INTRODUCTION TO CHEMISTRY: PRACTICE AND STUDY GUIDE E. KENNEDY, REVIEWED BY J. HONG, P. LIPPERT CONCEPTS INTRODUCTORY CONCEPTS: DEFINITIONS Intensive property property that is independent of the quantity observed, such as density. Extensive property property that is dependent of the quantity observed, such as mass. Molar mass (M m ) the weight of on mol of a substance, expressed in g per mol. These can be derived from the atomic weights on the periodic table of elements, or molecular weights for molecular mass. Law of Conservation of Mass states that the total mass of substances after a reaction is the same as before the reaction. Law of Constant Composition/Definite Proportions states that all samples of a compound have the same composition, that is, the same proportions by mass of the constituent elements. EQUATIONS Kelvin (K) from Celsius ( C): t(k) = t( C) + 273.15 Fahrenheit from Celsius ( F): t( F) = 9 5 t( C) + 32 PRACTICE ISOTOPE NOMENCLATURE: 1) Show the appropriate nomenclature and determine the number of protons, electrons and neutrons of a Calcium41 ion. Calcium is located under group 2 on the periodic table. Characteristic of this group, the atoms lose 2 electrons when they become ionized. Therefore, the correct nomenclature for this isotope is: 4120Ca 2+ and there are 20 protons, 21 neutrons (41 20) and 18 electrons (20 2). PRACTICE ISOTOPIC MASS: 2) Using data from a mass spectrometer, the mass of a Nitrogen15 atom is determined to be 1.25 times that of Carbon12 and 0.938 times that of Oxygen16. How many units is an Oxygen16 atom? We know that Carbon12 is always 12 units exactly, so we calculate Nitrogen to be 1.25 times that: Using this, we calculate Oxygen: 1.25 x 12 units = 15.00 units. mass of 16 O: mass of 15 N 0.938 = 15.00 0.938 = 15.99 units PRACTICE ATOMIC MASS: 3) The atomic mass of Silicon is 28.0855 u. What is the mass of 28 Si if its percent abundance is 92.23% and the mass/percent abundance of 29 Si and 30 Si are 28.9765 u/4.67% and 29.9738 u/3.10%, respectively?
The atomic mass of an element indicated on the periodic table is the sum of the mass of each isotope times its fractional abundance. Therefore, we can outline the equation as such: atomic mass = (fractional abundance of 28 Si * mass of 28 Si) + (fractional abundance of 29 Si * mass of 29 Si) + (fractional abundance of 30 Si * mass of 30 Si) and, plugging in the given values, we get: (0.9223 x M) + (0.0467 x 28.9765 u) + (0.0310 x 29.9738 u) = 28.0855 u Note that percent abundance becomes fractional abundance by dividing percent abundance by 100. Solving for M, we get the mass of 28 Si: 27.9263 u. CONCEPTS COMMON IONS: NOTES ON NAMING For naming binary compounds of a metal and nonmetal, the metal is named as is. The nonmetal follows, ending with an 'ide' suffix. For naming binary compounds of two nonmetals, the nonmetal with a positive oxidation state comes first. The nonmetal with a negative oxidation state follows, ending with an 'ide' suffix. Both begin with a Greek number prefix (mono, di, tri, etc...) corresponding to the number of atoms for that particular nonmetal in the molecule. the first nonmetal never begins with a mono prefix. SIMPLE IONS lithium ion Li + sodium ion Na + potassium ion K + rubidium ion Rb + cesium ion Cs + magnesium ion Mg 2+ calcium ion Ca 2+ strontium ion Sr 2+ barium ion Ba 2+ aluminum ion Al 3+ zinc ion Zn 2+ silver ion Ag + chromium (II) ion Cr 2+ chromium (III) ion Cr 3+ iron (II) ion Fe 2+ iron (III) ion Fe 3+ cobalt (II) ion Co 2+ cobalt (III) ion Co 3+ copper (I) ion Cu + copper (II) ion Cu 2+ mercury (I) ion 2+ Hg 2 mercury (II) ion Hg 2+ tin (II) ion Sn 2+ lead (II) ion Pb 2+ hydride ion H fluoride ion F chloride ion Cl bromide ion Br iodide ion I oxide ion O 2 sulfide ion S 2 nitride ion N 3 POLYATOMIC IONS ammonium ion + NH 4 acetate ion C 2 H 3 O 2 carbonate ion 2 CO 3 hydrogen carbonate ion HCO 3 hypochlorite ion ClO chlorite ion ClO 2 chlorate ion ClO 3 perchlorate ion ClO 4 chromate ion 2 CrO 4 dichromate ion 2 Cr 2 O 7 cyanide ion CN hydroxide ion OH nitrite ion NO 2 nitrate ion NO 3 oxalate ion 2 C 2 O 4 permanganate ion MnO 4 phosphate ion 3 PO 4 hydrogen phosphate ion 2 HPO 4 dihydrogen phosphate ion H 2 PO 4 sulfite ion 2 SO 3 hydrogen sulfite ion HSO 3 sulfate ion 2 SO 4 hydrogen sulfate ion HSO 4 thiosulfate ion S 2 O 2 3 1 [2]
PRACTICE MOLAR MASS: 4) What is the molar mass of anhydrous magnesium nitrate, Mg(NO 3 ) 2? To determine the molar mass, we first must extrapolate the molar mass of each participating element from the formula unit. The formula unit is given as Mg(NO 3 ) 2. molar mass of Mg(NO 3 ) 2 = atomic mass of Mg + 2(atomic mass of N + 3(atomic mass of O)). Note that this is set up just as any algebraic equation. Plugging in the values from the periodic table, we get: molar mass of Mg(NO 3 ) 2 : 24.305 + 2(14.007 + 3(15.999)) = 24.305 + 2(14.007 + 47.997) = 24.305 + 2(62.004) = 24.305 + 124.008 = 148.313 u. Units were left out until the end for the purpose of helping to visualize this in terms of an algebraic equation, but you should always leave units in for other problems. Also, this problem was done in stepbystep fashion to illustrate the mass of different constituents of the formula. For instance, the total mass of oxygen in NO 3 is 47.997 u, but the total mass of oxygen would be double this for Mg(NO 3 ) 2 because there are two units of NO 3 in every Mg(NO 3 ) 2. PRACTICE STOICHIOMETRY I: 5) How many ions are present in 0.5 mg of anhydrous magnesium nitrate Mg(NO 3 ) 2? To begin, we need to know what the problem is asking for. Ions are charged particles and if we look at Mg(NO 3 ) 2, an ionic compound, we can see that there are 3 ions present in each unit, one for Mg and two for each NO 3. We can look at any table of common ions and see that NO 1 3 is an ion with a negative 1 charge; that makes sense because two of these would be necessary to cancel out the positive 2 charge on Mg 2+ 1, an ionic metal from the group 2 elements. (The nitrogen and oxygen in NO 3 are not ions themselves but rather covalently bonded atoms.) The next step is to determine how to derive the total number of ions from the 0.5 mg given. In other words, how do we get ions from milligrams? We know that AMU values on the periodic table for each element represent grams per mol, and so the same applies to the formula mass. If we hadn t already done problem 4, we would have to calculate the molar mass of Mg(NO 3 ) 2. Our answer was 148.313 u and this is the same as 148.313 grams per mol of Mg(NO 3 ) 2. We can easily convert mg to g, and so the missing step is how to convert from number of ions to mols. To make this last conversion, we need to use Avogadro s number to convert from mols to molecules (the term molecule used loosely here for unit of ionic compound). This flow chart expresses the steps we need to take: mg g mol molecules ions This seems like a lot of steps, but the hard part is determining how to get from A to B. Plugging in our data, we get: 1 g # of ions: 0.5 mg Mg(NO 3 ) 2 x ( 1000 mg ) (1 mol Mg(NO 3) 2 148.313 g ) ( 6.0 x 1023 molecules Mg(NO 3 ) 2 3 ions 1 mol Mg(NO 3 ) ) ( 2 1 mol Mg(NO 3 ) ) = 6.1 x 10 18 ions 2 [3]
That s a large number, but does it make sense? We are asking for the number of tiny charged particles in a small amount of white anhydrous powder. As we begin our calculations, our number becomes smaller as we convert from mg to mols. This makes sense because if we compare the number of grams required for one mole of Mg(NO 3 ) 2, 148.313 g, we see that our starting amount, 0.5 mg, is much less than this. However, as we convert from mols to ions, our number becomes much larger. PRACTICE EMPIRICAL FORMULA: 6) An analysis of a sample of 2.43 g of ascorbic acid, AKA vitamin C, has determined that it is composed of 40.921% carbon, 4.582% hydrogen and 54.501% oxygen and the formula mass has been experimentally determined to be approximately 179 u. What are its empirical and molecular formulas? First, determine the empirical formula by determining the number of mols for each component. We could do this using the sample amount of 2.43 g, but it s easier to use 100 g because we can just use the percent mass composition values instead of taking the extra step of multiplying the sample amount by the mass percent for each component. Why can we do this? The amount of a substance isn t going to change its empirical formula because it is an intensive property: 1 mol C mols C: 40.921 g C x ( 12.011 g C ) = 3.407 mols C mols H: 4.582 g H x ( 1 mol H 1.008 g H ) = 4.546 mols H 1 mol O mols O: 54.501 g O x ( 15.999 g O ) = 3.407 mols O We can tentatively write the formula as C 3.407 H 4.546 O 3.407. Dividing each number by the smallest of the numbers, 3.407, gets us CH 1.334 O. Already we have two of the components in whole numbers. We can t divide any further and have whole units so we have to multiply. The 334 portion of 1.334 gives us a hint as to what to multiply by. Multiplying each number by 3 gets us the formula C 3 H 4.002 O 3, which we round to get our empirical formula. Each step is shown below: C 3.407 H 4.546 O 3.407 (divide by 3.407) CH 1.334 O (multiply by 3) C 3 H 4.002 O 3 (round off) C 3 H 4 O 3 Next, we derive our molecular formula from the empirical formula multiplying each of the component units by its mass number and adding these numbers up to get the molar mass: C: 3 x 12.0 u = 36.0 u H: 4 x 1.0 u = 4.0 u O: 3 x 16.0 u = 48.0 u 36.0 u + 4.0 u + 48.0 u = 88.0 u The mass of empirical formula is 88 u, yet we are given the experimentally determined value of approximately 179 u. 88 u goes into 179 u about two times, therefore we multiply each component by two to get our molecular formula: C 6 H 8 O 6. PRACTICE BALANCING EQUATIONS: 7) Show the balanced equation for the combustion of triethylene glycol, C 6 H 14 O 4. [4]
To balance an equation, we must know what the reactants and products are. When hydrocarbons combust by reacting with oxygen in the air, the byproducts formed are carbon dioxide and water. With this in mind, we can set up the unbalanced equation as follows: C 6 H 14 O 4 + O 2 CO 2 + H 2 O We begin by balancing by adding coefficients (called stoichiometric coefficients) in front of some of the products or reactants in order to get an equal amount of a common component on each side of the reaction. (In no case can we change a formula or add or remove a component in order to balance both sides; the products and reactants that take part in the reaction must be shown as they are.) Looking at the reaction, we notice that oxygen is present in every component as well as isolated by itself. Therefore, we will balance this last. It s usually easiest to balance components that are found by themselves last so that changing the coefficient in front of them doesn t change every other component. We begin by balancing carbon. There are 6 carbons on the left, so we will put a stoichiometric coefficient of 6 in front of carbon dioxide. Next, we will balance hydrogen by putting a 7 in front of hydrogen to get 14 carbons on each side: C 6 H 14 O 4 + O 2 6CO 2 + H 2 O C 6 H 14 O 4 + O 2 6CO 2 + 7H 2 O Next we balance oxygen. There are 6 oxygens on the left and 19 on the right. We need 13 more on the left and because oxygen is found by itself, we can adjust this stoichiometric coefficient by adding: C 6 H 14 O 4 + 15 2 O 2 6CO 2 + 7H 2 O Technically, this is balanced. However, it is not in proper form due to the fractional stoichiometric coefficient. We multiply each coefficient by 2 in order to get the final balanced equation: 2C 6 H 14 O 4 + 15O 2 12CO 2 + 14H 2 O PRACTICE STOICHIOMETRY II: 8) How many grams of water are produced by the combustion of 5.75 g of triethylene glycol with excess O 2? Excess O 2 means that the amount of oxygen consumed in this reaction does not limit the amount of products produced. It does not mean that an infinite quantity of oxygen is consumed; a definite quantity of oxygen is consumed but the limiting reactant is C 6 H 14 O 4. Most combustion reactions that have an excess of O 2 are open to the atmosphere. We ve already taken the first necessary step in problem 7 by finding the balanced equation. Next, we will find the molar mass of C 6 H 14 O 4 and H 2 O: molar mass of C 6 H 14 O 4 : (6 x 12.011) + (14 x 1.008) + (4 x 16.000) = 150.178 g mol 1 molar mass of H 2 O: (2 x 1.008) + (1 x 16.000) = 18.016 g mol 1 With the molar mass of C 6 H 14 O 4 and water determined, we can calculate the amount of water produced: 1 mol C 6 H 14 O 4 grams of H2O produced: 5.75 g C 6 H 14 O 4 x ( 150.178 g C 6 H 14 O ) 4 7 mol H 2 O ( 2 mol C 6 H 14 O ) ( 18.016 g H 2O 4 1 mol H 2 O ) = 2.41 g H 2O Note that in a combustion reaction, most of the water will be in the form of water vapor. [5]
CONCEPTS SOLUBILITY: DEFINITIONS Molarity (M) Concentration is a solution property defined as the number of moles solute per liter of solution. Precipitate An insoluble solid formed when aqueous cations and anions combine. EQUATIONS Solution Dilution: M i V i = n i = n f = M f V f SOLUBILITY RULES FOR COMMON SALTS IN WATER (NONSATURATED SOLUTION) Soluble: + Group 1 and NH 4 no exception Acetates, Chlorates, Nitrates no exception Chlorides, Bromides, Iodides except for Ag +, Pb 2+ 2+, Hg 2 Sulfates except for Ag +, Pb 2+, Hg 2+ 2, Ca 2+, Sr 2+, Ba 2+ Insoluble: Sulfides + except for Group 1, Group 2, NH 4 Hydroxides except for Group 1, NH + 4, Ca 2+, Sr 2+, Ba 2+ Carbonates, Chromates, Phosphates + except for Group 1, NH 4 PRACTICE SOLUTION DILUTION: 9) How many liters of 6 M NaOH must be diluted in water to obtain 1.2 L of 0.2005 M NaOH? How many liters of water are added to obtain this molarity? How many mols of NaOH are in the final solution? The solution dilution equation can be used to solve for this problem: (6 M NaOH) x (X L) = (0.2005 M NaOH) x (1.2 L). We get 0.0401 L, or 40.1 ml when solving for X. To determine the amount of water added, we subtract this amount from the final amount: 1.2 L 0.0401 L = 1.1599 L H 2 O = 1159.9 ml H 2 O The solution dilution equation indicates that the number of mols of NaOH will be the same in both initial and final solutions. (Note that the n being referred to in the solution dilution equation refers to the number of mols of solute and not solvent.) Therefore, we can calculate the mols as follows: 0.2005 mols NaOH mols of NaOH: 1.2 L NaOH x ( 1 L NaOH(aq) ) = 0.2406 mols NaOH PRACTICE PRECIPITATION: 10) 50.00 ml of 0.100 M K 2 CrO 4 (aq) is mixed with 150.00 ml of 0.150 M AgNO 3 (aq). How many grams of precipitate will form in the solution? First, we need to determine what precipitate will form. Using the concepts above, we see that potassium should always be soluble. Also, NO 3 1 will not form a precipitate with silver. Therefore, CrCO 4 2 will react with Ag 1+ to form Ag 2 CrO 4 (s). We can write out the reaction as follows: K 2 CrO 4 (aq) + 2AgNO 3 (aq) Ag 2 CrO 4 (s) + 2KNO 3 (aq) [6]
Next we need to determine the limiting reactant. We do this by solving for the number of mols of Ag 2 CrO 4 (s) produced for each of the reactants. The reactant that produces the least amount of mols is the limiting reactant. 1 L K 2 CrO 4 mols of Ag 2 CrO 4 (s) produced: 50.00 ml K 2 CrO 4 (aq) x ( 1000 ml K 2 CrO ) 4 ( 0.100 mol K 2CrO 4 1 L K 2 CrO ) ( 1 mol Ag 2CrO 4 4 1 mol K 2 CrO ) = 0.005 mols Ag 2 CrO 4 (s) 4 1 L AgNO 3 mols of Ag 2 CrO 4 (s) produced: 150.00 ml AgNO 3 (aq) x ( 1000 ml AgNO ) 3 ( 0.150 mols AgNO 3 (aq) 1 L AgNO 3 ) ( 1 mol Ag 2CrO 4 2 mol AgNO 3 ) = 0.01125 mols Ag 2 CrO 4 (s) K 2 CrO 4 produces the least amount of Ag 2 CrO 4 so it is the limiting reactant. We will use its initial quantity to find the amount of precipitate formed. To do this, we will need to convert from mols of Ag 2 CrO 4 into grams by finding its molar mass: molar mass of Ag 2 CrO 4 : (2 x 107.868) + (1 x 51.996) + (4 x 16.000) = 331.732 g mol 1 Finally, we multiply the mols of Ag 2 CrO 4 determined from the limiting reactant by its molar mass: grams of Ag 2 CrO 4 produced: 0.01125 mols Ag 2 CrO 4 (s) x ( 331.732 g Ag 2CrO 4 1 mol Ag 2 CrO 4 ) = 3.73 g Ag 2 CrO 4 (s) 11) 50.00 ml of 0.100 M K 2 CrO 4 (aq) is mixed with 150.00 ml of 0.150 M AgNO 3 (aq) to produce 3.41 g of Ag 2 CrO 4 (s). What is the percent yield for this process? Percent yield is found by dividing actual yield by theoretical yield and multiplying this value by 100. In problem 9, we found the theoretical yield to be 3.73 g of Ag 2 CrO 4 (s). Here, the actual yield is 3.41 g of Ag 2 CrO 4 (s). Therefore: percent yield of Ag 2 CrO 4 (s): ( 3.41 g Ag 2CrO 4 3.73 g Ag 2 CrO 4 ) x 100 = 91.4% CONCEPTS AQUEOUS REACTIONS: DEFINITIONS Spectator ion An ion that does not take part in a reaction. Net ionic equation Reaction equation written so that spectator ions have been removed and only participating reactants and products are left. Strong electrolyte An electrolyte that ionizes completely. All soluble ionic compounds are strong electrolytes. Weak electrolyte An electrolyte, including weak acid and weak bases, that partially dissociates to form some ions. Reactions are reversible (shown with double arrows). Most molecular compounds are either weak or nonelectrolytes. The proportions of ions to molecules remain fixed, but the predominant species is the original molecule, represented in aqueous form. Arrhenius acidbase theory States that acids produce H 3 O + and bases produce OH in aqueous solutions. Neutralization reaction A reaction where acids and bases react as electrolytes to form water and a salt. [7]
Salt metathesis Occurs when two compounds with two parts switch one part with each other to form new products. Also called double displacement when referring to precipitation reactions that take place between ions in a solution. Titration A method of determining either an unknown concentration of a known solution or unknown solution of known concentration, by slowly adding to it a reactive substance, the reagent, until both have reacted and neither is in excess. Equivalency point The point in a titration where the solution and reagent have reacted and neither is in excess and the ph is determined by the products. This point often corresponds to the endpoint, the point when an indicator that has been added to the reaction irreversibly changes color. STRONG ACIDS AND BASES (STRONGEST TO WEAKEST) Acids: HClO 4, HI, HBr, HCl, H 2 SO 4, H 3 O +, HNO 3 Bases: KOH, Ba(OH) 2, CsOH, NaOH, Sr(OH) 2, Ca(OH) 2, LiOH, RbOH, Mg(OH) 2 OXOACIDS: NAME, FORMULA, NONMETAL OXIDATION STATE, EXAMPLE SALT/ FORMULA hypochlorous acid HClO Cl + sodium hypochlorite NaClO chlorous acid HClO 2 Cl 3+ sodium chlorite NaClO 2 chloric acid HClO 3 Cl 5+ sodium chlorate NaClO 3 perchlorate HClO 4 Cl 7+ sodium perchlorate NaClO 4 nitrous acid HNO 2 N 3+ sodium nitrite NaNO 2 nitric acid HNO 3 N 5+ sodium nitrate NaNO 3 phosphorous acid H 3 PO 3 P 3+ sodium phosphite Na 3 PO 3 (note: diprotic only) phosphoric acid H 3 PO 4 P 5+ sodium phosphate Na 3 PO 4 sulfurous acid H 2 SO 3 S 4+ sodium sulfite Na 2 SO 3 sulfuric acid H 2 SO 4 S 6+ sodium sulfate Na 2 SO 4 (Note: for naming binary acids, start with a hydro prefix, modify the nonmetal to end with a ic suffix, and follow with the word 'acid') SPONTANEOUS AQUEOUS REACTIONS Precipitation reactions: salt 1 + salt 2 insoluble salt 3 + salt 4 Acidbase reactions: strong acid + strong base salt + H 2 O (aka neutralization reaction) strong acid + strong base weak acid + weak base Precipitate dissolving reactions: strong acid 1 + salt 1 weak acid 2 + salt 2 acid 1 + salt 1 acid 2 + insoluble salt 2 salt + NH 3 + H 2 O NH + 4 salt + metal OH Gas formation: reaction H 2 CO 3 H 2 O + CO 2 reaction H 2 SO 3 H 2 O + SO 2 reaction NH 4 OH H 2 O + NH 3 SINGLE REPLACEMENT REACTIVITY (MOST REACTIVE TO LEAST REACTIVE) reacts, replacing H in H 2 O(l), to produce H 2 (g): Li +, Rb +, K +, Cs +, Ba 2+, Sr 2+, Ca 2+, Na + reacts, replacing H in acids, to produce H 2 (g): Mg 2+, Al 3+, Mn 2+, Zn 2+, Cr 3+, Fe 2+, Cd 2+, Co 2+, Ni 2+, Sn 2+, Pb 2+ reference: H 2 (2H + ) will not produce H 2 (under normal conditions): Cu 2+, Ag +, Hg 2+, Pt 2+, Au 3+ Note: A simple way to remember this is that the ions that replace H in water to produce H 2 (g) are all group 1 and 2 metals. Magnesium is the only one that is not in the first series, but it is the most reactive in the next series. Besides magnesium, the next series contains only transition metals. The last series contains copper, silver, gold and platinum which have all been used as metal currency commodities. This series also contains mercury, which is also called quicksilver. The order of reactivity may vary depending on source. [8]
PRACTICE NET IONIC EQUATION: 12) Show the net ionic equation for the reaction of potassium chromate with silver nitrate. The reaction for these ionic compounds is shown in example 10: K 2 CrO 4 (aq) + 2AgNO 3 (aq) Ag 2 CrO 4 (s) + 2KNO 3 (aq) If we break this down to show all electrolytes where possible, the reaction equation would look like: 2K + (aq) + CrO 2 4 (aq) + 2Ag + (aq) + 2NO 3 (aq) Ag 2 CrO 4 (s) + 2K + (aq) + 2NO 3 (aq) Removing all spectator ions would result in the net ionic equation: CrO 2 4 (aq) + 2Ag + (aq) Ag 2 CrO 4 (s) PRACTICE ACIDS/BASE NEUTRALIZATION: 13) Show the net ionic equation for the reaction of solid aluminum hydroxide with hydrochloric acid. Hydrochloric acid is a strong acid and hydroxide is a strong base. neutralization reaction. We write the balanced equation as follows: Therefore, this reaction is a Al(OH) 3 (s) + 3HCl(aq) AlCl 3 (aq) + H 2 O(l) In the presence of a strong acid or base, water acts as a precipitate does in a precipitation reaction; it acts as the final product whose creation is spontaneous and nonreversible. The salt produced in this case however is soluble. Therefore, there are spectator ions that need to be removed to derive the net ionic equation: Al(OH) 3 (s) + 3H 3 O + (aq) Al 3+ (aq) + H 2 O(l) Note that H 3 O + is written instead of H +. It s a good idea to get in the habit of writing a hydrogen ion as the hydronium ion when part of an aqueous solution. PRACTICE TITRATION: 14) Store bought white vinegar is a mix of acetic acid and water. In order to determine how much of the vinegar is acetic acid, a solution of 10.0 ml white vinegar containing a few drops of the indicator phenolphthalein is titrated with a solution of 0.100 M NaOH(aq). After being titrated with 80.0 ml of the NaOH, the solution becomes a dark pink color. Assuming the density of the vinegar is the same as water (1.00 g ml 1 ), what is the percent by mass of acetic acid in the vinegar solution? Here we have an acid base neutralization reaction. Na + is a spectator ion and so we leave it out of the net ionic equation: HC 2 H 3 O 2 (aq) + OH (aq) C 2 H 3 O 2 (aq) + H 2 O(l) The change in color is due to the indicator reaching the endpoint. We can assume that this point corresponds to the equivalency point, the point when both reactants have neutralized each other and all [9]
that remains are the products. Looking at the net ionic equation, we can see that the molar ratio of the reactants is 1 to 1. Therefore, we can assume that the amount of NaOH (in mols) added to the solution is equal to the amount (in mols) of the acetic acid that reacts with it. Knowing this, we can calculate the amount of mols of acetic acid in the original solution: 0.100 mols NaOH mols of HC 2 H 3 O 2 : 80.00 ml NaOH (aq) x ( 1000 ml NaOH ) (1 mol HC 2H 3 O 2 1 mol NaOH ) = 0.008 mols HC 2 H 3 O 2 (aq) Note that we saved a step in conversion by recognizing that molarity can also be expressed as mols per thousand ml. Because the amount of mols that reacted with the solution is the same as the amount of mols in the original solution, we can now calculate the mass of HC 2 H 3 O 2 in the original solution. We begin by finding the molar mass of HC 2 H 3 O 2 : molar mass of HC 2 H 3 O 2 : 4(1.01) + 2(12.01) + 2(16.00) = 60.06 g mol 1 Next, using the given density, we find the percent by mass, first by dividing the number of mols by the initial volume of solution: percent by mass of HC 2 H 3 O 2 : ( 0.008 mols HC 2H 3 O 2 5.00 ml solution ) (60.06 g HC 2H 3 O 2 1 mol HC 2 H 3 O ) 2 1.00 ml solution ( 1.00 g solution ) x 100% = (0.096 g HC 2H 3 O 2 1.00g solution ) x 100% = 9.6% HC 2H 3 O 2 Thus, the percent by mass of acetic acid in the sample of store bought white vinegar is 9.6%. CONCEPT REDOX: DEFINITIONS Oxidation state The hypothetical charge of an atom in a molecule if it was ionic. Reductionoxidation reaction (Redox) A reaction where oxidation states change. Reduction A gain in electrons, decrease in oxidation state or negative charge of particle. Electrons appear on the left side in halfreaction. A reductant/reducer/reducing agent makes this possible by giving up electrons, becoming oxidized. Oxidation A loss of electrons, increase in oxidation state or positive charge of particle. Electrons appear on the right side in halfreaction. An oxidant/oxidizer/oxidizing agent makes this possible by taking electrons, becoming reduced. Halfreaction A reaction involving either the oxidation or reduction participants of a redox reaction. Disproportion reaction A reaction where the same substance is oxidized and reduced, for example: 2H 2 O 2 2H 2 O + O 2. GUIDELINES FOR DETERMINING OXIDATIVE STATE 1. Of these rules, when two contradict, follow the rule with the lowest numerical value. 2. The oxidation state (O.S.) of an atom in free element form is 0. (For example, neither Cl atoms have charge in Cl 2.) 3. The total O.S. of all the atoms in: a. a neutral species (molecules, neutral ionic compounds or single atoms) is 0. b. an ion is equal to the ionic charge. 4. In ionic compounds, group 1 metals have an O.S. of +1 and the group 2 metals have an O.S. of +2. 5. In ionic compounds, the O.S. of F is 1. 6. In ionic compounds, H usually has an O.S. of +1. [10]
7. In ionic compounds, O usually has an O.S. of 2. 8. In binary compounds with metals, group 15 elements have an O.S. of 3, group 16 elements have an O.S. of 2 and group 17 elements have an O.S. of 1. RULES FOR BALANCING A REDOX IN AN ACIDIC OR BASIC SOLUTION 1. Separate reaction into halfreactions. 2. Balance atoms of all elements except H and O. 3. Balance remaining O by adding H 2 O. 4. Balance remaining H by adding H +. 5. Balance charge using electrons. 6. Adjust coefficients so that the number of electrons are the same in each halfreaction. 7. Add halfreactions together and balance. 8. For acid reactions, change each H + to a H 3 O + and add one water for each change to the other side of the reaction. 9. For base reactions, add an OH to each side for each H +. 10. For base reactions, combine H + and OH into H 2 O and balance. PRACTICE REDOX REACTION: 15) Balance the equation for the reaction of dichromate ion with nitrite ion in an acidic solution to derive chromium(iii) ion and nitrate ion. We can set up the unbalanced reaction as follows: Unbalanced: Cr 2 O 7 2 (aq) + NO 2 (aq) Cr 3+ (aq) + NO 3 (aq) Note that the question is phrased naming only the ions involved in the reaction. In reality, there would be some cations or anions in the solution that would ultimately be removed from the balanced equation as spectator ions. Acid may be written over the arrow to indicate the conditions for the reaction. Next we have to identify which species undergo oxidation and which undergo reduction. The species undergoing oxidation is the nitrogen. How do we know? We can see that the number of negative oxygen atoms for the nitrogen containing ions increases (each with a negative two charge) yet the overall charge for the ion stays the same (negative 1). The reduction then must take place on the chromium. We separate these into two separate halfreactions: Oxidation: NO 2 (aq) NO 3 (aq) Reduction: Cr 2 O 7 2 (aq) Cr 3+ (aq) Next, we balance all of the elements except H and O. The oxidation halfreaction stays the same while the reduction halfreaction changes as follows: Reduction: Cr 2 O 7 2 (aq) 2Cr 3+ (aq) Next, we start adding water to the sides that need more O: and then we balance the remaining H: Oxidation: NO 2 (aq) + H 2 O(l) NO 3 (aq) Reduction: Cr 2 O 7 2 (aq) 2Cr 3+ (aq) + 7H 2 O(l) Oxidation: NO 2 (aq) + H 2 O(l) NO 3 (aq) + 2H + (aq) Reduction: Cr 2 O 7 2 (aq) + 14H + (aq) 2Cr 3+ (aq) + 7H 2 O(l) [11]
Next, we check the overall charge on each half of each halfreaction and balance it by adding electrons, denoted with e : Oxidation: NO 2 (aq) + H 2 O(l) NO 3 (aq) + 2H + (aq) +2e Reduction: Cr 2 O 7 2 (aq) + 14H + (aq) +6e 2Cr 3+ (aq) + 7H 2 O(l) That last part might seem a little tricky. The key is to simply add up the charges on each atom (using the oxidation state concepts) and look for the difference in charges on each side. Bring down positive charges with electrons to match the charge of the other side. Now that we have the balanced halfreactions, we need to make sure that the number of electrons are the same in each halfreaction by multiplying the whole reaction with a stoichiometric coefficient. In this case, we multiply the whole oxidation halfreaction by 3 so that the number of electrons becomes 6, to match the number of electrons in the reduction halfreaction: Oxidation: 3(NO 2 (aq) + H 2 O(l) NO 3 (aq) + 2H + (aq) +2e ) = 3NO 2 (aq) + 3H 2 O(l) 3NO 3 (aq) + 6H + (aq) +6e Now that the halfreactions have the same amount of free electrons, we can add them together. Note that the free electrons are always on opposite sides and will cancel out. Also, water and H + might cancel out: Oxidation: 3NO 2 (aq) + 3H 2 O(l) 3NO 3 (aq) + 6H + (aq) +6e Reduction: Cr 2 O 2 7 (aq) + 148H + (aq) +6e 2Cr 3+ (aq) + 74H 2 O(l) Overall: 3NO 2 (aq) + Cr 2 O 2 7 (aq) +8H + (aq) 3NO 3 (aq) + 2Cr 3+ (aq) + 4H 2 O(l) Finally, we change each hydrogen ion into a hydronium ion, adding an H 2 O to the other side of the reaction for each change, to get the balanced equation: Final overall: 3NO 2 (aq) + Cr 2 O 7 2 (aq) +8H 3 O + (aq) 3NO 3 (aq) + 2Cr 3+ (aq) + 12H 2 O(l) 16) Balance the equation for the reaction of cyanide ion with permanganate ion in a basic solution to derive cyanate ion and solid magnesium dioxide. We can use the same steps as above, simply changing out the last steps to accommodate a reaction in a basic solution. We begin by writing the unbalanced reaction and determining the half reactions: Unbalanced: MnO 4 (aq) + CN (aq) MnO 2 (s) + OCN (aq) Oxidation: CN (aq) OCN (aq) Reduction: MnO 4 (aq) MnO 2 (s) The nonoxygen and nonhydrogen atoms are already balanced, so next we balance by adding H + and H 2 O: Then we add and balance the electrons: Oxidation: CN (aq) + H 2 O(l) OCN (aq) + 2H + (aq) Reduction: MnO 4 (aq) + 4H + (aq) MnO 2 (s) + 2H 2 O(l) Oxidation: 3(CN (aq) + H 2 O(l) OCN (aq) + 2H + (aq) + 2e ) = 3CN (aq) + 3H 2 O(l) 3OCN (aq) + 6H + (aq) + 6e Reduction: 2(MnO 4 (aq) + 4H + (aq) + 3e MnO 2 (s) + 2H 2 O(l)) = 2MnO 4 (aq) + 8H + (aq) + 6e 2MnO 2 (s) + 4H 2 O(l) [12]
And then we add the equations together and balance: Oxidation: 3CN (aq) + 3H 2 O(l) 3OCN (aq) + 6H + (aq) + 6e Reduction: 2MnO 4 (aq) + 82H + (aq) + 6e 2MnO 2 (s) + 4H 2 O(l) Overall: 3CN (aq) + 2MnO 4 (aq) + 2H + (aq) 3OCN (aq) + 2MnO 2 (s) + H 2 O(l) Finally, we do the base specific steps by adding hydroxide ions to each side for every hydrogen ion, and then combining hydrogen and hydroxide ions where available and balancing: Overall: 3CN (aq) + 2MnO 4 (aq) + 2H + (aq) + 2OH (aq) 3OCN (aq) + 2MnO 2 (s) + H 2 O(l) + 2OH (aq) Final overall: 3CN (aq) + 2MnO 4 (aq) + H 2 O(l) 3OCN (aq) + 2MnO 2 (s) + 2OH (aq) [13]