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.

NH3(aq) + H3O+(l)

Acids & Bases H2O(l) + HCl(g)  H3O+(aq) + Cl1-(aq) In this reaction, water accepts a proton from HCl. Water is a base, and HCl is an acid. proton proton acceptor donor B-L base B-L acid

H2O(l) + HCl(g)  H3O+(aq) + Cl1-(aq) Acids & Bases Since this reaction goes to completion (note the one-way arrow), we classify HCl(aq) as a strong acid. H2O(l) + HCl(g)  H3O+(aq) + Cl1-(aq) proton proton acceptor donor

H2O(l) + HCl(g)  H3O+(aq) + Cl1-(aq) Acids & Bases Hydrochloric acid dissociates 100%, and exists as a solution of hydronium and chloride ions. H2O(l) + HCl(g)  H3O+(aq) + Cl1-(aq) proton proton acceptor donor Although a bottle may be labeled 1.0M HCl, it really contains 1.0 M H3O+(aq) and 1.0 M Cl1-(aq).

H2O(l) + HCl(g)  H3O+(aq) + Cl1-(aq) Acids & Bases Hydrochloric acid dissociates 100%, and exists as a solution of hydronium and chloride ions. H2O(l) + HCl(g)  H3O+(aq) + Cl1-(aq) proton 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), HNO3(aq), HClO4(aq) and H2SO4(aq)* *for the first proton only

HA(aq) + H2O(aq) ↔ H3O+(aq) + A-(aq) Acids & Bases Most acids are weak acids in which only a small percentage of the molecules dissociate to protonate water. HA(aq) + H2O(aq) ↔ H3O+(aq) + A-(aq) HA is a generic monoprotic weak acid such as HF, HCN or CH3COOH.

Weak Acids H-A + H2O ↔ H3O+ + A- 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 + H2O ↔ H3O+ + A-

Acids The equilibrium of acids in water can be viewed as a competition between the forward reaction and the reverse reaction. H-A + H2O ↔ H3O+ + 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 + H2O ↔ H3O+ + A-

Acids The forward reaction involves HA, a generic acid, and water. Water acts as a base by accepting a proton from the acid. Acid HA proton donor Base H2O proton acceptor H-A + H2O ↔ H3O+ + A-

Acids The reverse reaction involves A- accepting a proton and acting as a base. H3O+ donates a proton, and is an acid. Base A- proton acceptor Acid H3O+ proton donor Acid HA proton donor Base H2O proton acceptor H-A + H2O ↔ H3O+ + A-

Conjugate Acids & Bases The deprotonated acid is a base, and the protonated base is an acid. These acids and bases are called conjugate acids and bases. Base A- proton acceptor Acid H3O+ proton donor Acid HA proton donor Base H2O proton acceptor H-A + H2O ↔ H3O+ + A-

Acids For weak acids, the equilibrium lies to the left, indicating that A- is a stronger base than water. Base A- proton acceptor Acid H3O+ proton donor Acid HA proton donor Base H2O proton acceptor

Acids Acid HA and base A- are related, and differ only by the addition or removal of H+. Acid H3O+ Acid HA Base H2O Base A- remove H+ add H+

Acids HA and A- are called conjugate acid-base pairs. A- is the conjugate base of the acid HA. Acid H3O+ Acid HA Base H2O Base A- remove H+ add H+

Acids Likewise, H2O and H3O+ are related, and differ only by the addition or removal of H+. Acid H3O+ Acid HA Base H2O Base A- add H+ remove H+

Acids H2O and H3O+ are conjugate acid-base pairs. H2O is the conjugate base of H3O+ . Acid H3O+ Acid HA Base H2O Base A- add H+ remove H+

Conjugate Acid & Bases Provide the formulas for the conjugate bases of: H2SO4 HCN H2O CH3COOH

Conjugate Acid & Bases Provide the formulas for the conjugate acids of: NH3 OH1- HCO31- H2O

Strong and Weak Acids HX is a weak acid and forms only a small amount of H3O+ and X-

Strong and Weak Acids HY is a strong acid and dissociates completely to form H3O+ 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 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

Structure and Acids & Bases For binary acids of the general formula HnX (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 CH4), 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.

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 H-F 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 H2SO4, HNO3, HNO2, H3PO4 and HClO4. 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.

Organic Acids (carboxylic acids) Carboxylic acids contain the following structure: R represents an organic group or structure.

Organic Acids (carboxylic acids) Loss of the acidic proton results in the carboxylate ion, which has resonance, with the negative charge delocalized.

HA(aq) + H2O(l) ↔ H3O+(aq) + A-(aq) Ka Values Acid strength is determined by measuring the equilibrium constant for the following reaction: HA(aq) + H2O(l) ↔ H3O+(aq) + A-(aq) Ka = [H3O+][A-] [HA]

Ka Values HA(aq) + H2O(l) ↔ H3O+(aq) + A-(aq) Ka = [H3O+][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.

Ka Values

H2SO4(aq) + H2O(l)  H3O+(aq) + HSO4-(aq) Sulfuric Acid Sulfuric acid, H2SO4, 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. H2SO4(aq) + H2O(l)  H3O+(aq) + HSO4-(aq) This reaction goes 100% to the right.

H2SO4(aq) + H2O(l)  H3O+(aq) + HSO4-(aq) Sulfuric Acid H2SO4(aq) + H2O(l)  H3O+(aq) + HSO4-(aq) HSO4- can react with water to lose an additional proton.

Ka for HSO4- = [H3O+][SO42-] = 1.2 x 10-2 Sulfuric Acid HSO4-(aq) + H2O(l) ↔ H3O+(aq) + SO42-(aq) The double arrows indicate an equilibrium is established because HSO4- is a weak acid. Ka for HSO4- = [H3O+][SO42-] = 1.2 x 10-2 [HSO4-]

Ka for HSO4- = [H3O+][SO42-] = 1.2 x 10-2 Sulfuric Acid HSO4-(aq) + H2O(l) ↔ H3O+(aq) + SO42-(aq) Ka for HSO4- = [H3O+][SO42-] = 1.2 x 10-2 [HSO4-] 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, HNO3, H2SO4 or HClO4.

Ka Values

O2-(aq) + H2O(l)  2 OH-(aq) 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, O2-. Oxide reacts with water to become fully protonated. O2-(aq) + H2O(l)  2 OH-(aq)

Bases O2-(aq) + H2O(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. H2O(l) + H2O(l) ↔ H3O+(aq) + OH-(aq)

Autoionization of Water H2O(l) + H2O(l) ↔ H3O+(aq) + OH-(aq) Kw = [H3O+][OH-] = 1.0 x 10-14 at 25oC In pure water, the concentration of hydronium ion equals the concentration of hydroxide. Both ions have a concentration of 1.0 x 10-7M.

Autoionization of Water

[H3O+] 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.

[H3O+] 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.

[H3O+] and [OH-] in Aqueous Solution

pH = -log[H3O+] or –log[H+] The pH Scale A scale of acidity, the pH scale, is used to indicate the degree of acidity of aqueous solutions. pH = -log[H3O+] or –log[H+] The scale generally runs from 0-14, though negative pH values are possible. A neutral solution will have a pH = 7.00

The pH Scale A one unit change in pH is a ten-fold change in the concentration of hydronium ion. Acidic solutions have pH values less than 7.00, and basic solutions have pH values greater than 7.00

pH Values Most foods have pH values in the acidic range.

Determination of pH Universal Indicator

Determination of pH A portable pH meter

Problem: Calculation of pH Calculate the pH of 0.10M HCl. - Is it an acid or a base? - Is it strong or weak? - Write the appropriate chemical reaction(s). HCl(aq) + H2O(l)  H3O+(aq) + Cl-(aq) or HCl(aq)  H+(aq) + Cl-(aq)

Problem: Calculation of pH Calculate the pH of 0.10M HCl. HCl(aq) + H2O(l)  H3O+(aq) + Cl-(aq) [H3O+] = 0.10M pH = - log [H3O+] = - log( 0.10) = 1.00 For every significant digit in the concentration, there is a place after the decimal in the pH.

Problem: Calculation of pH Calculate the pH of 0.10M HCl. pH = - log [H3O+] = - log( 0.10) = 1.00 - Does your answer make sense? The pH of a fairly concentrated strong acid should be much less than 7. Yes, the answer makes sense.

Question: pH Can you have a negative value for pH? Under what circumstances? If the hydronium concentration is > 1.0M, the pH will be negative. The pH will be zero if the hydronium concentration equals 1.0M.

Problem: pH Calculate the pH of 1.5 x 10-10M HClO4.

Ca(OH)2(aq)  Ca2+(aq) + 2 OH-(aq) Problem: pH Calculate the pH of 0.0050M 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)  Ca2+(aq) + 2 OH-(aq)

Ca(OH)2(aq)  Ca2+(aq) + 2 OH-(aq) Problem: pH Calculate the pH of 0.0050M Ca(OH)2. Ca(OH)2(aq)  Ca2+(aq) + 2 OH-(aq) 0.0050M 0.0050M 2(0.0050M) [OH-] = 2(0.0050M) = 0.0100M Since [OH-] and [H3O+] are related by Kw, the pH can be calculated.

Problem: pH from [OH-] [OH-] = 2(0.0050M) = 0.0100M You can use: [H3O+] [OH-] = 1.0 x 10-14 to calculate hydronium concentration and the pH, or: pH + pOH = -log Kw=14.00 (@25oC)

HCN(aq) + H2O(l) ↔ H3O+(aq) + CN-(aq) Problem: pH of 0.50M HCN Calculate the pH of 0.50M HCN. - Acid or base? - Weak or strong? HCN is a weak acid, with Ka = 6.2 x 10-10. HCN(aq) + H2O(l) ↔ H3O+(aq) + CN-(aq)

HCN(aq) + H2O(l) ↔ H3O+(aq) + CN-(aq) Problem: pH of 0.50M HCN Calculate the pH of 0.50M HCN. HCN is a weak acid, with Ka = 6.2 x 10-10. HCN(aq) + H2O(l) ↔ H3O+(aq) + CN-(aq) - Write the equilibrium constant expression. Ka = 6.2 x 10-10 = [H3O+][CN-] [HCN]

Problem: pH of 0.50M HCN Make a table of initial, change and equilibrium concentrations. HCN(aq) + H2O(aq) ↔ H3O+(aq) + CN-(aq) HCN H3O+ CN- initial 0.50 change -x +x equil. 0.50-x x

Problem: pH of 0.50M HCN Substitute and solve. initial 0.50 change -x +x equil. 0.50-x x Ka = 6.2 x 10-10 = [H3O+][CN-] [HCN] Ka = 6.2 x 10-10 = [x] [x] [.50-x]

Problem: pH of 0.50M HCN - State any assumptions. Solve the problem. Check your assumptions. Answer the question. Does your answer make sense? Check your results, if possible.

pH of Weak Acids If the same question were asked concerning a 0.50 M solution of chlorous acid, HClO2 (Ka=1.2 x 10-2), you would need to use the quadratic formula to answer the question. Generally, if the acid concentration is .10M or larger, and K is 10-5 or smaller, you won’t need to use the quadratic formula.

HA(aq) + H2O(l) ↔ H3O+(aq) + A-(aq) Problem A 0.0875M solution of an unknown monoprotic acid has a pH of 2.28. Calculate the Ka of the acid. - Write the chemical reaction. HA(aq) + H2O(l) ↔ H3O+(aq) + A-(aq) - Write the equilibrium constant expression. Ka = [H3O+][A-] [HA]

Problem A 0.0875M solution of an unknown monoprotic acid has a pH of 2.28. Calculate the Ka of the acid. - Make a table of concentrations.

Problem A 0.0875M solution of an unknown monoprotic acid has a pH of 2.28. Calculate the Ka of the acid. [HA] [H3O+] [A-] init .0875 chg equil

Problem A 0.0875M solution of an unknown monoprotic acid has a pH of 2.28. Calculate the Ka of the acid. The pH can be used to calculate how much H3O+ and A- form. [HA] [H3O+] [A-] init .0875 chg equil

Problem A 0.0875M solution of an unknown monoprotic acid has a pH of 2.28. Calculate the Ka of the acid. To calculate [H3O+] from pH, enter the pH, change its sign, and take the inverse log. You will get a [H3O+] = 5.2 x10-3. This is the equilibrium concentration of H3O+ (and A-).

Problem A 0.0875M solution of an unknown monoprotic acid has a pH of 2.28. Calculate the Ka of the acid. [HA] [H3O+] [A-] init .0875 chg equil 5.2 x10-3

Problem A 0.0875M solution of an unknown monoprotic acid has a pH of 2.28. Calculate the Ka of the acid. [HA] [H3O+] [A-] init .0875 chg - .0052 + 5.2 x 10-3 equil 5.2 x10-3

Problem A 0.0875M solution of an unknown monoprotic acid has a pH of 2.28. Calculate the Ka of the acid. [HA] [H3O+] [A-] init .0875 chg - .0052 + 5.2 x 10-3 equil .0823 5.2 x10-3

Problem A 0.0875M solution of an unknown monoprotic acid has a pH of 2.28. Calculate the Ka of the acid. [HA] [H3O+] [A-] equil .0823 5.2 x10-3 Substitute into the equilibrium constant expression and solve for Ka.

Problem A 0.0875M solution of an unknown monoprotic acid has a pH of 2.28. Calculate the Ka of the acid. [HA] [H3O+] [A-] equil .0823 5.2 x10-3 Ka = [H3O+] [A-]/ [HA] = (5.2 x10-3) (5.2 x10-3) (0.0823) Ka = 3.3 x 10-4

Problem: What is the pH of 0.15M NH3? NH3 is a weak base, and NH4+ is its conjugate acid. Kb = [OH-][NH4+] [NH3]

Values of Kb for Some Common Weak Bases

NH3(aq) + H2O(l) ↔ OH-(aq) + NH4+(aq) What is the pH of 0.15M NH3? Kb = [OH-][NH4+] = 1.8 x 10-5 [NH3] NH3(aq) + H2O(l) ↔ OH-(aq) + NH4+(aq) NH3 OH- NH4+ initial 0.15 change -x +x equil. .15-x x

Polyprotic Acids Acids that have more than one proton that can be donated are called polyprotic acids. Examples are H2SO4, H3PO4, H2S and H2CO3. The loss of protons can be viewed as a step-wise process.

Polyprotic Acids H3A(aq) + H2O(l) ↔ H3O+(aq) + H2A-(aq) Ka1 = [H3O+][H2A-]/[H3A] H2A-(aq) + H2O(l) ↔ H3O+(aq) + HA2-(aq) Ka2 = [H3O+][HA2-]/[H2A-] HA2-(aq) + H2O(l) ↔ H3O+(aq) + A3-(aq) Ka3 = [H3O+][A3-]/[HA2-]

Polyprotic Acids For all polyprotic acids, Ka1>Ka2>Ka3. 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 pH of a solution of the polyprotic acid. The subsequent steps are usually negligible.

Polyprotic Acids

Problem: Polyprotic Acids Calculate the pH and the concentration of all species present in 0.40M H2CO3. (Ka1 = 4.3 x 10-7, Ka2 =5.6 x 10-11). Since Ka1>>Ka2, the pH can be obtained by considering only the first dissociation.

Problem: Polyprotic Acids Calculate the pH and concentration of all species present in 0.40M H2CO3. (Ka1 = 4.3 x 10-7, Ka2 =5.6 x 10-11). H2CO3(aq) + H2O(l) ↔ H3O+(aq) + HCO31-(aq) Ka1=[H3O+][HCO31-] = 4.3 x 10-7 [H2CO3]

pH 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.

The “neutral” anions are: Cl-, Br-, I-, NO3- and ClO4- pH of Salt Solutions Since the conjugate bases of strong acids have no tendency to accept protons, these ions will have no effect on pH. The “neutral” anions are: Cl-, Br-, I-, NO3- and ClO4- (Although HSO41- isn’t basic, it is acidic, and will donate a proton to form hydronium ion.)

pH of Salt Solutions The cations of strong bases also have no effect on pH 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 Al3+, Cr3+ or Fe3+ produce slightly acidic solutions.

pH of Salt Solutions Are the following aqueous solutions acidic, basic or neutral? NaNO3, KCN, NH4Cl, Ca(HSO4)2, LiCH3CO2

pH of Salt Solutions Calculate the pH of 0.10M NaCH3CO2. NaCH3CO2(aq)  Na+(aq) + CH3CO2-(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. CH3CO2-(aq) + H2O(l) ↔ CH3CO2H(aq) + OH-(aq) The solution should be basic with the pH>7.0

CH3CO2-(aq) + H2O(l) ↔ CH3CO2H(aq) + OH-(aq) pH of Salt Solutions Calculate the pH of 0.10M NaCH3CO2. CH3CO2-(aq) + H2O(l) ↔ CH3CO2H(aq) + OH-(aq) Kb = [CH3CO2H][OH-] [CH3CO2-] The value of Kb for acetate ion can be calculated from the value of Ka for acetic acid.

pH 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

CH3CO2-(aq) + H2O(l) ↔ CH3CO2H(aq) + OH-(aq) pH of Salt Solutions Calculate the pH of 0.10M NaCH3CO2. CH3CO2-(aq) + H2O(l) ↔ CH3CO2H(aq) + OH-(aq) Kb = [CH3CO2H][OH-] = 5.6 x 10-10 [CH3CO2-] [CH3CO2-] [CH3CO2H] [OH-] initial 0.10 change -x +x equil. .10-x x