+ NH4 , Ba and Sr ions; Ca(OH)2 is just slightly soluble
2- + 6. S : all sulfides are insoluble, except those with group I, group II and NH4 ions
Combination of KNO3(aq) with BaCl2(aq) gives a solution with the following free
+ - 2+ - ions: K (aq) + NO3 (aq) + Ba (aq) + Cl (aq)
- + 2+ 2+ Table: all nitrates and chlorides are soluble, exceptions for Cl : Ag , Pb , Hg2 but they are not present. Thus no precipitation reaction here because all possible combinations of cations with anions are soluble.
Combination of Na2SO4(aq) with Pb(NO3)2(aq) gives a solution with the following free + 2- 2+ - ions: Na (aq) + SO4 (aq) + Pb (aq) + NO3
Na2SO4(aq) and Pb(NO3)2(aq) are in solution from the beginning and also the table tells that they are soluble. Also the table tells that all nitrates are soluble. But PbSO4 is told to be insoluble. So we get a precipitation of PbSO4(s) when we mix the solutions and NaNO3 stays in solution.
Molecular equation: Na2SO4(aq) + Pb(NO3)2(aq) PbSO4(s) + 2 NaNO3(aq) + 2- 2+ - Complete ionic equation: 2 Na (aq) + SO4 (aq) + Pb (aq) + 2 NO3 (aq) PbSO4(s)
+ - + 2 Na (aq) + 2 NO3 (aq)
2+ 2- Net ionic equation: Pb (aq) + SO4 (aq) PbSO4(s) (cancel the spectator ions)
Combination of KOH(aq) with Fe(NO3)3(aq) gives a solution with the following free
+ - 3+ - ions: K (aq) + OH (aq) + Fe (aq) + NO3 (aq)
KOH(aq) and Fe(NO3)3(aq) are in solution from the beginning and also the table tells that they are soluble. Also the table tells that all nitrates are soluble. But
Fe(OH)3 is told to be insoluble. So we get a brown precipitation of Fe(OH)3(s) when we mix the solutions and KNO3 stays in solution.
Molecular equation: 3 KOH(aq) + Fe(NO3)3(aq) Fe(OH)3(s) + 3 KNO3(aq)
+ - 3+ - Complete ionic equation: 3 K (aq) + 3 OH (aq) + Fe (aq) + 3 NO3 (aq)
+ - Fe(OH)3(s) + 3 K (aq) + 3 NO3 (aq)
3+ - Net ionic equation: Fe (aq) + 3 OH (aq) Fe(OH)3(s)
Stoichiometry in precipitation reactions
Calculate the mass of BaCrO4(s) formed when 200 mL of a 0.650 M solution of
K2CrO4(aq) is precipitated completely with a Ba(NO3)2(aq) solution. Use the net ionic equation for this.
2+ 2- Ba (aq) + CrO4 (aq) BaCrO4(s)
The molar weights are g MM 2- = MM Cr + 4 MM O = 116.00 CrO4 mol
g MM Ba = MM 2+ + MM 2- = 253.30 CrO4 Ba CrO4 mol
+ 2- In a K2CrO4 solution we have the dissociation K2CrO4(aq) 2 K (aq) + CrO4 (aq) 2- and thus the number of moles (n) of CrO4 is mol n = n 2- = M Cr V = 0.650 0.200 L = 0.130 mol K 2CrO4 CrO4 K 2 O4 L
= nBaCrO4
Then the mass (m) of the precipitated BaCrO4 is g mBaCr = n MM = 0.130 mol 253.30 = 32.9 g O4 BaCrO4 BaCrO4 mol
2+ For analysis: Take your unknown solution of Ba ions and precipitate with K2CrO4 solution of known concentration until all BaCrO4 is precipitated.
Filtrate, wash, dry and weigh the precipitation and you can calculate how much Ba2+ was in the original solution.
Important, because Ba2+ is a very dangerous poison.
Limiting reactant
Calculate the mass of Fe(OH)3(s) formed when 50.0 mL of a 0.120 M solution of
Fe(NO3)3 is reacted with 120. mL of a 0.100 M solution of KOH.
3+ - The net ionic equation is Fe (aq) + 3 OH (aq) Fe(OH)3(s)
The masses are: 1 mol Fe3+(aq): 55.85 g
3 mol OH-(aq): 3 mol x (16.00 + 1.008) g/mol = 51.03 g
1 mol Fe(OH)3(s): 106.88 g
Thus the number of moles (n) of Fe3+ is
mol -3 n 3+ = M 3+ V = 0.120 0.050 L = 6.0 10 mol Fe Fe L From the net ionic equation we can get the unit factor that gives the amount of
3+ Fe(OH)3 that would be formed if all Fe can react: 1 mol Fe(OH ) 6.0 10-3 mol Fe3+ 3 = 6.00 10-3 mol Fe(OH ) 1 mol Fe3+ 3 So, if ALL Fe3+ present in our solution can precipitate (enough OH- ions), we would
-3 get 6.00 x 10 mol Fe(OH)3(s).
In a similar way we get the amount of OH- ions present:
mol -3 n - = M KOH V = 0.100 0.120 L = 12.0 10 mol OH L
- and how much Fe(OH)3(s) can precipitate, if all OH can be used: 1 mol Fe(OH ) 12.0 10-3 mol OH - 3 = 4.00 10-3 mol Fe(OH ) 3 mol OH - 3
-3 - Only 4.00 x 10 mol Fe(OH)3(s) can be precipitated when ALL OH ions present are used and thus some of the Fe3+ ions stay in solution after precipitation.
Thus OH- is the limiting reactant, because there are not enough OH- ions there to precipitate all the Fe3+ ions.
Thus the mass of Fe(OH)3(s) obtained in the reaction is
-3 g Fe(OH )3 4.00 10 mol Fe(OH )3 106.88 = 0.428 g Fe(OH )3 mol Fe(OH )3 Objectives: Acid-base reactions and titration
Acid-Base reactions
:Bröّnsted-Lowry
An acid is a proton, H+, donor
A base is a proton acceptor, where OH- (hydroxide ion) is a very good one
Strong acid and strong base
strong acids: HCl, HNO3, H2SO4, HClO4
strong bases: NaOH, KOH, RbOH (group I), Ba(OH)2
Acids or bases which are strong electrolytes (100 % ionized) are strong acids and bases. strong acid + strong base gives a salt which is a strong electrolyte
Molecular equation:
HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l) here all elctrolytes are 100 % ionized. Complete ionic equation:
+ - + - + - H (aq) + Cl (aq) + Na (aq) + OH (aq) Na (aq) + Cl (aq) + H2O(l)
Cancellation of the spectator ions (Na+(aq) and Cl-(aq)): net ionic equation:
+ - H (aq) + OH (aq) H2O(l) Neutralization
Weak acids and strong bases
weak acids, e.g.: HC2H3O2, acetic acid, HF(aq), hydrofluoric acid (hydrogen fluoride,
HF(g), in water) or H2C2O4, oxalic acid weak acid + strong base gives also a salt which is a strong electrolyte. Acetic acid,
HAc, is a weak electrolyte which is to over 90 % undissociated. However, the salt formed in the reaction, sodium acetate, NaAc, is again a strong electrolyte and 100
% dissociated.
The weak acid is not 100 % dissociated, but the neutralization reaction happens completely, taking all H+ out of equilibrium. To keep the dissociation equilibrium more weak acid must dissociate and so on, until all is dissociated and thus a salt which is a strong electrolyte is formed by the neutralization.
Molecular equation:
HC2H3O2(aq) + NaOH(aq) NaC2H3O2(aq) + H2O(l)
or: HAc(aq) + NaOH(aq) NaAc(aq) + H2O(l)
Because the weak acid is not dissociated, in the complete ionic equation we write it as undissociated acid molecule in solution (aq):
+ - + - HC2H3O2(aq) + Na (aq) + OH (aq) Na (aq) + C2H3O2 (aq) + H2O(l)
+ - + - or: HAc(aq) + Na (aq) + OH (aq) Na (aq) + Ac (aq) + H2O(l)
Thus for the net ionic equation only the spectator ion Na+(aq) can be cancelled:
- - HC2H3O2(aq) + OH (aq) C2H3O2 (aq) + H2O(l)
- - or: HAc(aq) + OH (aq) Ac (aq) + H2O(l)
Strong acid and weak base
weak bases e.g.: NH3, ammonia, C2H5NH2, ethyl amine Molecular equation:
HCl(aq) + NH3(aq) NH4Cl(aq)
In this example no water is formed, but the strong electrolyte ammonium chloride, NH4Cl.
Complete ionic equation:
+ - + - H (aq) + Cl (aq) + NH3(aq) NH4 (aq) + Cl (aq)
+ NH4 is called the ammonium ion.
Net ionic equation (cancel Cl-(aq)):
+ + H (aq) + NH3(aq) NH4 (aq)
Weak acid and weak base
+ - HC2H3O2(aq) + NH3(aq) NH4 (aq) + C2H3O2 (aq)
HAc is a weak acid and NH3 a weak base, but ammonium acetate, NH4Ac is again a strong electrolyte, because again the neutralization (here not involving water) goes to completion.
Acid - Base titrations
This is called volumetric analysis, which can be also used for precipitation reactions and redox reactions. titration: from a buret a measured volume of a solution of known concentration (the titrant) is dropped into a solution that has to be analyzed (analyte). the titrant must be chosen such, that it reacts fast, complete and in a known manner with the analyte. When all analyte is used up, that point in the titration is called the endpoint, the stoichiometric point, and also the equivalence point.
An indicator has to be added to the analyte which changes e.g. the color of the solution, when additional titrant is added which cannot react anymore, because the analyte is used up.
If the analyte is a base: the titrant must be a strong acid
If the analyte is an acid: the titrant must be a strong base
An indicator can be for example phenolphthalein which has no color in acidic solution but becomes pink in basic solution.
If an acid is titrated with a strong base as titrant then the solution is colorless before the endpoint, because all OH-(aq) dropped in reacts with H+(aq) to water.
After the endpoint all H+(aq) in the analyte has reacted to water and when more OH-
(aq) from the titrant is added, the solution gets basic, and changes its color to pink.
Standardization of a NaOH solution
1.3009 g of KHP (KHC8H4O4, potassium hydrogenphthalate, that has 1 acidic proton; the H4 are not acidic) which has a molar mass of MMKHP = 204.22 g/mol is dissolved in water and gives an acidic solution.
This acidic solution has a well defined [H+] (square brackett for molarity), because we know there is 1 acidic proton per KHP molecule and the mass of KHP dissolved as well as its molar mass are known. + + A good measure for [H ] is the pH value of a solution: pH = - log10 ([H ]/M)
Meaning: take [H+] in units of molarity (M), remove the unit and take the neagtive decadic
log of the number: pH < 7: acidic, pH > 7: basic, pH = 7: neutral (pure water)
If we have a NaOH solution of unknown concentration, we can use it to titrate the
KHP solution to obtain the concentration of our titrant. This is called standardization of a NaOH solution.
With 41.20 mL NaOH solution the endpoint of the titration is reached (change of the color of the KHP solution with phenolphthalein from colorless to pink). Thus all H+ ions from the 1.3009 g KHP are consumed.
What is the molarity MNaOH of the NaOH solution?
KHP is an electrolyte and dissociates when dissolved:
H2O + - KHC8H4O4 K (aq) + HC8H4O4 (aq)
- HC8H4O4 (aq) is an acid where the first H is acidic the H4 part not.
Thus when OH-(aq) is added to the solution neutralization happens:
- - 2- HC8H4O4 (aq) + OH (aq) H2O(l) + C8H4O4 (aq)
Thus we know that in 41.20 mL of the NaOH solution (titrant) there are as many OH- ions as there are H+ ions coming from 1.3009 g of KHP.
Number of moles of KHP: 1 mol KHP 1.3009 g KHP = 6.3710 10-3 mol KHP 204.22 g KHP
Thus 6.3701 x 10-3 mol of OH- ions are needed to neutralize all the 6.3710 x 10-3 mol of H+ ions from the KHP.
Thus at the endpoint 6.3170 x 10-3 mol NaOH are in 41.20 mL NaOH solution:
6.3710 10-3 mol NaOH M NaOH = = 0.1546 M 4.120 10-2 L solution
Now the NaOH standard solution can be used to titrate other unknown acidic analyte solutions.
Example:
Waste from industry contains CCl4 and benzoic acid, HC7H5O2 (carcinogenic weak acid with 1 acidic proton).
0.3518 g of waste is shaken with water (to get the acid from the CCl4 phase into the water) and in the titration 10.59 mL of 0.1546 M standard NaOH solution was needed to reach the endpoint. How much % weight of benzoic acid is in the aqueous solution?
- - HC7H5O2(aq) + OH (aq) H2O(l) + C7H5O2 (aq) moles of OH- ions needed to react all the protons from benzoic acid: 1 L NaOH(aq) 0.1546 mol NaOH 10.59 mL NaOH(aq) 1000 mL NaOH(aq) 1 L NaOH(aq)
-3 -3 = 1.637 10 mol NaOH = 1.637 10 mol HC7 H 5 O2
Molar weight, MM, of benzoic acid, HC7H5O2(aq):
MM = (1.008 + 7 x 12.01 + 5 x 1.008 + 2 x 16.00) g/mol = 122.12 g/mol mass of benzoic acid:
-3 g 1.637 10 mol 122.12 = 0.1999 g HC7 H 5 O2 mol Since we used 0.3518 g of waste, the % weight of benzoic acid is
0.1999 g % weight = 100 % = 56.82 % 0.3518 g
Example:
An acetic acid, HAc, solution is titrated with a 0.100 M Ba(OH)2 (a strong base) solution using phenolphthalein as indicator. The endpoint was reached after adding
39.4 mL of the Ba(OH)2 solution to the analyte. How many g HAc was there in the analyte?
moles of Ba(OH)2 needed to reach the endpoint:
39.4 mL x 10-3 L/mL x 0.100 mol/L = 3.94 x 10-2 L x 0.100 mol/L = 3.94 x 10-3 mol
Ba(OH)2 reaction equation (neutralization):
Ba(OH)2(aq) + 2 HAc(aq) 2 H2O(l) + Ba(Ac)2 thus (HAc: HC2H3O2, 1 acidic proton)
-3 2 mol HAc 3.94 10 mol Ba(OH )2 1 mol Ba(OH )2
= 7.88 10-3 mol HAc
MMHAc = (4 x 1.008 + 2 x 12.01 + 2 x 16.00) g/mol = 60.372 g/mol mass of HAc in the analyte:
-3 mHAc = 60.372 g/mol x 7.88 x 10 mol = 0.476 g HAc
Example:
A phosphoric acid, H3PO4, solution is titrated with a 0.100 M Ba(OH)2 (a strong base) solution using phenolphthalein as indicator. The endpoint was reached after adding 49.8 mL of the Ba(OH)2 solution to the analyte. How many g phosphoric acid was there in the analyte?
moles of Ba(OH)2 needed to reach the endpoint:
49.8 mL x 10-3 L/mL x 0.100 mol/L = 4.98 x 10-2 L x 0.100 mol/L = 4.98 x 10-3 mol
Ba(OH)2 reaction equation (neutralization), H3PO4 has 3 acidic protons:
3 Ba(OH)2(aq) + 2 H3PO4(aq) 6 H2O(l) + Ba3(PO4)2
Rule: similar to the writing of the formula for a charge neutral electrolyte: coefficient (index)
2+ at the cation is the number of the negative charges on the anion, (Ba )3, and the coefficient
3- at the anion is the number of positive charges on the cation, (PO4 )2.
Here: the stoichiometric factor at the base is the number of acidic protons in the acid,
- 3 Ba(OH)2, that at the acid is the number of hydroxide, OH , ions in the base, 2 H3PO4 thus:
-3 2 mol H 3 PO4 4.98 10 mol Ba(OH )2 3 mol Ba(OH )2
-3 = 3.32 10 mol H 3 PO4 molar mass of H3PO4: MM = (3 x 1.008 + 30.97 + 4 x 16.00) g/mol =
97.994 g/mol mass of H3PO4 in the analyte:
-3 m(H3PO4) = 97.994 g/mol x 3.32 x 10 mol = 0.325 g H3PO4 Objectives: Balancing of Redox reactions
Examples for redox reactions:
+ - 2 Na(s) + Cl2(g) 2 NaCl(s) (solid having Na and Cl ions)
+ 2+ Zn(s) + 2 H (aq) Zn (aq) + H2(g)
These redox reactions contain reduction (gain of electrons) and oxidation (loss of electrons) parts.
Separate oxidations (loss of electrons):
Zn Zn2+ + 2 e-
2 Na 2 Na+ + 2 e-
The reactant that is oxidized, Zn, Na, is called the REDUCING AGENT, because it reduces an other reactant in the reaction.
Separate reductions (gain of electrons):
+ - 2 H + 2 e H2
- - Cl2 + 2 e 2 Cl
+ The reactant that is reduced, H , Cl2, is called the OXIDIZING AGENT, because it oxidizes an other reactant in the reaction.
oxidations and reductions occur ALWAYS together, because there are no free electrons flying or swimming around in chemistry.
When a reactant gives up electrons there must be always another one who takes them.
Thus we do not observe reductions or oxidations alone, but ALWAYS both together in form of REDOX reactions, where the final reaction equation does never contain electrons.
When we combine the above oxidation with the corresponding reduction equation there is no net change in the number of electrons, because they always occur together.
To balance redox equations, first we must identify oxidation and reduction part.
For that we must introduce the concept of oxidation numbers.
Rules:
1. The oxidation numbers of the atoms in a pure element are always 0:
0 0 0 H2, Cl2, Fe
2. The oxidation numbers of mono-atomic ions are equal to the charge:
+1 +2 -1 Na+, Ba2+, Cl-
3. The oxidation numbers of an element in a compound
+1-1 group I (alkaline) element: always +1 (except H, can be -1): NaCl
Note: the sum of all oxidation numbers in a compound must be equal to the outside charge (when there is AB2, then the oxidation number of B counts twice in the sum).
+2-1 group II (earth alkaline) element: always +2: BaCl2
+2-2 Oxygen, O, in a compound mostly has the oxidation number -2: CaO,
-1 +2 except in peroxides, where it is -1: H2O2, and in oxygen fluoride where it is +2: OF2
Fluorine has -1 in all molecules, except F2.
+1 +1 Hydrogen has in compounds mostly the oxidation number +1, H2SO4, HF, -1 except in hydrides (matals and hydrogen), where it has -1, like in NaH.
4. The sum of the oxidation numbers of all atoms in a molecule must be equal to the charge of the molecule.
Oxidation = increase in oxidation number
Reduction = decrease in oxidation number
Oxidation numbers in CaSO4? the oxidation number of Ca is +2, that of O -2 and that of S is x (unknown). the charge is 0 and therefore 1 x Ca + 1 x S + 4 x O = 0: 2 + x - 8 = x - 6 = 0 and thus the oxidation number of S, x = + 6.
- Oxidation numbers in MnO4 ? the oxidation number of O is -2 and that of Mn is x (unknown). the charge is -1 and therefore 1 x Mn + 4 x O = -1: x - 8 = -1 and thus the oxidation number of Mn, x = + 7.
When you write an oxidation or a reduction equation alone, then the charges on the left and right side of the equation must be the same and it is balanced by electrons: oxidation: oxidation number of Zn increases from 0 to +2:
Zn Zn2+ charge on the left: 0, charge on the right +2: add 2 e- on the right and the charge is balanced: Zn Zn2+ + 2 e- reduction: oxidation number of H decreases from +1 to 0: + - 2 H H2 charge on the left: +2, charge on the right 0: add 2 e on the left and the
+ - charge is balanced: 2 H + 2 e H2
Types of Redox reactions
1. Combination reactions: A + B C
example: S + O2 SO2 oxidation number of oxygen decreases from 0 in the element to -2 in the compound (reduction), that of sulfur increases from 0 in the element to
+4 in the compound (oxidation).
2. Decomposition reactions: C A + B in KClO3 the oxidation number of K is +1 (group I), that of O is -2 and that of Cl is x: 1 + x - 6 = x - 5 = 0 (charge). So the oxidation number of Cl, x = +5.
Decomposition: 2 KClO3 2 KCl + 3 O2. The oxidation number of Cl decreases from
+5 to -1 in KCl (reduction) and that of O increases from -2 to 0 in the element.
3. Displacement reactions: A + BC AC + B
a) Hydrogen displacement: 2 Na + H2O 2 NaOH + H2 sodium, Na, is oxidized from 0 in the element to +1 in NaOH and hydrogen, H, is reduced from +1 in water to 0 in the element.
b) Metal displacement: Zn + CuSO4 Cu + ZnSO4, Zn is oxidized from 0 in the element to +2 in the salt, Cu is reduced from +2 in the salt to 0 in the element.
2- salt: SO4 , thus the metals must be 2+ cations.
Cu(s) + 2 Ag+(aq) Cu2+(aq) + Ag(s): Cu is oxidized, silver reduced.
Always: the more noble metal is reduced from the cation to the metal, the less noble metal is dissolved, i.e. oxidized from the metal to the cation. c) Halogen displacement: Cl2 + 2 KBr 2 KCl + Br2 rules:
F2 displaces Cl2, Br2, I2 from its compounds
Cl2 displaces Br2, I2 from its compounds
Br2 displaces I2 from its compounds
The halogen higher up in the group displaces all the lower ones.
4. Disproportionation reactions: in basic medium:
- - - Cl2 +2 OH Cl + ClO + H2O one element is both oxidized and reduced.
Here: one Cl is reduced from 0 in the element to -1 in Cl- and the other Cl is oxidized from 0 in the elment to +1 in hypochlorite, ClO-.
To get the right balance, you write oxidation and reduction separate, and after the material balance is finished (same no. of atoms of one kind left and right) you balence the charges with electrons.
Then you add the two equations such that all electrons cancel out:
When the reduction involves x electrons and the oxidation involves y electrons, then you multiply the reduction equation by y and the oxidation equation by x and add them up. Then you get xy electrons left and right and they cancel out.
- - Reduction: (1/2) Cl2 + e Cl : Cl from 0 to -1 - - - Oxidation: (1/2) Cl2 + 2 OH ClO + H2O + e : Cl from 0 to +1 you must add 2 OH- on the left because you need 1 O for ClO- and 1 O for water on the right. Since both equations have 1 e-, you can directly add them up and cancel the
- - - electron to get the redox equation: Cl2 + 2 OH ClO + Cl + H2O
Balancing is different in acidic or in basic medium. Acidic medium is more easy, because there you can simply use H+ ions and water molecules to balance the oxygen.
3+ - Example: balance Cr (aq) + Cl (aq) Cr(s) + Cl2(g) in acidic medium
As first step you balance the two half reactions (having electrons makes it easier):
Reduction: Cr3+(aq) Cr(s) Cr from +3 to 0 you have 3 positive charges left and 0 charge right, so add 3 e- on the left to get 0 charge on both sides, the atom balance is already ok: Cr3+(aq) + 3 e- Cr(s)
- Oxidation: Cl (aq) Cl2(g) Cl from -1 to 0
- For the material balance you need 2 Cl on the left, because you have Cl2(g) on the right.
Since now you have 2 negative charges on the left, you add 2 e- on the right to get -2
- - on both sides: 2 Cl (aq) Cl2(g) + 2 e
To get the final redox equation you must have the same number of e- on both sides.
Here you have 3 e- in the reduction and 2 e- in the oxidation, so you multiply the reduction by 2 and the oxidation by 3 to get 6 e- on both sides when you add them:
Reduction x 2: 2 Cr3+(aq) + 6 e- 2 Cr(s)
- - Oxidation x 3: 6 Cl (aq) 3 Cl2(g) +6 e So in the redox equation as redox = reduction x 2 + oxidation x 3 the 6 e- cancel as
3+ - they must: 2 Cr + 6 Cl (aq) 2 Cr(s) + 3 Cl2(g)
Now the atom and charge balance is fine and there are no electrons in the final equation.
- 2+ 2+ 3+ Example: balance MnO4 (aq) + Fe (aq) Mn (aq) + Fe (aq) in acidic medium remember that Mn has the oxidation number +7 in permanganate: give Mn the
- oxidation number x, each O has -2, and MnO4 has 1 negative charge: x - 8 = -1 and the oxidation number x of Mn is x = +7. Thus the reduction involves Mn going from 2+ - 2+ +7 to +2 in Mn : MnO4 Mn
- + To balance the O, we form for each O in MnO4 one H2O, by adding 8 H of the left
+ + (each O needs 2 H to form H2O): So add 8 H on the left and form 4 H2O on the
- + 2+ right: MnO4 (aq) + 8 H (aq) Mn (aq) + 4 H2O(l)
Now we have a charge of 8 - 1 = +7 on the left and +2 on the right. So when we add 5 e- on the left we get a charge of +2 on both sides and the reduction half
- + - 2+ reaction is ok: MnO4 (aq) + 8 H (aq) + 5 e Mn (aq) + 4 H2O(l)
The oxidation Fe2+(aq) Fe3+(aq), Fe goes from +2 to +3. The balance is simple, because the atoms are already ok, and we only need to add 1 e- on the right to make the charge +2 on both sides: Fe2+(aq) Fe3+(aq) + 1 e-.
Since we have 5 e- in the reduction half reaction and 1 e- in the oxidation we must add the reduction to 5 x oxidation and cancel the 5 e- from both sides:
- + 2+ 3+ 2+ MnO4 (aq) + 8 H (aq) + 5 Fe (aq) 5 Fe (aq) + Mn (aq) + 4 H2O(l)
- - - Example: balance MnO4 (aq) + I (aq) IO3 (aq) + MnO2(s) in basic medium Simple rule for balancing in basic medium: Do all steps to get the balanced half equations like if the reaction would be in the acidic medium and then add to the half reaction equations on both sides of each one 1 OH- for each H+ and combine them to
H2O. Finally cancel the H2O molecules occurring on both sides and add the half reaction equations again such that the electrons cancel out.
- Mn is +7 in MnO4 again. In MnO2 we give x to Mn and each O has -2. Since there is no charge we have x - 4 = 0 and thus the oxidation number x of Mn is x = 4. So
- Mn is reduced from +7 to +4: MnO4 (aq) MnO2(s).
Do like if you are in acidic medium, so to balance the oxygen, you must add 4 H+ on
- the left to change 2 O in MnO4 to 2 H2O:
- + MnO4 (aq) + 4 H (aq) MnO2(s) + 2 H2O(l); now with a charge of +3 on the left and 0 on the right we must add 3 e- on the left to get 0 charge on both sides:
- - + 3 e + MnO4 (aq) + 4 H (aq) MnO2(s) + 2 H2O(l)
Because we are in basic medium (no H+) we must add on both sides an OH- ion for each H+ ion, thus here 4 OH- ions:
- - + - - 3 e + MnO4 (aq) + 4 H (aq) + 4 OH (aq) MnO2(s) + 2 H2O(l) + 4 OH (aq)
+ - Now on the left we make 4 H2O out of 4 H + 4 OH :
- - - 3 e + MnO4 (aq) + 4 H2O(l) MnO2(s) + 2 H2O(l) + 4 OH (aq)
As last step for the reduction we cancel 2 H2O on both sides:
- - - Reduction: 3 e + MnO4 (aq) + 2 H2O MnO2(s) + 4 OH (aq)
- in IO3 we assign x to I, while O has -2 and the charge is -1, so x - 6 = -1 and thus the oxidation number of I, x = +5 - - oxidation I (aq) IO3 (aq) where I goes from -1 to +5
+ In acicic medium we would add 6H on the right and put 3 H2O to the left:
- - + I (aq) + 3 H2O(l) IO3 (aq) + 6 H (aq)
Now we have a charge of -1 on the left and of +5 on the right, thus we must add
6 e- on the right to get -1 on both sides:
- - + - I (aq) + 3 H2O(l) IO3 (aq) + 6 H (aq) + 6 e
Again, because we are in basic medium, we must add 6 OH- ions on both sides to make 6 + H2O on the right from the 6 H :
- - - - I (aq) + 3 H2O(l) + 6 OH (aq) IO3 (aq) + 6 H2O(l) + 6 e
- - - - Now cancel 3 H2O: I (aq) + 6 OH (aq) IO3 (aq) + 3 H2O(l) + 6 e
To cancel 6 e- when adding the half equations, we add reduction x 2 + oxidation:
- - - - - 2MnO4 (aq)+4H2O(l)+I (aq)+6OH (aq) 2MnO2(s)+8OH (aq)+IO3 (aq)+3H2O(l)
- Finally we can cancel 3 H2O and 6 OH :
- - - - 2 MnO4 (aq) + I (aq) + H2O(l) 2 MnO2(s) + 2 OH (aq) + IO3 (aq)
Note, that I- which gets oxidized is called the REDUCING AGENT, because it
- - reduces MnO4 and that MnO4 which gets reduced is called the OXIDIZING AGENT, because it oxidizes I-.
- - balance Ag(s) + CN (aq) + O2(g) Ag(CN)2 (aq) + H2O(l) in basic medium
do as if acidic: reduction: O2(g) H2O(l); O goes from 0 to -2
+ to balance O we need 2 H2O, and then 4 H on the left to balance H:
+ O2(g) + 4H (aq) 2 H2O (l) With a charge of +4 on the left and charge 0 on the
- - + right, we must add 4 e on the left: 4 e + O2(g) + 4H (aq) 2 H2O (l)
- Now we add 4OH on both sides to make 4 H2O on the left: - - 4 e + O2(g) + 4H2O(l) 2 H2O (l) + 4 OH (aq); now we can cancel 2 H2O:
- - 4 e + O2(g) + 2H2O(l) 4 OH (aq)
- In Ag(CN)2 , Ag has as usual oxidation number +1, so Ag goes from 0 to +1 and is
- - thus oxidized: Ag(s) + 2 CN (aq) Ag(CN)2 (aq) With a - charge on the right and a 2- charge on the left we must add 1 e- on the right:
- - - Ag(s) + 2 CN (aq) Ag(CN)2 (aq) + e oxidation
Thus reduction + 4 x oxidation cancels the 4 electrons:
- - - O2(g) + 2 H2O(l) + 4 Ag(s) + 8 CN (aq) 4 OH (aq) + Ag(CN)2 (aq)
2+ 2- 3+ 3+ balance Fe (aq) + Cr2O7 (aq) Fe (aq) + Cr (aq) in acidic medium
Oxidation: Fe2+(aq) Fe3+(aq) Fe goes from +2 to +3; atom balance is ok, for the charge balance we must add 1 e- to the right: Fe2+(aq) Fe3+(aq) + e-
2- In dichromate, Cr2O7 , we assign x to Cr, O has -2, and the charge is -2; thus 2x - 14 = -2 and the oxidation number of Cr goes from x = + 6 to +3.
2- 3+ + Thus the reduction is: Cr2O7 (aq) Cr (aq) when we add 14 H to the left, we can
3+ make 7 H2O on the right and we must have 2 Cr :
2- + 3+ Cr2O7 (aq) + 14 H (aq) 2 Cr (aq) + 7 H2O(l); with a charge of 12+ on the left and one of 6+ on the right we must add 6 e- on the left:
- 2- + 3+ 6 e + Cr2O7 (aq) + 14 H (aq) 2 Cr (aq) + 7 H2O(l) reduction
Thus reduction + 6 x oxidation will cancel the 6 e- to give the final redox equation:
2- + 2+ 3+ 3+ Cr2O7 (aq) + 14 H (aq) + 6 Fe (aq) 6 Fe (aq) + 2 Cr (aq) + 7 H2O(l)
2- OXIDIZING AGENT: Cr2O7 (gets reduced) REDUCING AGENT: Fe2+ (gets oxidized) Figure 2 Figure 3 Figure 4 Figure 5