Unit 4a Acids, Bases, and Salts Theory

Unit 4a Acids, Bases, and Salts Theory Chemistry 12 Arrhenius Theory of Acids and Bases The first theory that was proposed to explain the actions of acids and bases was by Svante Arrhenius. It is still valuable in many circumstances. His theory: An ACID is any substance which releases H + (aq) in water. A BASE is any substance which releases OH - (aq) in water. A SALT is the neutralization product which results when an acid and a base react: HCl (aq) + NaOH (aq) NaCl (aq) + H 2 O (l) ACID BASE SALT Acids: react with bases. are electrolytes. act on some metals to produce H 2(g) : Mg (s) + 2 HCl (aq) MgCl 2(aq) + H 2(g) turn litmus paper RED. taste SOUR (vinegar, lemon juice). 1

A strong acid is any ionic species whose formula starts with an H. HCl HNO 3 H 2 SO 4 (Hydrochloric Acid) (Nitric Acid) (Sulphuric Acid) Bases: react with acids. are electrolytes. feel slippery. turn litmus paper BLUE. taste BITTER (baking soda). Common Acids and Bases Homework: You should be aware of some common acids and bases and where they are used. Base used in oven cleaner: NaOH (aq) Read: Pages 115-119 Do: #1-9 Rust remover: HCl (aq) See your Hebden book (Page #112). 2

The True Nature of H + (aq) The Hydrogen atom consists of a proton surrounded by a single electron. Removing the electron creates H +. We are left with a naked proton (no electron). This tiny nuclear particle has an enormous charge concentration. The highly concentrated positive charge is very strongly attracted to any region where negative charges exist. A water molecule has its electrons distributed as follows: A region of space called a non-bonding orbital, in which are stored the electrons not being used in bonds, are very vulnerable to attack by H +. All the H + present in water will attach itself to water molecules: H + (aq) + H 2 O (l) H 3 O + (aq) H + (aq) is called the PROTON. H 3 O + (aq) is called the HYDRONIUM ION or the hydrated proton. Therefore, what we previously have been calling H + (aq) is actually H 3 O + (aq). Use H 3 O + (aq) instead of H + (aq) when showing the IONIZATION of an acid. 3

For Example: THE BRØNSTED-LOWRY THEORY OF ACIDS AND BASES When HCl (g) added to water to produce hydrochloric acid solution, HCl (aq), we would previously write: HCl (g) H + (aq) + Cl - (aq) But we should now write: The Bronsted-Lowry theory is a modification of the Arrhenius theory which allows for the effect of equilibrium reactions. It permits us to extend our idea of acids and bases to a wider range of species. HCl (g) + H 2 O (l) H 3 O + (aq) + Cl - (aq) Let s look at a second example: The Theory: An ACID is a substance which donates a proton to another substance (gives away an H + ). A BASE is a substance which accepts a proton from another substance (gets an H + ). CH 3 COOH (aq) + H 2 O (l) CH 3 COO - (aq) + H 3 O + (aq) ACID BASE BASE ACID Here is a typical Bronsted-Lowry acid-base reaction equation: NH 3(aq) + H 2 O (l) NH + 4 (aq) + OH - (aq) BASE ACID ACID BASE 4

AMPHIPROTIC SUBSTANCES In the Arrhenius theory, recall that: ACID + BASE SALT + WATER Now, in the Bronsted-Lowry theory: ACID + BASE BASE + ACID In every case of a Bronsted-Lowry reaction we will have an acid and a base on both sides of the equation. Mono = one, di = two, tri = three, etc. Acids can have different numbers of protons available to donate to a base. We can call them MONOprotic, DIprotic, TRIprotic, etc., depending on how many they have. A general term for an acid that has more than one proton is called POLYprotic. An AMPHIPROTIC substance can act either as an acid or a base. In other words, it can either donate or accept protons. Examples include: H 2 O, H 2 PO 4-, HS -, HCO 3 - Apart from H 2 O, the species are ions which have lost at least one proton from a POLYPROTIC ACID. Hence, the resulting ion can either lose another proton, or regain one. 5

For Example: Acting a BASE an ACID as To summarize If a substance: Possesses a negative charge, and H 2 S (aq) HS - (aq) S 2- (aq) still has an easily removable hydrogen H H 2 PO - 3 PO 4(aq) 4 (aq) HPO 2-4 (aq) PO 3-4 (aq) H 3 O + (aq) H 2 O (l) OH - (aq) Remember exception: H 2 O (l)!!! Homework: Conjugate Acids and Bases Read: Pages 119-125 Do: #10-14 A pair of chemical species which differ by only one proton. A CONJUGATE ACID is the member of the conjugate pair which has the extra proton. A CONJUGATE BASE is the member of the conjugate pair which lacks the extra proton. 6

For Example: In the equilibrium reaction: CH 3 COOH (aq) + H 2 O (l) CH 3 COO - (aq) + H 3 O + (aq) What are the two conjugate pairs? Conjugate Pair Conjugate Acid Conjugate Base CH 3 COOH (aq) /CH 3 COO - (aq) CH 3 COOH (aq) CH 3 COO - (aq) H 2 O (l) /H 3 O + (aq) H 3 O + (aq) H 2 O (l) Strong and Weak Acids and Bases STRONG acids and bases are 100% ionized in solution. NaOH (s) Na + (aq) + OH - (aq) HCl (aq) + H 2 O (l) H 3 O + (aq) + Cl - (aq) WEAK acids and bases are less than 100% ionized in solution. NH 3(aq) + H 2 O (l) NH + 4 (aq) + OH - (aq) HF (aq) + H 2 O (l) H 3 O + (aq) + F - (aq) Note the difference in arrows! A couple of things to keep in mind Weak acids and bases are involved in equilibrium reactions, strong acids and bases are not. Acids that are 99.9% ionized will be weak, while an acid which is 100% ionized is strong. 7

The Strong Acids The strong acids are the top six acids listed on the table of Relative Strengths of Acids and Bases (see page 6 of your data table): HClO 4, HI, HCl, HNO 3, HBr, H 2 SO 4 NOTE: H 2 SO 4 is only strong for the first ionization: H 2 SO 4(aq) + H 2 O (l) HSO 4 - (aq) + H 3 O + (aq) The Strong Bases The hydroxide ion, OH -, found on the lower right side of the table, is a strong base. Metal hydroxides are 100% dissociated in solution, and thus are strong bases. NaOH, Mg(OH) 2, Fe(OH) 3, KOH, Ca(OH) 2, Zn(OH) 2 The last two bases, on the right hand side of the table are also Strong Bases : O 2- and NH 2 - The Weak Acids The Weak Bases All the acids listed on the left hand side of the table from IODIC to WATER are considered weak. The last two species, OH - and NH 3, at the bottom left side of the table are such weak acids that they will NEVER act as acids in aqueous solutions. The weak bases are found on the right hand side of the table from WATER to MONOHYDROGEN PHOSPHATE ION. The last six species on the top right side of the table are such weak bases that they will NEVER act as bases in aqueous solutions. 8

We must now be more precise with our terminology: The terms weak and strong refer to the percentage of ionization. The terms dilute and concentrated refer to the molarity of a solution. 10.0 M HF (aq) is concentrated and weak. 0.001 M HCl (aq) is dilute and strong. Some Relationships Found on the Table The higher an acid is on the left hand side of the table, the stronger the acid. The lower a base is on the right hand side of the table, the stronger the base. The stronger an acid, the weaker its conjugate base, and vice-versa. Now it is easy to understand why H 2 O is amphiprotic: The Leveling Effect If it is mixed with a stronger acid than itself, the water will act as a base. H 2 CO 3(aq) + H 2 O (l) <-> HCO 3 - (aq) + H 3 O + (aq) If it is mixed with a stronger base than itself, the water will act as an acid. Consider the six strong acids on the left side of the table. There is a hierarchy of strengths from HClO 4 to H 2 SO 4. HI is a stronger acid than HCl. CN - (aq) + H 2 O (l) <-> HCN (aq) + OH - (aq) 9

Whenever we place these acids in an aqueous environment (water), the whole thing changes: HI (l) + H 2 O (l) I - (aq) + H 3 O + (aq) HCl (l) + H 2 O (l) Cl - (aq) + H 3 O + (aq) We can say that both these acids have leveled themselves to the strength of H 3 O + (aq). Therefore, the strongest acid that can exist in an aqueous environment is H 3 O + (aq). Which H 3 O+ (aq) is stronger? Trick question They are both the same!!! Homework: K w Read: Pages 126-128 Do: #15 27 Study for the quiz!!! A NEUTRAL solution is a solution having: [H 3 O + ] = [OH - ] An ACIDIC solution is a solution having: [H 3 O + ] > [OH - ] A BASIC solution is a solution having: [H 3 O + ] < [OH - ] 10

When a strong acid and a strong base react, a great deal of heat is released: HCl (aq) + NaOH (aq) NaCl (aq) + H 2 O (l) + 59 kj In total ionic form: H + (aq) + Cl - (aq) + Na + (aq) + OH - (aq) Na + (aq) + Cl - (aq) + H 2 O (l) + 59kJ In net ionic form becomes: H + (aq) + OH - (aq) H 2 O (l) + 59kJ Hence, the SELF-IONIZATION of water can be written as: H 2 O (l) + 59 kj H + (aq) + OH - (aq) The equilibrium expression is: K w = [H + ][OH - ] = 1.0 x 10-14 (at 25 o C) [H 2 O] is a constant and is not incorporated in to the K w expression. We can also write the self ionization of water as follows: 2 H 2 O (l) + 59 kj H 3 O + (aq) + OH - (aq) K w = [H 3 O + ][OH - ] = 1.0 x 10-14 (at 25 o C) The equilibrium expression tells us that the product of [H 3 O + ] and [OH - ] is a small constant. As [H 3 O + ] increases, [OH - ] decreases, and vice versa. The [H 3 O + ] and [OH-] are very low in neutral water: [H 3 O + ] = [OH - ] = 1.0 x 10-7 M Even in acidic solutions, a small amount of OH - will be present. The equilibrium expression for water will play a major role in much of the remaining material in this section. One of the most common usages for the expression will be to solve for [H 3 O + ] once we know [OH - ], and vice versa. Unless you are told otherwise, you are allowed to assume that the temperature is 25 o C, and therefore the value of K w is 1.0 x 10-14. 11

K a and K b When a weak acid such as CH 3 COOH reacts with water we find: CH 3 COOH (aq) + H 2 O (l) CH 3 COO - (aq) + H 3 O + (aq) This process is called ACID IONIZATION. The equilibrium expression for the ionization is given by: [ K a = CH 3 COO ]H [ 3 O + ] = 1.8x10 5 CH 3 COOH The value of K a is called the ACID IONIZATION CONSTANT. When a weak base such as NH 3 reacts with water we find: NH 3(aq) + H 2 O (l) NH + 4 (aq) + OH - (aq) This process is called BASE IONIZATION. The equilibrium expression for the ionization is given by: K b = NH + [ 4]OH [ ] 1.8x10 5 NH 3 The value of K b is called the BASE IONIZATION CONSTANT. The LARGER the K a the STRONGER the acid. For a strong acid the value of K a is not defined since the acid is 100% ionized and the concentration of the non-ionized acid in the denominator of the K a expression is zero. The LARGER the K b the STRONGER the base. Lastly You may compare K a s and K b s for amphiprotic substances: If K a > K b, the species will act as an acid. If K a < K b, the species will act as a base. The table does not list K b values but we shall soon see how to calculate the K b values for bases using K a values in the table. 12

Read: Pages 129-133 Do: #28 34 Homework: The Relationship Between K a and K b for a Conjugate Pair Experimentally, it is found that for the acid ionization: We have: NH + 4 (aq) + H 2 O (l) NH 3(aq) + H 3 O + (aq) + [ NH 4 ] [ K a = NH 3]H 3 O + While for the base ionization: We have: NH 3(aq) + H 2 O (l) NH + 4 (aq) + OH - (aq) K b = NH + [ 4]OH NH 3 Since both equations involve NH 3 and NH 4+, there is a relationship between K a and K b for the species: + [ NH 4 ] [ K a K b = NH 3]H 3 O + For a conjugate pair we get: NH + [ 4]OH [ NH 3 ] K w =K a K b = [ H 3 O + ]OH 13

For Example: DO NOT USE THE WRONG K A VALUE! If you are required to find K b for a species, then you must look on the right-hand side of the table where it is acting as a BASE. Relative Strengths of Acids and Bases Recall that in a Brønsted-Lowry acid/base equilibrium we have: acid + base acid + base When there are two acids in an equilibrium, both of them are trying to donate a proton. We end up having a proton competition between them. Reasonably, the stronger of the two acids involved will be more successful in donating a proton. CO 2-3 (aq) + H 2 PO - 4 (aq) ACID STRONGEST HCO - 3 (aq) + HPO 2-4 (aq) ACID Products are favoured. 14

Since the above reaction is an equilibrium, we can write an equilibrium expression for it: k eq = HCO 2 [ 3]HPO 4 2 [ CO 3 ]H 2 PO 4 = 1.3 x 103 A general method to calculate the K eq value is for a Brønsted-Lowry equilibrium is: k eq = k a (reactant acid) k a (product acid) The K eq value agrees with our prediction: there should be more products than reactants if K eq > 1. For Example: For Example: 15

Read: Pages 134-139 Do: #35-46 Homework: The ph Scale ph stands for the potential of Hydrogen. ph s of solutions vary from 0 14 (although <0 and >14 is possible.) On the scale: ph < 7 is acidic ph = 7 is neutral ph > 7 is basic The ph scale is logarithmic (a change of 10x in the [H + ] results in a change of 1 in the ph scale) ph and poh A solution is: BASIC when its ph > 7 and poh < 7 ACIDIC when its ph < 7 and poh > 7 NEUTRAL solution has ph = poh = 7. Anywhere on the scale ph + poh = 14 and [H 3 O + ][OH - ] = 1.0 x 10-14 M. The ph and poh is defined as: ph = - log H 3 O + poh = - log OH 16

Converting From [H 3 O + ] and [OH-] to ph and poh If [H 3 O + ] = 3.94 x 10-4 M, what is the ph? If [OH-] = 9.51 X 10-12 M, what is the poh? Converting ph and poh to [H 3 O+] and [OH-] This process is a little more complicated than the process in the above two examples. Since: ph = -log[h 3 O + ] We must rearrange the above equation and solve for [H 3 O + ]. Remove the negative sign by changing the sign of the entire equation: -ph = log[h 3 O + ] Remove the log by taking the antilog of the entire equation: antilog(-ph) = antilog (log[h 3 O + ]) For Example: If ph = 3.405, what is [H 3 O + ]? If poh = 11.682, what is [OH-]? antilog(-ph) = [H 3 O + ] We have our new equation: [H 3 O + ]=10 ph or [OH ]=10 poh 17

There is a very important relationship between with the above material and the K w expression. Recall: [H 3 O + ][OH - ] = 1.0 x 10-14 Taking the logarithm of both sides of the equation we get: log ( [H 3 O + ][OH - ] ) = log (1.0 x 10-14 ) But, log (a b) = log (a) + log (b), so that: log [H 3 O + ] + log [OH - ] = log (1.0 x 10-14 ) log (1.0 x 10-14 ) = -14.0 We get: log [H 3 O + ] + log [OH - ] = -14.0 Multiplying the equation through by -1 we have: -log [H 3 O + ] log [OH - ] = 14.0 Finally, since ph = -log[h 3 O + ], and poh = -log[oh - ] we have: ph+ poh= 14.0 We can now calculate values back and forth between any of [H 3 O + ], [OH - ], ph, and poh as shown in the following scheme: For Example: If [H 3 O + ] = 2.47 x 10-5 M, what is the poh? [H 3 O + ] = 10 -ph [OH - ] = 10 -poh If poh = 2.35, what is the [H 3 O + ]? Calculate the ph of 0.25 M Sr(OH) 2. 18

Homework: Mixtures of Strong Acids and Bases Read: Pages 140-143 Do: #47 57 Study for your Quiz! When we mix an acid and a base, we can obtain an acidic, basic, or neutral solution depending on the relative amounts of reactants involved. Since the species are both strong, the reaction goes to 100% completion. Whatever is left over (excess) will determine the ph of the final solution. For Example: Homework: Do: #58 68 19