Acid Base Equilibria maybe do packet 2 before packet 3? Combine concepts from 1 and 3?

Arrhenius Acids & Bases (1 st year chem definition!) An Arrhenius acid is a substance that, when dissolved in water, increases the concentration of H + (needs H + in it) Example: HCl (monoprotic) H 2 SO 4 (diprotic) An Arrhenius base is a substance that, when dissolved in water, increases the concentration of OH – Example: NaOH (monobasic) Ca(OH) 2 (dibasic)

Brønsted-Lowry Acids and Bases Brønsted-Lowry acid is a species that donates H + (proton) Brønsted-Lowry base is a species that accepts H + (proton) Brønsted-Lowry definition of a base does not mention OH – and the reaction does not need to be aqueous.

The H + Ion in Water The H + (aq) ion is simply a proton with no surrounding valence electrons. In water, clusters of hydrated H + (aq) ions form.

The simplest cluster is H 3 O + (aq) We call this a hydronium ion. Larger clusters are also possible (such as H 5 O 2 + and H 9 O 4 + ). Generally we use H + (aq) and H 3 O + (aq) interchangeably.

Proton-Transfer Reactions Consider NH 3 (aq) + H 2 O(l)  NH 4 + (aq) + OH – (aq) H 2 O donates a proton to ammonia. –Therefore, water is acting as an acid. NH 3 accepts a proton from water. –Therefore, ammonia is acting as a base.

Amphoteric substances Can behave as acids and bases. Water is an example of an amphoteric species. Water as an acid (proton donor) H 2 O + NH 3  NH OH - Water as a base (proton acceptor) H 2 O + HNO 2  H NO 2 -

Conjugate Acid-Base Pairs A conjugate acid is the substance formed by adding a proton to the base. A conjugate base is the substance left over after the acid donates a proton. Within a pair the acid has more hydrogen!

Strong Acids & Bases Strong AcidsStrong Bases sulfuric acidH 2 SO 4 Lithium hydroxideLiOH Nitric acidHNO 3 Sodium hydroxideNaOH Perchloric acidHClO 4 Potassium hydroxideKOH Chloric acidHClO 3 Rubidium hydroxideRbOH Hydrochloric acidHClCesium hydroxideCsOH Hydrobromic acidHBrCalcium hydroxideCa(OH) 2 Hydroiodic acidHIStrontium hydroxideSr(OH) 2 Barium hydroxideBa(OH) 2 Less common: O 2- H - CH 3 - All other acids and bases are weak!

Ions Neutral anionsNeutral cations Hydrogen sulfateHSO 4 - LithiumLi + NitrateNO 3 - SodiumNa + PerchlorateClO 4 - PotassiumK+K+ ChlorateClO 3 - RubidiumRb + ChlorideCl - CesiumCs + BromideBr - CalciumCa 2+ IodideI-I- StrontiumSr 2+ BariumBa 2+ Most anions are weak bases Most cations are weak acids Anions of strong acids and cations of strong bases are neutral

Strengths of Acids and Bases Strong acids completely ionize in water. HCl + H 2 O  H 3 O + + Cl - HCl  H + + Cl - Essentially no un-ionized molecules remain in solution so the equation usually does not contain equilibrium arrows. Keq >>1 Their conjugate bases have negligible tendencies to become protonated Cl - + H +  HCl. The conjugate base of a strong acid is a neutral anion.

Strengths of Acids and Bases Strong bases completely dissociate in water NaOH(aq)  Na + (aq) + OH - (aq) Essentially no undissociated compound remains in solution so the equation usually does not contain equilibrium arrows. Keq >>1 The ions have negligible tendencies to attract OH - in solution. Na + (aq) + OH - (aq)  NaOH(aq) The cation of a strong base is a neutral cation.

All other acids are Weak acids. They only partially dissociate in aqueous solution. HC 2 H 3 O 2 (aq) + H 2 O(l)  H 3 O + (aq) + C 2 H 3 O 2 - (aq) They exist in solution as a mixture of molecules and component ions. (usually mostly molecules in equilibrium) Their conjugate bases are weak bases. –Besides non-neutral anions, weak bases tend to be nitrogen containing organic compounds Example: Acetic acid is a weak acid; acetate ion (conjugate base) is a weak base. acid conjugate base base conjugate acid

Non-acid compounds with hydrogen Not all compounds containing hydrogen are acidic. These are extremely weak acids…so much that we don’t consider them acids at all. Their conjugate bases are strong bases! Negligible acidity: OH - H 2 CH 4 Strong bases:O 2- H - CH 3 -

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The stronger an acid is, the weaker its conjugate base will be. In acid-base reactions, the reaction favors the transfer of a proton from the stronger acid to the stronger base to form a weaker acid and weaker base. We need a more specific way to determine acid strength!

Where does that acid/base ranking come from? Keq!!! Since weak acids and bases are in equilibrium…we can write equilibrium constant expressions! When looking at the reaction of a weak acid with water we label the equilibrium constant K a : HF(aq) + H 2 O(l)  H 3 O + (aq) + F - (aq) When looking at the reaction of a weak base with water we label the equilibrium constant K b : NH 3 (aq) + H 2 O(l)  NH 4 + (aq) + OH - (aq)

K a and K b The value of K a or K b indication the extent to which the weak acid or base ionizes/dissociates. –Larger K a or K b means more products! Weak acids with larger K a values are stronger. Weak bases with larger K b values are stronger. (Values in Appendix D)

19 Percent Ionization Percent ionization is another method to assess acid strength. For the reaction: HA(aq)  H + (aq) + A – (aq) The higher the percent ionization, the stronger the acid.

Polyprotic Acids Polyprotic acids have more than one ionizable proton. The protons are removed in successive steps. Consider the weak acid, H 2 SO 3 (sulfurous acid): 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) K a2 = 6.4 x 10 –8 It is always easier to remove the first proton in a polyprotic acid than the second. K a1 > K a2 > K a3, etc

The Autoionization of Water In pure water the following equilibrium is established: H 2 O(l) + H 2 O(l)  H 3 O + (aq) + OH – (aq) acid base acid base This process is called the autoionization of water. We can write an equilibrium constant expression for the autoionization of water: K w = [H 3 O + ] [OH – ] = 1.0*10 K w is called the “ion-product constant” K w = K a *K b for conjugate pair

The Ion Product Constant This applies to pure water as well as to aqueous solutions. (for our purposes…) A solution is neutral if [OH – ] = [H 3 O + ]. If the [H 3 O + ] > [OH – ], the solution is acidic. If the [H 3 O + ] < [OH – ], the solution is basic. In a neutral solution at 25 o C, [H + ] = [OH - ] = 1.0 x M

Give the conjugate base of the following Bronsted-Lowry acids: (a) HIO 3 (b) NH 4 +1 (c) H 2 PO 4 -1 (d) HC 7 H 5 O 2 remove H + (a) IO 3 -1 (b) NH 3 (c) HPO 4 -2 (d) C 7 H 5 O 2 -

24 Designate the Bronsted-Lowry acid and the Bronsted-Lowry base on the left side of each of the following equations, and also designate the conjucate acid and base on the right side: (a) NH 4 +1 (aq) + CN -1 (aq)  HCN(aq) + NH 3 (aq) acid base acid base (b) (CH 3 ) 3 N(aq) + H 2 O  OH -1 (aq) + (CH 3 ) 3 NH +1 (aq) base acid base acid (c) HCHO 2 (aq) + PO 4 -3 (aq)  HPO 4 -2 (aq) + CHO 2 -1 (aq) acid base acid base

(a) The hydrogen oxalate ion (HC 2 O 4 -1 ) is amphoteric. Write a balanced chemical equation showing how it acts as an acid toward water and another equation showing how it acts as a base towards water. (b) What is the conjugate acid and base of HC 2 O 4 -1 ? (a)Acid: HC 2 O 4 -1 (aq) + H 2 O(l)  C 2 O 4 -2 (aq) + H 3 O +1 (aq) acid base conj base conj acid Base: HC 2 O 4 -1 (aq) + H 2 O(l)  H 2 C 2 O 4 (aq) + OH -1 (aq) base acid conj acid conj base (b)

26 Label each of the following as being a strong acid, a weak acid, or a species with negligible acidity. In each case write the formula of its conjugate base, and indicate whether the conjugate base is a strong base, a weak base or a species with negligible basicity: (a) HNO 2 (b) H 2 SO 4 (c) HPO 4 -2 (d) CH 4 (e) CH 3 NH 3 +1 (a)HNO 2, (b)H 2 SO 4, (c)HPO 4 -2, (d)CH 4, (e)CH 3 NH 3 +1, NO 2 -1, HSO 4 -1, PO 4 -3, CH 3 -1, CH 3 NH 2, ACID BASE strong negligible weak negligible strong IMPORTANT weak acid weak base strong acid negligible base weak acid weak base negligible acid strong base weak acid weak base

(a)Which of the following is the stronger acid, HBrO or HBr? (b)Which is the stronger base, F -1 or Cl -1 ? Briefly explain your choices. (a)HBr - It is one of the seven strong acids (b)F -1 HF is a weak acid so F- is a weak base HCl is a strong acid so Cl- is neutral

Predict the products of the following acid-base reactions & determine whether equilibrium lies to the right or left: (a) O -2 (aq) + H 2 O(l)  (b) CH 3 COOH(aq) + HS -1 (aq)  (c) NO 3 -1 (aq) + H 2 O(l)  * You will need a chart like this or Ka/Kb values to determine equilibrium!

(a)O -2 (aq) + H 2 O(l)  OH -1 (aq) + OH -1 (aq) base acid acid base *OH- is the weaker acid so equilibrium lies to the right (b)CH 3 COOH(aq) = HC 2 H 3 O 2 (aq) HC 2 H 3 O 2 (aq) + HS -1 (aq)  C 2 H 3 O 2 -1 (aq) + H 2 S(aq) acid base base acid *H 2 S is the weaker acid so equilibrium lies to the right (c)NO 3 -1 (aq) + H 2 O(l)  HNO 3 (aq) + OH -1 (aq) base acid acid base *H 2 O is the weaker acid so equilibrium lies (far!) to the left

The “p” Function p is short for “– log 10 ” pH = -log 10 [H + ] = -log[H + ] pOH = -log 10 [OH - ] = -log [OH - ] Note that this is a logarithmic scale. Thus a change in [H + ] by a factor of 10 causes the pH to change by 1 unit. Most pH values fall between 0 and 14.

In neutral solutions at 25 o C, [H + ] = 1.0 x pH = -log[1.0 x ] = 7.00 Lower pH = more acidic Higher pH = more basic [H+]pH Acidic> 1.0 x 10 –7 < 7.00 Neutral1.0 x Basic< 1.0 x 10 –7 > 7.00

Another one: pK We can use a similar system to describe the equilibrium constant pK = - log[K] The value of K w at 25 o C is 1.0 x 10 –14 pK w = - log (1.0 x 10 –14 )= 14.0 K w =[H + ][OH - ] pK w = pH + pOH = 14.0

The pH Loop pH pOH [H + ] [OH - ] [H + ]=10 -pH pH = -log[H + ] [OH - ]=10 -pOH pOH = -log[OH - ] pH + pOH=14.0 [H + ] [OH - ]=10 -14

Measuring pH The most accurate method to measure pH is to use a pH meter. Acid Base Indicators: certain dyes change color as pH changes. Indicators are less precise than pH meters. –Many indicators do not have a sharp color change as a function of pH. Most acid-base indicators can exist as either an acid or a base. –These two forms have different colors. –The relative concentration of the two different forms is sensitive to the pH of the solution. Thus, if we know the pH at which the indicator turns color, we can use this color change to determine whether a solution has a higher or lower pH than this value.

Example 1: Calculate [H +1 ] for each of the following solutions and indicate whether the solution is acidic, basic or neutral. (a) [OH -1 ] = M (b) a solution where [OH -1 ] is 100 times greater than [H +1 ]

Example 2: (a) If NaOH is added to water, how does the [H +1 ] change? How does pH change? (b) If [H +1 ] = M, what is the pH of the solution? Is the solution acidic or basic? (c) If pH = 6.3, what are the molar concentrations of H +1 (aq) and OH -1 (aq) in the solution? (a)K w = [H + ] [OH - ]. the [OH - ] will increase and the [H + ] will decrease. [H + ] decreases, pH increases. (b)pH = - log [H + ] = - log (0.005) = 2.3 acidic (c)pH = 6.3 [H + ] = 10 -pH = = 5 x M pOH = 14 – pH = 14 – 6.3 = 7.7 [OH - ] = 10 -pOH = = 2 x M

Strong Acid Calculations In solution the strong acid is usually the only source of H + The pH of a solution of a monoprotic acid may usually be calculated directly from the initial molarity of the acid. Caution: If the molarity of the acid is less than 10 –6 M then the autoionization of water needs to be taken into account.

Example 3: Calculate the pH of each of the following strong acid solutions: (a) 1.8  M HBr (b) 1.02 g HNO 3 in 250 mL of solution (c) 2.00 mL of M HClO 4 diluted to 50.0 mL (d) a solution formed by mixing 10.0 mL of M HBr with 20.0 mL of 2.5  M HCl (a)1.8 x M HBr = 1.8 x M H + pH = -log(1.8 x ) = 3.74 (b) M HNO 3 = M H + pH = -log (.0647) = 1.19 (c)M 1 V 1 = M 2 V 2 (0.500 M)( L) = (x M)( L) x =.0200 M HCl (d) pH = -log(.0050) = 2.30 [H + ] =.0200 M pH = -log(.0200) = 1.70

Strong Base Calculations Strong bases are strong electrolytes and dissociate completely in solution. For example: NaOH(aq)  Na + (aq) + OH – (aq) The pOH (and thus the pH) of a strong base may be calculated using the initial molarity of the base.

Example 4: Calculate [OH -1 ] and pH for (a) 3.5  M Sr(OH) 2 (b) 1.50 g LiOH in 250 mL of solution (c) 1.00 mL of M NaOH diluted to 2.00 L (d) a solution formed by adding 5.00 mL of M KOH to 15.0 mL of 3.5  M Ca(OH) 2 (a)[OH - ] = 2[Sr(OH) 2 ] = 2( M) = M OH - pOH = -log(.00070) = 3.15 pH = 14 – 3.15 = (b) pOH = -log (.251) = pH = = (c) M 1 V 1 = M 2 V 2 (0.095 M)( L) = (x M)(2.00 L) x = M NaOH = [OH - ] pOH = -log( ) = 4.32 pH = 14 – 4.32 = 9.68 (d) pOH = -log(.0079) = 2.1 pH = 14 – 2.1 = 11.9

Weak Acid Calculations Weak acids are only partially ionized in aqueous solution. Therefore, weak acids are in equilibrium: HA(aq) + H 2 O(l)  H 3 O + (aq) + A – (aq) OR HA(aq)  H + (aq) + A – (aq) We can write Ka for this dissociation:

Calculating Ka from pH for weak acids In order to find the value of K a, we need to know all of the equilibrium concentrations. (ICE Chart) The pH gives the equilibrium concentration of H +. We then substitute these equilibrium concentrations into the equilibrium constant expression and solve for K a.

Example 4: A 0.20 M solution of niacin (a monoprotic weak acid) has a pH of What is the K a for niacin? HA  H+H+ A- I C E =

Using K a to Calculate pH for weak acids Write the balanced chemical equation clearly showing the equilibrium. Write the equilibrium expression. Look up the value for K a (in a table). Write down the initial and equilibrium concentrations for everything except pure water. (ICE table) We usually assume that the equilibrium concentration of H + is x. Substitute into the equilibrium constant expression and solve. Remember to convert x to pH if necessary.

Polyprotic Acids Polyprotic acids have more than one ionizable proton. 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) K a2 = 6.4 x 10 –8 The majority of the H + (aq) at equilibrium usually comes from the first ionization If the successive K a values differ by a factor of 10 3, we can usually get a good approximation of the pH of a solution of a polyprotic acid by considering the first ionization only. If not, then we have to account for the successive ionizations

Example 5: The acid dissociation constant for benzoic acid, HC 7 H 5 O 2 is 6.5 x Calculate the equilibrium concentrations of H 3 O +, C 7 H 5 O 2 -, and HC 7 H 5 O 2. The initial concentration of HC 7 H 5 O 2 is 0.050M. HC 7 H 5 O 2 (aq)  H + (aq) + C 7 H 5 O 2 - (aq) I C -xxx E(.050-x)xx x 2 + (6.5*10 -5 )x – (3.25*10 -6 ) = 0 x=.0018M = [H + ]=[C 7 H 5 O 2 - ] [HC 7 H 5 O 2 ] = =.048 M

What if I don’t have a quadratic equation program? Algebra Shortcut: Assume x is much smaller (less than 5% of.050) To simplify:.050-x =.050 Now you don’t need the quadratic equation! x=.0018M = [H + ]=[C 7 H 5 O 2 - ] [HC 7 H 5 O 2 ] = =.048 M If you made this assumption you need to check and make sure it’s valid – if this answer isn’t less than 5% use the quadratic equation!

Example 6: Calculate the pH of the following solution (K a and K b values are in Appendix D) M hydrazoic acid, HN 3 (a) HN 3 (aq)  H + (aq) + N 3 - (aq) I0.175M 0 0 C-x x x E0.175-x x x x 2 + (1.9*10 -5 )x – (3.325*10 -6 ) = 0 x=.0018M H + pH = -log(0.0018) = 2.74

Example 7: Calculate the percent ionization of M hydrazoic acid, HN 3, solution. HN 3 (aq)  H + (aq) + N 3 - (aq) I0.400M 0 0 C-x x x E0.400-x x x x 2 + (1.9*10 -5 )x – (7.6*10 -6 ) = 0 x=.0028M H +

Example 8: Citric acid, which is present in citrus fruits, is a triprotic acid. Calculate the pH of a M solution of citric acid. H 3 C 6 H 5 O 7 (aq)  H + (aq) + H 2 C 6 H 5 O 7 -1 (aq) K a = 7.4*10 -4 H 2 C 6 H 5 O 7 -1 (aq)  H + (aq) + HC 6 H 5 O 7 -2 (aq) K a = 1.7*10 -5 HC 6 H 5 O 7 -2 (aq)  H + (aq) + C 6 H 5 O 7 -3 (aq) K a = 4.0*10 -7 To calculate the pH of a.050M solution, assume initially that only the first ionization is important H 3 C 6 H 5 O 7 (aq) H + (aq) + H 2 C 6 H 5 O 7 -1 (aq) I C -x x x E x x x x = =[H + ] = [H 2 C 6 H 5 O 7 -1 ]

Does the second ionization have any effect? H 2 C 6 H 5 O 7 -1 (aq)  H + (aq) + HC 6 H 5 O 7 -2 (aq) K a = 1.7*10 -5 I C-yy y E y y y y = This value is small compared to (think SDs) Total [H + ] = = which indicates the 2 nd (and any subsequent) ionizations can be ignored pH = -log(0.0057) = 2.24

Weak Base Calculations Weak bases remove protons from substances. There is an equilibrium between the base and the resulting ions: Example: NH 3 (aq) + H 2 O(l)  NH 4 + (aq) + OH – (aq). The base-dissociation constant, K b, is The larger K b, the stronger the base.

16.8 Relationship Between K a and K b Generally only either Ka or Kb for a conjugate pair is reported in tables. If you know one you can find the other! Consider the following equilibria: NH 4 + (aq)  NH 3 (aq) + H + (aq) NH 3 (aq) + H 2 O(l)  NH 4 + (aq) + OH – (aq) We can write equilibrium expressions for these reactions:

If we add these equations together: NH 4 + (aq)  NH 3 (aq) + H + (aq)Ka NH 3 (aq) + H 2 O(l)  NH 4 + (aq) + OH – (aq)Kb The net reaction is the autoionization of water. H 2 O(l)  H + (aq) + OH – (aq)Kw When we add equations we multiply K. K w = K a x K b Alternatively, we can express this as: pK a + pK b = pK w = (at 25 o C)

Example 9: Calculate the molar concentration of OH -1 ions in a M solution of hydrazine, H 2 NNH 2, K b = 1.3  What is the pH of this solution?

Example 10:Although the acid dissociation constant for phenol, C 6 H 5 OH, is listed in Appendix D, the base dissociation constant for the phenolate ion, C 6 H 5 O -1 is not. (a) Explain why it is not necessary to list both. (b) Calculate the K b for the phenolate ion. (c) Is the phenolate ion a weaker or stronger base than ammonia, NH 3 ? Appendix D

Acid-Base Properties of Salt Solutions All soluble salts are strong electrolytes. - In solution, they exist nearly entirely of ions. - Acid-base properties of salts are due to the reactions of their ions in solution. Many ions can react with water to form OH – or H +. This process is called hydrolysis.

Remember this? Neutral anionsNeutral cations Hydrogen sulfateHSO 4 - LithiumLi + NitrateNO 3 - SodiumNa + PerchlorateClO 4 - PotassiumK+K+ ChlorateClO 3 - RubidiumRb + ChlorideCl - CesiumCs + BromideBr - CalciumCa 2+ IodideI-I- StrontiumSr 2+ BariumBa 2+ Most anions are weak bases Most cations are weak acids Anions of strong acids and cations of strong bases are neutral

Summary Anions – Of strong acids are neutral. Example: Cl - of HCl – Cl - + H 2 O  X – Of weak acids are basic. Example: F - of HF – F - + H 2 O  HF + OH - – With ionizable protons are amphoteric. Example: HSO 4 –

Summary Cations – Of strong bases are neutral. Example: Na + of NaOH – Na + + H 2 O  X –All other cations are weak acids Example: Fe 3+

Cation hydrolysis reaction: Smaller and more highly charged ions = stronger (weak) acids

(All are 1.00 M solutions)

The pH of a solution may be qualitatively predicted: Salts from a weak acid and weak base can be either acidic or basic. Compare K a of the cation and K b of the anion - If the K a is larger the solution will be acidic - If the K b is larger the solution will be basic For example, consider NH 4 CN. The K b of CN -1 is larger than the K a of NH 4 +1 so the solution will be basic. CationAnionSolution is:Example Neutral (from strong base) Neutral (from strong acid) NeutralNaCl Neutral (from strong base) Basic (from weak acid) BasicNaF Acidic (from a weak base) Neutral (from a strong acid) AcidicFeCl 3 NH 4 Cl Acidic (from a weak base) Basic (from a weak acid) Depends!NH 4 CN

Example 1: Predict whether aqueous solutions of the following compounds are acidic, basic or neutral. (a)NH 4 Br (b) FeCl 3 (c) Na 2 CO 3 (d) KClO 4 (e) NaHC 2 O 4 K a for acid HC 2 O 4 -1 = 6.4  K b for base HC 2 O 4 -1 = 1.7  acidic basic neutralacidic

Example 2: Using data from Appendix D, calculate [OH -1 ] and pH for the following solution 0.10 M NaCN K a for HCN (given in Appendix D) = 4.9x10 -10

16.10 Acid-Base Behavior and Chemical Structure Acidity is directly related to the strength of attraction for a pair of electrons to a central atom. 4 situations to consider: 1.Ions Ionic Charge and Size When comparing ions of similar structure: More positive ions are stronger acids. tie breaker: Smaller ion is stronger acid Acid strength: Na + < Ca 2+ < Cu 2+ < Al 3+ PO 4 3- < HPO 4 2- < H 2 PO 4 - < H 3 PO 4

69 Example 3: Predict which member of each pair produces the more acidic aqueous solution: (a) K +1 or Cu +2 (b) Fe +2 or Fe +3 (c) Al +3 or Ga +3 (a)Cu +2 has higher charge (and K +1 is neutral) (b) Fe +3 has higher charge (c) Al +3 has a smaller size

2.Binary Acids: Bond Polarity (Electronegativity) & Strength The H–X bond strength is important in determining relative acid strength in any group in the periodic table. – The weaker the bond the easier it will break –The H–X bond strength tends to decrease down a group - acid strength increases down a group H–X bond polarity is important in determining relative acid strength in any period of the periodic table. –The H-X bond polarity tends to increase across a period - acid strength increases (from left to right) across a period

71 more polar bond – stronger acid larger atom weaker bond stronger acid

3. Oxyacids (Acids with oxygen) with different central atoms Generally, the larger the electronegativity of the central atom the stronger the acid. –The stronger the pull on electrons the less tightly the H is held Acid Strength: H 3 BO 3 < H 2 CO 3 < HNO 3

The higher EN of central atom means more electron density shift away from H - H is easier to remove – stronger acid

4. Oxyacids with the same central atom Generally, the more oxygens attached to the central atom the stronger the acid. –The more atoms pulling on electrons the less tightly the H is held Acid Strength: HClO < HClO 2 < HClO 3 < HClO 4

More O atoms means more electron density shift away from H and H is easier to remove – stronger acid

76 Example 4:Explain the following observations: (a) HNO 3 is a stronger acid than HNO 2 (b) H 2 S is a stronger acid than H 2 O (c) H 2 SO 4 is a stronger acid than HSO 4 -1 (d) H 2 SO 4 is a stronger acid than H 2 SeO 4 (e) CCl 3 COOH is a stronger acid than CH 3 COOH (a)more oxygen atoms - electron density shifts away from H bond - easier to remove H (b) bond strength decreases down a group - H easier to remove from S (c)More positive ion is stronger (d)S has higher electronegativity than Se – pulls electrons from H – easier to remove H (e)the more electronegative 3 Cl atoms (as opposed to the 3 H atoms) pull electron density more weakens the O-H bond and makes H easier to remove

16.11 Lewis Acids and Bases A Brønsted-Lowry acid is a proton donor. Lewis proposed a new definition of acids and bases that emphasizes the shared electron pair. A Lewis acid is an electron pair acceptor. A Lewis base is an electron pair donor. Note: Lewis acids and bases do not need to contain protons. Therefore, the Lewis definition is the most general definition of acids and bases.

78 Typical Lewis acid-base reaction

79 What types of compounds can act as Lewis acids? Lewis acids must have a vacant orbital (into which the electron pair can be donated). Lewis acids sometimes have an incomplete octet (e.g., BF 3 ). Transition-metal ions can be Lewis acids (empty d orbitals) Compounds with multiple bonds can act as Lewis acids.

Example 5:Identify the Lewis acid and Lewis base in each of the following reactions: (a) Fe(ClO 4 ) H 2 O  Fe(H 2 O) ClO 4 -1 (b) CN -1 + H 2 O  HCN + OH -1 (c) (CH 3 ) 3 N + BF 3  (CH 3 )NBF 3 (d) HIO + NH 2 -1  NH 3 + IO -1 Acid Base H donorH acceptor e acceptor e donator empty orbitals/mult. bondshas lone pairs incomplete octet/cationoften contains N (a)Fe(ClO 4 ) 3 or Fe +3 H 2 O (b)H 2 OCN -1 (c)BF 3 (CH 3 ) 3 N (d)HIONH 2 -1