1. AP CHEMISTRY

2.  Acids ◦ Sour, can corrode metals, cause certain dyes to change colors  Bases ◦ Bitter taste, feel slippery, usually used in cleaning products  Arrhenius Acid ◦ substances when dissolved in water, increase the concentration of H + ions  Arrhenius Base ◦ Substances, when dissolved in water, increase the concentration of OH - ions

3.  Arrhenius definition is restricted to aqueous solutions  Brønsted-Lowery Acid ◦ Substance (molecule or ion) that can donate a proton to another substance  Brønsted-Lowery Base ◦ Substance that can accept a proton

4.  In any acid-base equilibrium, both the forward and reverse reactions involve proton transfers.  An acid and base in an equation that differ only in the presence or absence of a proton are called conjugate acid-base pairs. ◦ Every acid has a conjugate base and every base has a conjugate acid

5.  What is the conjugate base of each of these acids? ◦ HClO 4, PH 4 +, HNO 2 ◦ ClO 4 -, PH 3, NO 2 -  What is the conjugate acid of each of these bases? ◦ CN -, H 2 O, HCO 3 - ◦ HCN, H 3 O +, H 2 CO 3  Identify the conjugate base pairs. NH 3(aq) + H 2 O (l) ⇆ NH 4 + (aq) + OH - (aq)

6.  Some acids/bases are better proton donors/acceptors than others.  The stronger the acid, the weaker its conjugate base (and the stronger the base, the weaker the conjugate base.  Strengths of acids and bases ◦ Strong Acids completely transfer their protons to water. (Their conjugate base have negligible tendency to be protonated.) ◦ Weak Acids only partly dissociate and therefore exist as a mixture of acid molecules and their ions (Conjugate base is a weak base as well.) ◦ Negligible Acidity contain hydrogen but do not demonstrate any acidic behavior in water. (Conjugate bases are strong bases, reacting completely in water)

7.  Relative Strengths of Conjugate Acid-Base pairs

8.  Think of proton-transfer reactions as being governed by the relative abilities of two bases to abstract protons HC 2 H 3 O 2(aq) + H 2 O (l) ⇆ H 3 O + (aq) + C 2 H 3 O 2 - (aq)  C 2 H 3 O 2 - is a stronger base than H 2 O and therefore abstracts the proton from H 3 O +.  In every acid-base reaction the position of the equilibrium favors transfer of the proton to the stronger base.

9.  One of the most important chemical properties of water is its ability to act as either a Brønsted-Lowery acid or base.  This is called the autoionization of water ◦ Reaction is rapid and no individual molecules remains ionized for long (only 2 in every 10 9 molecules is ionized at any moment)

10.  Because water ionizes, we can write its equilibrium constant.  K c = [H 3 O + ] [OH - ] ◦ K w = 1.0 x 10 -14 (at 25 ⁰ C) ◦ Ion-product constant ◦ Commit this to memory!!  Important because this equilibrium applies to any dilute aqueous solution and can be used to calculate the [H + ] or [OH - ] ◦ The product of [H + ] and [OH - ] = 1.0 x 10 -14 ◦ [H + ] = [OH - ] is said to be neutral. ◦ [H + ] exceeds [OH - ] is acidic ◦ [OH - ] exceeds [H + ] is basic  [H 3 O + ] [OH - ] = 1.0 x 10 -14

12.  The molar [H + ] is very small in aqueous solutions so we express [H + ] in terms of pH, which is the negative logarithm in base 10 of [H + ].  pH = -log[H + ]

14.  What is the pH for a solution that has [H + ] = 5.6 x 10 -6 M?  pH = -log(5.6 x 10 -6 )  = 5.25  A sample of apple juice has a pH of 3.76. Calculate the [H + ].  pH = -log[H + ] = 3.76  [H+] = 10 -3.76 = 1.7 x 10 -4 M

15.  pOH = -log[OH - ]  pH + pOH = 14 (at 25˚C)

16.  7 Common Strong Acids ◦ 6 Monoprotic: HCl, HBr, HI, HNO 3, HClO 3, and HClO 4 ◦ 1 Diprotic: H 2 SO 4  Strong acids completely dissociate into their ions ◦ HNO 3(aq) + H 2 O(l) H 3 O + (aq) + NO 3 - (aq)  [H + ] = the original concentration of acid. ◦ If 0.20 M HNO 3 was used, [H + ] = 0.20 M ◦ pH = -log(0.20) = 0.7 ◦ (diprotic acid is a little more complex)

17.  Common Soluble Strong Bases ◦ Alkali metal hydroxides and heavier alkaline earth metal hydroxides  LiOH, NaOH, KOH, Ca(OH) 2, Sr(OH) 2, and Ba(OH) 2  Strong Bases also completely dissociate so calculating pH is also straightforward. ◦ What is pH of 0.028 M solution of NaOH?  pOH = -log(0.028) = 1.55  pH = 14.00 – 1.55 = 12.45 OR  [H + ] = 1.0 x 10 -14 = 2.35 x 10 -13 0.028  pH = -log(3.57 x 10 -13 ) = 12.45

18.  Weak acids only partially ionize in aqueous solutions  We can use the equilibrium constant for the ionization reaction to express the extent to which an acid ionizes.  Ionization reaction can be written in two ways: ◦ HA (aq) + H 2 O (l) ⇌ H 3 0 + (aq) + A - (aq) ◦ HA (aq) ⇌ H + (aq) + A - (aq)

19.  HA (aq) + H 2 O (l) ⇌ H 3 0 + (aq) + A - (aq)  HA (aq) ⇌ H + (aq) + A - (aq)  K a = [H 3 0 + ][A - ] or [H + ][A - ] [HA]  K a is acid-dissociation constant  The larger the value of K a, the stronger the acid

20.  A 0.10 M solution of formic acid (HCHO 2 ) has a pH of 2.38 at 25 ⁰ C. Calculate the K a for formic acid at this temperature. ◦ HCHO 2(aq) ⇌ H + (aq) + CHO 2 - (aq) ◦ K a = [H + ][CHO 2 - ] [HCHO 2 ] ◦ pH = -log[H + ] = 2.38 ◦ [H + ] = 10 -2.38 = 4.2 x 10 -3 M ◦ 0.10 -4.2 x 10 -3 M ≈ 0.10 ◦ K a = [H + ][CHO 2 - ] = (4.2 x 10 -3 M)(4.2 x 10 -3 M) = 1.8 x 10 -4 [HCHO 2 ] 0.10 HCHO 2 H+H+ CHO 2 - I0.10 M00 C-4.2 x 10 -3 M+4.2 x 10 -3 M E 0.10 -4.2 x 10 -3 M 4.2 x 10 -3 M

21.  Knowing the value of K a and the initial concentration of the weak acid, we can calculate the [H + ] in a solution of a weak acid.  Calculate the pH of a 0.30 M solution of acetic acid. 1.Write the ionization for acetic acid: HC 2 H 3 O 2 ⇆ H + (aq) + C 2 H 3 O 2 - (aq) 2.Write the acid-dissociation constant expression and calculate the value: K a = [H + ][C 2 H 3 O 2 - ] = 1.8 x 10 -5 (from table 16.2 page 628) [HC 2 H 3 O 2 ] 3.Use an ICE box to express equilibrium concentrations

22. 4.Substitute equilibrium concentrations into acid dissociation constant expression (0.30 – x ≈ 0.30) K a = [H + ][C 2 H 3 O 2 - ] = (x)(x) = 1.8 x 10 -5 [HC 2 H 3 O 2 ] 0.30 5.Solve for x x 2 = (0.30)(1.8 x 10 -5 ) = 5.4 x 10 -6 x = 2.3 x 10 -3 6.Solve for [H + ] [H + ] = x = 2.3 x 10 -3 M 7.Solve for pH pH = -log(2.3 x 10 -3 ) = 2.64 HC 2 H 3 O 2 H+H+ C2H3O2-C2H3O2- I0.30 M00 C-x M+x M E(0.30 –x) Mx Mx Mx Mx M

24.  In polyprotic acids, H atoms ionize in successive steps: ◦ H 2 SO 4(aq) ⇌ H + + HSO 3 - (aq) Ka = 1.71 x 10 -2 ◦ HSO 3 - (aq) ⇌ H + (aq) + SO 3 2- (aq) Ka = 6.4 x 10 -8  It is always easier to remove the first proton from a polyprotic acid than the second.  As long as successive K a values differ by a factor of 10 3 or more, it is possible to obtain a satisfactory estimate of the pH of polyprotic acid solutions by considering K a1

25.  The most commonly encountered weak base is NH 3 NH 3(aq) + H 2 0 ⇌ NH 4 + (aq) + OH - (aq)  Base-dissociation constant  The constant K b always refers to the equilibrium in which a base reacts with H 2 O to form the corresponding conjugate acid and OH -  Table 16.4 on page 636 provides formulas and K b values for several weak bases in water.

26.  Weak Bases fall into 2 general categories 1.Neutral substances that have an atom with a nonbonding pair of electrons that can serve as a proton acceptor. 2.Anions of weak acids ClO - (aq) + H 2 O (l) ⇌ HClO (aq) + OH - K b = 3.33 x 10 -7

27.  The product of the acid-dissociation constant for and acid and the base-dissociation constant for its conjugate base is the ion- product for water ◦ (K a )(K b ) = K w AcidKaKa BaseKbKb HNO 3 (strong acid)NO 3 - Negligible HF6.8 x 10 -4 F-F- 1.5 x 10 -11 HC 2 H 3 O 2 1.8 x 10 -5 C2H3O2-C2H3O2- 5.6 x 10 -10 H 2 CO 3 4.3 x 10 -7 HCO 3 - 2.3 x 10 -8 NH 4 + 5.6 x 10 -10 NH 3 1.8 x 10 -5 HCO 3 - 5.6 x 10-11 CO 3 2- 1.8 x 10 -4 OH - NegligibleO 2- (Strong base)

28.  Sometimes acid- and base- dissociation constants are expressed as pK a or pK b ◦ pK a = -log K a ◦ pK b = -log K b ◦ pK a + pK b = pK w = 14.00 at 25˚C

29.  Salts completely dissociate in water  Many ions are able to react with water to generate H + or OH - ions. ◦ This type of reaction is called hydrolysis ◦ The pH of an aqueous salt solution can be predicted by considering the ions of which the salt is composed.

30.  In general, an anion, X -, in solution can be considered the conjugate base of an acid. ◦ Whether or not an anion will react with water to produce OH - depends on the strength of the acid the anion would create. ◦ To identify the acid, add a proton to the anion ◦ X - + a proton = HX ◦ If the acid is a strong acid, the anion will have a negligible tendency to abstract a proton from water. ◦ If the acid is a weak acid, X - will react to a small extent. He pH would be higher (more basic) ◦ If the ion has an ionizable proton like HSO 3 -, it is amphoteric. The behavior is determined by the magnitude of K a and K b

31.  Most metal ions can react with water to decrease the pH of an aqueous solution ◦ Ions of alkali metals and of the heavier alkaline earth metals do not react with water and therefore do not affect pH. (same cations that make strong bases) ◦ Generally, the higher the charge on the cation, the lower the pH of the solution.  The positive charge attracts the unpaired electrons on oxygen. The larger the charge, the higher the attraction. The O-H bonds become weaker and the H+ can be transferred from the hydrated water molecules to the solvent water molecules.

32. 1. An anion that is conjugate base of a strong acid will not affect the pH of solution 2. An anion that is conjugate base of a weak acid will cause an increase in pH 3. A cation that is the conjugate acid of a weak base will cause a decrease in pH 4. With the exception of group 1A and heavy 2B, metal ions will decrease pH 5. When a solution contains both the conjugate base of a weak acid and the conjugate acid of a weak base, the ion with the largest ionization constant will have the greatest affect on pH

33.  Predict whether the salt Na 2 HPO 4 will form an acidic or basic solution on dissolving in water. ◦ Na 2 HPO 4 ⇆ 2Na + + HPO 4 2- ◦ HPO 4 2- can act like an acid or base  1. HPO 4 2- ⇆ H + + PO 4 3- (acid)  2. HPO 4 2- + H 2 O ⇆ H 2 PO 4 - + OH - (base)  The reaction with the larger ionization constant will determine whether it is acidic or basic  K a = 4.2 x 10 -13 (Table 16.3)  (K a )(K b ) = K w  (6.2 x 10 -8 )(K b2 ) = 1.0 x 10 -14  K b = 1.6 x 10 -7  Since K b > K a, solution is basic

34.  Factors that Affect Acid Strength 1.A molecule containing H will transfer a proton only if the H-X bond is polarized H-X  In ionic hydrides (such as NaH) H has a negative charge and acts as a proton acceptor 2.Strong bonds (Table 8.4) are more difficult to dissociate than weak bonds (H-F has very high bond strength so it is a weak acid) 3.The greater the stability of the acid’s conjugate base, the stronger the acid ** The strength of an acid is often a combination of all 3 factors.

35.  For binary acids in the same group, the strength of the H-X bond is the most important factor in determining acid strength  In the same period, bond polarity is the major factor

36.  Acids in which OH groups and possibly additional oxygen atoms are bound to a central atom are called oxyacids.  Consider -Y-O-H ◦ When Y is a metal, they are sources of OH - ions and behave as bases ◦ When Y is a nonmetal, the OH bond is more polar so it will donate a H + and behave as an acid

37.  For oxyacids that have the same number of OH groups and the same number of O atoms, acid strength increases with increasing electronegativity of the central atom. ◦ Example: HNO 3 is a stronger acid than H 2 CO 3  For oxyacids that have the same central atom Y, acid strength increases as the number of oxygen atoms attached to Y increases. ◦ Example: the strength of the oxyacids of chlorine increases from HClO to HClO 2 to HClO 3 to HClO 4

38.  Another group of acids are those with a carboxyl group, often written COOH ◦ Example: acetic acid HC 2 H 3 O 2 can also be written CH 3 COOH  Other examples:  Formic AcidBenzoic Acid

39.  The acid strength of carboxylic acids increase as the number of electronegative atoms in the acid increase. ◦ Example: trifluoroacetic acid (CF 3 COOH) is stronger than acetic acid (CH 3 COOH)

40.  Many acids have more than one ionizable H atom. ◦ H 2 SO 3 (aq) ⇋ H + (aq) + HSO 3 - (aq) K a1 = 1.7 x 10 -2 ◦ HSO 3 -(aq) ⇋ H + (aq) + SO 3 2- (aq)Ka 2 = 6.4 x 10 -8  It is always easier to remove the first proton from a polyprotic acid then to remove the second.  If Ka values differ by a factor of 10 3 or more, it is possible to obtain an accurate pH by treating the acids as if they were monoprotic and considering only K a1

41.  For a substance to be a proton acceptor (base), it must have an unshared pair of electrons for binding a proton.  G.N. Lewis noticed this and proposed definitions for acid and base ◦ A Lewis acid is an electron-pair acceptor ◦ A Lewis base is an electron-pair donor ◦ **Not using this definition in AP Chem.**