Acids and Bases Acids Bases The BronstedLowry model

Acids and Bases

Acids & Bases The Bronsted-Lowry model defines an acid as a proton donor. A base is a proton acceptor. Note that this definition is based on the transfer of a proton from the acid to the base.

Acids & Bases H 2 O(l) + HCl(g) H 3 O+(aq) + Cl 1(aq) In this reaction, water accepts a proton from HCl. Water is a base, and HCl is an acid. H 2 O(l) + HCl(g) H 3 O+(aq) + Cl 1 -(aq) proton acceptor donor B-L base B-L acid

Acids & Bases Since this reaction goes to completion (note the one-way arrow), we classify HCl(aq) as a strong acid. H 2 O(l) + HCl(g) H 3 O+(aq) + Cl 1 -(aq) proton acceptor donor

Acids & Bases Hydrochloric acid dissociates 100%, and exists as a solution of hydronium and chloride ions. H 2 O(l) + HCl(g) H 3 O+(aq) + Cl 1 -(aq) proton acceptor donor Although a bottle may be labeled 1. 0 M HCl, it really contains 1. 0 M H 3 O+(aq) and 1. 0 M Cl 1 -(aq).

Acids & Bases Hydrochloric acid dissociates 100%, and exists as a solution of hydronium and chloride ions. H 2 O(l) + HCl(g) H 3 O+(aq) + Cl 1 -(aq) proton acceptor donor There is no reverse reaction because chloride has no tendency to accept a proton to form HCl.

Acids & Bases There are only a few common strong acids. They are: HCl(aq), HBr(aq), HI(aq), HNO 3(aq), HCl. O 4(aq) and H 2 SO 4(aq)* *for the first proton only

Acids & Bases Most acids are weak acids in which only a small percentage of the molecules dissociate to protonate water. HA(aq) + H 2 O(aq) ↔ H 3 O+(aq) + A-(aq) HA is a generic monoprotic weak acid such as HF, HCN or CH 3 COOH.

Weak Acids Weak acids in water form an equilibrium with hydronium ion and the deprotonated anion of the acid. The reaction does not go to completion. H-A + H 2 O ↔ H 3 O + + A-

Acids The equilibrium of acids in water can be viewed as a competition between the forward reaction and the reverse reaction. H-A + H 2 O ↔ H 3 O + + A-

Acids The forward reaction involves HA, a generic acid, and water. Water acts as a base by accepting a proton from the acid. H-A + H 2 O ↔ H 3 O + + A-

Acids The forward reaction involves HA, a generic acid, and water. Water acts as a base by accepting a proton from the acid. Base H 2 O Acid HA proton donor proton acceptor H-A + H 2 O ↔ H 3 O + + A-

Acids The reverse reaction involves Aaccepting a proton and acting as a base. H 3 O+ donates a proton, and is an acid. Base A Acid H 3 O+ proton donor acceptor Base H 2 O Acid HA proton donor proton acceptor H-A + H 2 O ↔ H 3 O + + A-

Ka Values Acid strength is determined by measuring the equilibrium constant for the following reaction: HA(aq) + H 2 O(l) ↔ H 3 O+(aq) + A-(aq) Ka = [H 3 O+][A-] [HA]

Ka Values HA(aq) + H 2 O(l) ↔ H 3 O+(aq) + A(aq) Ka = [H 3 O+][A-] [HA] Water is left out of the equilibrium constant expression because it is a pure liquid (with constant concentration). The “a” subscript stands for acid.

Conjugate Acids & Bases The deprotonated acid is a base, and the protonated base is an acid. These acids and bases are called conjugate Base A acids and bases. Acid H O+ Base H O Acid HA 2 proton donor proton acceptor H-A + H 2 O 3 proton donor acceptor ↔ H 3 O + + A-

Acids For weak acids, the equilibrium lies to the left, indicating that A- is a stronger base than water. Base H 2 O Acid HA proton donor proton acceptor Base A Acid H 3 O+ proton donor acceptor

Acids Acid HA and base A- are related, and differ only by the addition or removal of H+. Acid HA Base H 2 O Acid H 3 O+ remove H+ add H+ Base A-

Acids HA and A- are called conjugate acidbase pairs. A- is the conjugate base of the acid HA. Acid HA Base H 2 O Acid H 3 O+ remove H+ add H+ Base A-

Acids Likewise, H 2 O and H 3 O+ are related, and differ only by the addition or removal of H+. Acid HA Base H 2 O Acid H 3 O+ add H+ remove H+ Base A-

Acids H 2 O and H 3 O+ are conjugate acidbase pairs. H 2 O is the conjugate base of H 3 O +. Acid HA Base H 2 O Acid H 3 O+ add H+ remove H+ Base A-

Conjugate Acid & Bases n Provide the formulas for the conjugate bases of: H 2 SO 4 HCN H 2 O CH 3 COOH

Conjugate Acid & Bases n Provide the formulas for the conjugate acids of: NH 3 OH 1 HCO 31 H 2 O

Structure and Acid Strength For binary acids of the general formula Hn. X (where X is a non-metal), the acidity reflects the H-X bond strength and polarity. If the bond is strong and non-polar, such as with carbon (in CH 4), the compound is non-acidic, and the C-H bonds remain intact.

Structure and Acids & Bases The effects of bond polarity can be seen in the hydrogen halides: HF, HCl, HBr and HI. HF has the most polar bond, yet it is the weakest acid of the group. The HF bond is the strongest of the group, and this very high bond strength results in a lower tendency for dissociation in water.

Binary acid strength depends upon the electronegativity of the non-metal, and also the bond strength of the bond with hydrogen.

The bond strength is greatest when there is a good match in energy between the 1 s orbital on hydrogen, and the bonding orbital on the non -metal.

Structure and Acids & Bases The acids HCl, HBr and HI are all strong acids. In all cases, the bonds are polar, but they are also weaker than the HF bond, and are more easily broken in aqueous solution.

Structure and Acids & Bases

The Oxyacids The oxyacids typically have one or more oxygen atoms attached to a central non-metal. Attached to one or more of the oxygen atoms are acidic hydrogens. Examples include H 2 SO 4, HNO 3, HNO 2, H 3 PO 4 and HCl. O 4. It is important to remember that in all cases, the acidic hydrogen is attached to oxygen (X-O-H).

The Oxyacids

Oxyacids For a given central atom, the greater the number of oxygen atoms attached, the more acidic the acid. The larger number of oxygen atoms polarizes and weakens the O-H bonds.

Oxyacids The greater the number of oxygen atoms, the weaker the O-H bonds, and the stronger the acid.

Oxyacids The nature of the central atom also affects the acidity of the acid. The more electronegative X is (in X-O-H), the weaker the O-H bond, and the stronger the acid.

Oxyacids The nature of the central atom also affects the acidity of the acid. The more electronegative X is (in X-O-H), the weaker the O-H bond, and the stronger the acid.

Strong and Weak Acids HX is a weak acid and forms only a small amount of H 3 O+ and X-

Strong and Weak Acids HY is a strong acid and dissociates completely to form H 3 O+ and Y-

Strong and Weak Acids

Acid Strength The conjugate bases of strong acids have no tendency to pick up a proton. There is no reverse reaction. The conjugate bases of infinitely strong acids will not act as a base and accept a proton.

Ka Values

Bases Strong bases are very effective at accepting protons. The most common strong bases are the soluble group IA and IIA metal hydroxides. In general, the metal ion is non-reactive, and serves as a spectator ion. Another strong base is oxide ion, O 2 -. Oxide reacts with water to become fully protonated. O 2 -(aq) + H 2 O(l) 2 OH-(aq)

Bases O 2 -(aq) + H 2 O(l) 2 OH-(aq) In this reaction, water is donating a proton, and hence acting as an acid. In previous reactions, water accepted a proton and served as a base. Substances that can behave as either an acid or a base are called amphoteric.

Amphoteric Nature of Water Depending upon its environment, water may donate a proton, acting as an acid, or accept a proton, and act as a base. This behavior is characteristic of amphoteric substances. Pure water molecules can react with each other, to a very small extent, to form hydronium and hydroxide ions.

Autoionization of Water This process is called autoionization or selfionization. One water molecule donates a proton to another. The result is the formation of equal amounts of hydronium and hydroxide ions. H 2 O(l) + H 2 O(l) ↔ H 3 O+(aq) + OH-(aq)

Autoionization of Water H 2 O(l) + H 2 O(l) ↔ H 3 O+(aq) + OH-(aq) Kw = [H 3 O+][OH-] = 1. 0 x 10 -14 at 25 o. C In pure water, the concentration of hydronium ion equals the concentration of hydroxide. Both ions have a concentration of 1. 0 x 10 -7 M.

[H 3 O+] and [OH-] in Aqueous Solution The product of the hydroxide and hydronium concentration in any aqueous solution must equal Kw. As a result, when a solution is acidic, the hydronium concentration increases, and the hydroxide concentration decreases. Kw = [H 3 O+][OH-] = 1. 0 x 10 -14 at 25 o. C

[H 3 O+] and [OH-] in Aqueous Solution Likewise, in basic solutions, the hydroxide ion concentration is greater than the hydronium ion concentration. There is always some hydroxide ion and some hydronium ion present in any aqueous solution.

[H 3 O+] and [OH-] in Aqueous Solution

The p. H Scale A scale of acidity, the p. H scale, is used to indicate the degree of acidity of aqueous solutions. p. H = -log[H 3 O+] or –log[H+] The scale generally runs from 0 -14, though negative p. H values are possible. A neutral solution will have a p. H = 7. 00

The p. H Scale A one unit change in p. H is a ten-fold change in the concentration of hydronium ion. Acidic solutions have p. H values less than 7. 00, and basic solutions have p. H values greater than 7. 00

p. H Values Most foods have p. H values in the acidic range.

Determination of p. H Universal Indicator

Determination of p. H A portable p. H meter

Problem: Calculation of p. H n Calculate the p. H of 0. 10 M HCl. - Is it an acid or a base? - Is it strong or weak? - Write the appropriate chemical reaction(s). HCl(aq) + H 2 O(l) H 3 O+(aq) + Cl-(aq) or HCl(aq) H+(aq) + Cl-(aq)

Problem: Calculation of p. H n Calculate the p. H of 0. 10 M HCl(aq) + H 2 O(l) H 3 O+(aq) + Cl(aq) [H 3 O+] = 0. 10 M p. H = - log [H 3 O+] = - log( 0. 10) = 1. 00 For every significant digit in the concentration, there is a place after the decimal in the p. H.

Problem: Calculation of p. H n Calculate the p. H of 0. 10 M HCl. p. H = - log [H 3 O+] = - log( 0. 10) = 1. 00 - Does your answer make sense? The p. H of a fairly concentrated strong acid should be much less than 7. Yes, the answer makes sense.

Problem: p. H of 0. 50 M HCN n Calculate the p. H of 0. 50 M HCN. - Acid or base? - Weak or strong? HCN is a weak acid, with Ka = 6. 2 x 10 -10. HCN(aq) + H 2 O(l) ↔ H 3 O+(aq) + CN(aq)

Problem: p. H of 0. 50 M HCN n Calculate the p. H of 0. 50 M HCN is a weak acid, with Ka = 6. 2 x 1010. HCN(aq) + H 2 O(l) ↔ H 3 O+(aq) + CN(aq) - Write the equilibrium constant expression. Ka = 6. 2 x 10 -10 = [H 3 O+][CN-] [HCN]

Problem: p. H of 0. 50 M HCN - Make a table of initial, change and equilibrium concentrations. HCN(aq) + H 2 O(aq) ↔ H 3 O+(aq) + CN(aq) HCN H 3 O+ CNinitial 0. 50 chang -x e equil. 0. 50 -x 0 0 +x +x x x

Problem: p. H of 0. 50 M HCN - Substitute and solve. initial HCN H 3 O + CN- 0. 50 0 0 +x +x x x chang -x e equil. 0. 50 -x Ka = 6. 2 x 10 -10 = [H 3 O+][CN-] [HCN] Ka = 6. 2 x 10 -10 = [x] [. 50 -x]

Problem: p. H of 0. 50 M HCN - State any assumptions. - Solve the problem. - Check your assumptions. - Answer the question. - Does your answer make sense? - Check your results, if possible.

p. H of Weak Acids If the same question were asked concerning a 0. 50 M solution of chlorous acid, HCl. O 2 (Ka=1. 2 x 10 -2), you would need to use the quadratic formula to answer the question. Generally, if the acid concentration is. 10 M or larger, and K is 10 -5 or smaller, you won’t need to use the quadratic formula.

Problem n A 0. 0875 M solution of an unknown monoprotic acid has a p. H of 2. 28. Calculate the Ka of the acid. - Write the chemical reaction. HA(aq) + H 2 O(l) ↔ H 3 O+(aq) + A-(aq) - Write the equilibrium constant expression. Ka = [H 3 O+][A-] [HA]

Problem n A 0. 0875 M solution of an unknown monoprotic acid has a p. H of 2. 28. Calculate the Ka of the acid. - Make a table of concentrations.

Problem n A 0. 0875 M solution of an unknown monoprotic acid has a p. H of 2. 28. Calculate the Ka of the acid. [HA] [H 3 O+] [A-] init chg equil . 0875 0 0

Problem n A 0. 0875 M solution of an unknown monoprotic acid has a p. H of 2. 28. Calculate the Ka of the acid. The p. H can [HA] [H 3 O+] [A-] be used to init. 0875 0 0 calculate how + much H O chg 3 and A- form. equil

Problem n A 0. 0875 M solution of an unknown monoprotic acid has a p. H of 2. 28. Calculate the Ka of the acid. To calculate [H 3 O+] from p. H, enter the p. H, change its sign, and take the inverse log. You will get a [H 3 O+] = 5. 2 x 10 -3. This is the equilibrium concentration of H 3 O+ (and A-).

Problem n A 0. 0875 M solution of an unknown monoprotic acid has a p. H of 2. 28. Calculate the Ka of the acid. [HA] [H 3 O+] [A-] init . 0875 0 0 chg equil 5. 2 x 10 -3 3

Problem n A 0. 0875 M solution of an unknown monoprotic acid has a p. H of 2. 28. Calculate the Ka of the acid. [HA] [H 3 O+] [A-] init . 0875 0 0 chg -. 0052 + 5. 2 x 10 -3 equil 5. 2 x 10 -3 3

Problem n A 0. 0875 M solution of an unknown monoprotic acid has a p. H of 2. 28. Calculate the Ka of the acid. [HA] [H 3 O+] [A-] init . 0875 0 0 chg -. 0052 + 5. 2 x 10 -3 equil. 0823 5. 2 x 10 -3 3

Problem n A 0. 0875 M solution of an unknown monoprotic acid has a p. H of 2. 28. Calculate the Ka of the acid. [HA] [H 3 O+] [A-] equil . 0823 5. 2 x 10 -3 3 Substitute into the equilibrium constant expression and solve for Ka.

Problem n A 0. 0875 M solution of an unknown monoprotic acid has a p. H of 2. 28. Calculate the Ka of the[HA] acid. [H O+] [A-] 3 equil . 0823 5. 2 x 10 -3 3 Ka = [H 3 O+] [A-]/ [HA] = (5. 2 x 10 -3) (0. 0823) Ka = 3. 3 x 10 -4

Percent Ionization Sometimes the ionization of a weak acid solution is expressed in % ionization. For a given solution, % ionization = (100%)[H 3 O+]eq/[HA]init

Problem: p. H n Calculate the p. H of 0. 0050 M Ca(OH)2. - Acid or base? - Strong or weak? - Write the reaction(s). Ca(OH)2 in water is a strong base. Ca(OH)2(aq) Ca 2+(aq) + 2 OH-(aq)

Problem: p. H n Calculate the p. H of 0. 0050 M Ca(OH)2(aq) Ca 2+(aq) + 2 OH-(aq) 0. 0050 M 2(0. 0050 M) [OH-] = 2(0. 0050 M) = 0. 0100 M Since [OH-] and [H 3 O+] are related by Kw, the p. H can be calculated.

Problem: p. H from [OH-] = 2(0. 0050 M) = 0. 0100 M You can use: [H 3 O+] [OH-] = 1. 0 x 10 -14 to calculate hydronium concentration and the p. H, or: p. H + p. OH = -log Kw=14. 00 (@25 o. C)

Problem: What is the p. H of 0. 15 M NH 3? NH 3 is a weak base, and NH 4+ is its conjugate acid. Kb = [OH-][NH 4+] [NH 3]

Values of Kb for Some Common Weak Bases

What is the p. H of 0. 15 M NH 3? Kb = [OH-][NH 4+] = 1. 8 x 10 -5 [NH 3] NH 3(aq) + H 2 O(l) ↔ OH-(aq) + NH 4+(aq) NH 3 OH- NH 4+ initial 0. 15 0 0 chang e -x +x +x equil. . 15 -x x x

p. H of Salt Solutions When salts (ionic compounds) dissolve in water, they dissociate into separate hydrated ions. Since many ions are conjugate acids or bases, they may react to form hydronium or hydroxide ions.

Acid Strength The conjugate bases of strong acids have no tendency to pick up a proton. There is no reverse reaction. The conjugate bases of infinitely strong acids are infinitely weak.

Conjugate Acid. Base Strength The weaker the acid, the stronger its conjugate base. As the acid gets weaker, the reverse reaction with water becomes more significant.

Conjugate Acid-Base Strength

p. H of Salt Solutions Since the conjugate bases of strong acids have no tendency to accept protons, these ions will have no effect on p. H. The “neutral” anions are: Cl-, Br-, I-, NO 3 and Cl. O 4(Although HSO 41 - isn’t basic, it is acidic, and will donate a proton to form hydronium ion. )

p. H of Salt Solutions The cations of strong bases also have no effect on p. H when dissolved in water. As a result, the group IA and group IIA metal ions will produce neutral solutions. Some small, highly positively charged metal ions, such as Al 3+, Cr 3+ or Fe 3+ produce slightly acidic solutions.

p. H of Salt Solutions n Are the following aqueous solutions acidic, basic or neutral? Na. NO 3, KCN, NH 4 Cl, Ca(HSO 4)2, Li. CH 3 CO 2

p. H of Salt Solutions n Calculate the p. H of 0. 10 M Na. CH 3 CO 2(aq) Na+(aq) + CH 3 CO 2 -(aq) Acetate ion is the conjugate base of the weak acid, acetic acid. So, acetate will act as a base and accept protons from water. CH 3 CO 2 -(aq) + H 2 O(l) ↔ CH 3 CO 2 H(aq) + OH-(aq) The solution should be basic with the p. H>7. 0

p. H of Salt Solutions n Calculate the p. H of 0. 10 M Na. CH 3 CO 2 -(aq) + H 2 O(l) ↔ CH 3 CO 2 H(aq) + OH-(aq) Kb = [CH 3 CO 2 H][OH-] [CH 3 CO 2 -] The value of Kb for acetate ion can be calculated from the value of Ka for acetic acid.

p. H of Salt Solutions For any weak conjugate acid-base pairs, Ka(Kb) = Kw (Ka for acetic acid) (Kb for acetate ion)= 1. 0 x 10 -14 Kb for acetate ion=1. 0 x 10 -14/1. 8 x 10 -5 = 5. 6 x 10 -10

p. H of Salt Solutions n Calculate the p. H of 0. 10 M Na. CH 3 CO 2 -(aq) + H 2 O(l) ↔ CH 3 CO 2 H(aq) + OH-(aq) Kb = [CH 3 CO 2 H][OH-] = 5. 6 x 10 -10 [CH 3 CO 2 -] initial chang e equil. 0. 10 [CH 3 CO 2 H [OH-] ] 0 0 -x +x +x . 10 -x x x

Polyprotic Acids that have more than one proton that can be donated are called polyprotic acids. Examples are H 2 SO 4, H 3 PO 4, H 2 S and H 2 CO 3. The loss of protons can be viewed as a step-wise process.

Sulfuric Acid Sulfuric acid, H 2 SO 4, can lose two protons. It is called a diprotic acid. Sulfuric acid is also one of the common strong acids. This applies to loss of the first proton only. H 2 SO 4(aq) + H 2 O(l) H 3 O+(aq) + HSO 4(aq) This reaction goes 100% to the right.

Sulfuric Acid H 2 SO 4(aq) + H 2 O(l) H 3 O+(aq) + HSO 4(aq) HSO 4 - can react with water to lose an additional proton.

Sulfuric Acid HSO 4 -(aq) + H 2 O(l) ↔ H 3 O+(aq) + SO 42(aq) The double arrows indicate an equilibrium is established because HSO 4 - is a weak acid. Ka for HSO 4 - = [H 3 O+][SO 42 -] = 1. 2 x 10 -2 [HSO 4 -]

Sulfuric Acid HSO 4 -(aq) + H 2 O(l) ↔ H 3 O+(aq) + SO 42(aq) Ka for HSO 4 - = [H 3 O+][SO 42 -] = 1. 2 x 10 -2 [HSO 4 -] The value of Ka indicates that hydrogen sulfate ion is a relatively strong weak acid. It is stronger than most weak acids, but weaker than the strong acids HCl, HNO 3, H 2 SO 4 or HCl. O 4.

Polyprotic Acids H 3 A(aq) + H 2 O(l) ↔ H 3 O+(aq) + H 2 A(aq) Ka 1 = [H 3 O+][H 2 A-]/[H 3 A] H 2 A-(aq) + H 2 O(l) ↔ H 3 O+(aq) + HA 2 -(aq) Ka 2 = [H 3 O+][HA 2 -]/[H 2 A-] HA 2 -(aq) + H 2 O(l) ↔ H 3 O+(aq) + A 3 -(aq) Ka 3 = [H 3 O+][A 3 -]/[HA 2 -]

Polyprotic Acids For all polyprotic acids, Ka 1>Ka 2>Ka 3. It becomes progressively more difficult to remove protons as the acid becomes more negative in charge. Since the first dissociation usually predominates, it can often be used to determine the p. H of a solution of the polyprotic acid. The subsequent steps are usually negligible.

Polyprotic Acids

Problem: Polyprotic Acids n Calculate the p. H and the concentration of all species present in 0. 40 M H 2 CO 3. (Ka 1 = 4. 3 x 10 -7, Ka 2 =5. 6 x 10 -11). Since Ka 1>>Ka 2, the p. H can be obtained by considering only the first dissociation.

Problem: Polyprotic Acids n Calculate the p. H and concentration of all species present in 0. 40 M H 2 CO 3. (Ka 1 = 4. 3 x 10 -7, Ka 2 =5. 6 x 10 -11). H 2 CO 3(aq) + H 2 O(l) ↔ H 3 O+(aq) + HCO 31(aq) Ka 1=[H 3 O+][HCO 31 -] = 4. 3 x 10 -7 [H 2 CO 3]

Lewis Acids & Bases The Lewis definition of acids and bases is broader than the Bronsted-Lowry definition. It defines a Lewis base as an electron pair donor. Thus, all Lewis bases have at least one lone pair of electrons. A Lewis acid is an electron pair acceptor. H+(aq) + H 2 O(l) H 3 O+(aq)

Lewis Acids & Bases A Lewis acid had an empty orbital than can accept an electron pair. Metal ions can act as Lewis acids: Ag+(aq) + 2 : NH 3(aq) Ag(NH 3)2+(aq) The product is often referred to as an adduct.