Unit 10- Acids & Bases

14.1- The Nature of Acids & Bases 1.Arrhenius Concept  Acids produce hydrogen ions (protons)  H +  Ex. HCl  H + + Cl -  Bases produce hydroxide ions  OH -  Ex. NaOH  Na + + OH -

2.Brønsted-Lowry Model  An acid is a proton (H + ) donor  A base is a proton (H + ) acceptor

3. Hydronium Ion H3O+H3O+  This is how H + exists in water

4. Conjugate Acid-Base Pair  Consists of 2 substances related to each other by the donating & accepting of a single proton (H + ) HA(aq) + H 2 O(l)  H 3 O + (aq) + A - (aq) AcidBase Conj. A Conj. B

5. Acid Dissociation Constant (K a )  Represented by the equilibrium expression (K eq ) for an acid HA(aq) + H 2 O(l)  H 3 O + (aq) + A - (aq)  Why is H 2 O not included?  Because it is a liquid!

Ex  Write the simple dissociation (ionization) reaction (omitting water) for each of the following acids: a)HCl(aq)  b)HC 2 H 3 O 2 (aq)  c)NH 4 + (aq)  d)C 6 H 5 NH 3 + (aq)  e)[Al(H 2 O) 6 ] +3 (aq) 

14.2- Acid Strength 1.A strong acid has a large K a  Equilibrium lies far to the right & nearly all acid molecules dissociate & donate an H +  Strong acids yield weaker conjugate bases  Strong Acids  HCl, HBr, HI, H 2 SO 4, HNO 3, HClO 4

2.A weak acid has a small K a  Equilibrium lies far to the left & only a small amount of acid dissociates & donates an H +  Weak acids yield stronger conjugate bases  The weaker the acid, the stronger its conjugate base  Examples  HF, HC 2 H 3 O 2 (acetic acid), H 3 PO 4, H 2 CO 3

3.Monoprotic Acid  An acid having one acidic proton (H + )  Ex. HI, HBr, HCl, HF (All Halogens) 4.Polyprotic Acids a)Diprotic Acid  An acid with 2 acidic protons (H + )  Ex. H 2 SO 4, H 2 CO 3 b)Triprotic Acid  An acid with 3 acidic protons (H + )  Ex. H 3 PO 4

5.Oxyacids  The acidic proton is on an oxygen atom  Ex. H 3 PO 4, H 2 SO 4, HNO 2, etc. 6.Organic Acids  Acids with a carbon backbone & a carboxyl group (-COOH):

Ex  Using the information below, arrange the following species according to their strengths as bases: H 2 O, F -, Cl -, NO 2 -, CN -  K a (HF) = 7.2 x  K a (HNO 2 ) = 4.0 x  K a (HCN) = 6.2 x  K a (HCl) = very large

7.H 2 O  Is amphoteric- can act as an acid or a base  Autoionization of H 2 O: H 2 O(l) + H 2 O(l)  H 3 O + (aq) + OH - (aq) K eq = [H 3 O + ][OH - ] K w = [H + ][OH - ]  Where K w is the Dissociation Constant for H 2 O

 At 25°C, [H + ] = [OH - ] = 1.0 x M  So:K w = (1.0 x )(1.0 x ) K w = 1.0 x  The importance of this:  In aqueous solutions at 25°C, no matter what it contains, the product of [H + ] and [OH - ] will always be 1.0 x

Ex  Calculate [H + ] or [OH - ] as required for each of the following solutions at 25°C, and state whether the solution is neutral, acidic, or basic. a)1.0 x M OH - b)1.0 x M OH - c)10.0 M H +

14.3- The pH Scale 1.A scale to measure acidity from 0  14  0  acidic7  neutral14  basic 2.pH Equation: pH = -log[H + ]  Example:  If [H + ] = 1.0 x then,  pH = -log(1.0 x )  pH = 5

3.pOH Equation: pOH = -log[OH - ]  Measures basicity  0  basic7  neutral14  acidic 4.Relation to Equilibrium Constant: pK = -log(K)

Ex  Calculate the pH and pOH for each of the following solutions at 25°C a)1.0 x M OH - a)1.0M H +

Ex  The pH of a sample of human blood was measured to be 7.41 at 25°C. Calculate pOH, [H + ], and [OH - ] for the sample.

14.4- pH of Strong Acids 1.Strong acids dissociate completely so the concentration of H + is the same as the concentration of the acid  Example  If [HCl] = 0.1M  Then [H + ] = 0.1M  Because HCl  H + + Cl - 0.1M 0.1M

Ex a)Calculate the pH of 0.10M HNO 3 b)Calculate the pH of 1.0 x M HCl

14.5- pH of Weak Acids  They do not ionize completely  1.Write dissociation reaction 2.Write equilibrium expression 3.List initial concentrations 4.Set up ICE Box in terms of M (set up row E in terms of x) 5.Sub equilibrium concentrations into the equilibrium expression 6.Eliminate x-factor (because K a is small) and solve for x 7.Calculate [H + ] and pH

Example  Calculate the pH of a 1M solution of HF (K a = 7.2 x )

Ex  Calculate the pH of a 0.100M aqueous solution of hypochlorous acid (HOCl, K a = 3.5 x )

 If 2 weak acids are present, the stronger of the two (the one with the larger K a ) will dominate & control the pH  The addition of H + ions by the weaker acid is insignificant in regards to effecting the pH

Ex  Calculate the pH of a solution that contains 1.00M HCN (K a = 6.2 x ) and 5.00M HNO 2 (K a = 4.0 x )

14.6- Bases 1.Strong Bases  Completely dissociate (ionize) in H 2 O  Group 1 & 2 Metal Hydroxides  KOH, NaOH, LiOH, Mg(OH) 2, Ca(OH) 2, etc.

2. pH of Strong Bases  Ex  Calculate the pH of a 0.050M NaOH solution.

3.Weaker bases that do not involve a hydroxide ion can create one in H 2 O: NH 3 + H 2 O  NH OH - Weak base 4.Base Dissociation Constant (K b ) B(aq) + H 2 O(l)  BH + (aq) + OH - (aq)

5. pH of Weak Bases  Ex  Calculate the pH for a 15.0M solution of NH 3 (K b = 1.8 x )

Ex  Calculate the pH of a 1.0M solution of methylamine (CH 3 NH 2 ). (K b = 4.38 x )

14.7- Polyprotic Acids 1.They dissociate in a stepwise manner  Example  H 2 CO 3 (aq)  H + (aq) + HCO 3 - (aq) K a1 = 4.3 x  HCO 3 - (aq)  H + (aq) + CO 3 -2 (aq) K a2 = 5.6 x K a1 > K a2

 As an H + is taken, the following H + is always less acidic  Since the first K a is larger it will dominate the pH (the amount of H + given off by the second step is insignificant compared to the amount given off by the first)

Ex  Calculate the pH of a 5.0M H 3 PO 4 solution & the equilibrium concentrations of the species H 3 PO 4, and H 2 PO 4 -.

Ex  Calculate the pH of a solution 1.0M H 2 SO 4 solution.

14.8- Salts 1.Salt- an ionic compound 2.Salts that produce neutral solutions  Consist of the conjugates of strong acids & bases  They have no affinity for, nor do they produce any H + or OH -  Examples  Cl - (from HCl)  NO 3 - (from HNO 3 )  Na + (from NaOH)

3.Salts that produce Basic Solutions  They are the conjugate bases of weak acids  Example: NaC 2 H 3 O 2 C 2 H 3 O 2 - (aq) + H 2 O(l)  HC 2 H 3 O 2 (aq) + OH - (aq) Acts as a base and creates

Crazy Cool Derivation!

Ex  Calculate the pH of a 0.30M NaF solution. The K a value for HF is 7.2 x

4.Salts that produce Acidic Solutions  They are the conjugate acids of weak bases  Example: NH 4 Cl NH 4 + (aq) + H 2 O(l)  NH 3 (aq) + H 3 O + (aq) Acts as an acid and creates

Ex  Calculate the pH of a 0.10M NH 4 Cl solution. The K b value for NH 3 is 1.8 x

The Acid & Base Properties of Oxides 1.Acid Oxides  Covalent & form acidic solutions  Examples: SO 2, CO 2, NO 2, etc. CO 2 (g) + H 2 O(l)  H 2 CO 3 (aq)  H + (aq) + HCO 3 - (aq)  CO 2 forms an acidic solution when dissolved in water!

2.Basic Oxides  Ionic and form basic solutions  Examples: CaO, K 2 O, etc.  CaO(s) + H 2 O(l)  Ca(OH) 2 (aq)  Because:  CaO(s)  Ca +2 (aq) + O -2 (aq)  O -2 (aq) + H 2 O(l)  OH - (aq) + OH - (aq)

The Lewis Model 1.Lewis Acid  An electron pain acceptor 2.Lewis Base  An electron pair donor

Ex  For each reaction, identify the Lewis Acid and Lewis Base a)Ni +2 (aq) + 6NH 3 (aq)  Ni(NH 3 ) 6 +2 (aq) a)H + (aq) + H 2 O(l)  H 3 O + (aq)

4.8- Acid-Base Reactions 1.Neutralization Reaction:  Acid + Base  Salt + Water 2.Titrations a)Volumetric Analysis- technique for determining the amount of a certain substance by doing a titration b)Titration- involves delivery (from a buret) of a measured volume of a solution of known concentration (the titrant) into a solution of the substance being analyzed

c)Equivalence (Stoichiometric) Point- the point where enough titrant has been added to react exactly with the substance being analyzed (the analyte) d)Indicator- often marked the equivalence point by a change in color (added in the beginning) e)Endpoint- when indicator changes color  Goal: choose an indicator that matches the endpoint with the equivalence point

Before Titration During Titration Equivalence Point Too Far!!

Ex  What volume of a 0.100M HCl solution is needed to neutralize 25.0mL of 0.350M NaOH?

Ex  28.0mL of 0.250M HNO 3 and 53.0mL of 0.320M KOH are mixed. Calculate the amount of water formed in the resulting reaction. What is the concentration of H + or OH - ions in excess after the reaction goes to completion.

Ex  g of acidic KHC 8 H 4 O 4 (MW = g/mol) is titrated with 41.20mL of NaOH solution until the endpoint is reached. Calculate the concentration of the basic solution used.

Ex  A g sample of waste containing an unknown amount of benzoic acid (HC 7 H 5 O 2, MW = g/mol) was titrated with 10.50mL of M NaOH solution until endpoint was reached. Calculate the mass percent of HC 7 H 5 O 2 in the sample.

15.1- Common Ions with Acids & Bases  Use the same procedure as before  Set up an ICE Box with Molarity  Plug in equilibrium concentrations and solve for x

Ex  Calculate [H + ] in a solution containing 1.0M HF (K a = 7.2 x ) and 1.0M NaF.

15.2- Buffered Solutions 1.A buffered solution is one that resists a change in pH 2.Buffer- a weak conjugate acid-base pair  A weak acid and its conjugate base OR a weak base and its conjugate acid  Examples  HF & NaF  NH 3 & NH 4 Cl  Think of a buffer like a sponge that absorbs H + and OH - so that the pH doesn’t change too much

Example: Weak Acid/Conjugate Base Buffer

Ex  A buffered solution contains 0.50M acetic acid (HC 2 H 3 O 2, K a = 1.8 x ) and 0.50M sodium acetate (NaC 2 H 3 O 2 ). Calculate the pH of the solution.

Ex  Calculate the change in pH that occurs when mol solid NaOH is added to 1.0 L of the buffered solution described in Ex Compare this pH change with that which occurs when mol solid NaOH is added to 1.0L of water.

2. Henderson-Hasselbalch Equation

Ex  Calculate the pH of a solution containing 0.75M lactic acid (K a = 1.4 x ) and 0.25M sodium lactate.

Ex  A buffered solution contains 0.25M NH 3 (K b = 1.8 x ) and 0.40M NH 4 Cl. Calculate the pH of this solution.

Ex  Calculate the pH of the solution that results when 0.10mol gaseous HCl is added to 1.0L of the buffered solution from Ex

15.3- Buffering Capacity  Represents the amount of H + or OH - the buffer can absorb without a significant change in pH  The pH of a buffered solution is dependent on [A - ] / [HA]  When [A - ] = [HA]:  pH = pK a + log (1)  pH = pK a  and THIS is the most effective buffer!

Ex  Calculate the change in pH that occurs when 0.010mol gaseous HCl is added to 1.0L of each of the following solutions:  Solution A:5.00M HC 2 H 3 O 2 & 5.00M NaC 2 H 3 O 2  Solution B:0.050M HC 2 H 3 O 2 & 0.050M NaC 2 H 3 O 2 For acetic aced, K a = 1.8 x 10 -5

 One should choose a buffer with a pK a closest to the desired pH  Ex  A chemist needs a solution buffered at pH 4.30 and can choose from the following acids and their sodium salts. Which choice should be made? a)Chloroacetic acid (K a = 1.35 x ) b)Propanoic acid (K a = 1.3 x ) c)Benzoic acid (K a = 6.4 x ) d)Hypochlorous acid (K a = 3.5 x )

15.4- Titrations and pH Curve 1.pH (Titration) Curve  A plot of solution pH vs. Volume of titrant added during a titration 2.Strong Acid – Strong Base Titration  The Net Ionic Reaction is always: H + (aq) + OH - (aq)  H 2 O(l)  Small amounts are used sometimes, so we can use mmol  1 mmol = mol  mol = 5 mmol  M = mmol / mL

Example  50.0mL of 0.2M HNO 3 titrated with 0.1M NaOH Solve for the pH at Various Points of the Titration A.No NaOH has been added:

B. 10mL of 0.1M NaOH added:

C.100mL 0.1M NaOH added: (Equivalence Point)  The pH at the equivalence point for a strong acid – strong base titration is always 7!

D. 150mL of 0.1M NaOH added:  pH will just keep increasing

Graph: Strong Acid – Strong Base Titration

3. Weak Acid – Strong Base Titration  Example: 50.0mL of 1.0M acetic acid (HC 2 H 3 O 2, K a = 1.8 x ) with 0.1M NaOH Solve for the pH at Various Points of the Titration A.No NaOH has been added:

B. 10mL of 0.1M NaOH added:

C. 25mL 0.1M NaOH added: (Halfway Point)  The pH will always equal the pK a at the Halfway Point

D. 50mL 0.1M NaOH added: (Equivalence Point)  We must find the pH using the K b of C 2 H 3 O 2 - :

 The pH at the equivalence point of a weak acid – strong base titration is always greater than 7.

E. 60mL 0.1M NaOH added:

Graph: Weak Acid – Strong Base Titration

4. Weak Base – Strong Acid Titration  Same procedure as for weak acid – strong base titration, just opposite  The pH of the solution at the Halfway Point is equal to the pK b  The pH of the solution at the Equivalence point is always less than 7.

Graph: Weak Base – Strong Acid Titration

Ex  Hydrogen cyanide gas (HCN, K a = 6.2 x ) is dissolved in water. If a 50.0mL sample of 0.10M HCN is titrated with 0.10M NaOH, calculate the pH of the solution: a)After 8.00mL of 0.10M NaOH is added b)At the halfway point of the titration c)At the equivalence point of the titration

15.5- Indicators 1.Two methods for determining the equivalence point: a)pH Meter- monitor pH & plot the titration curve b)Indicator- changes color at specific pH’s

2.Indicators  Complex molecules that are actually themselves weak acids HIn(aq)  H + (aq) + In - (aq) (Color 1) (Color 2)  Ex. Phenolphthalein HPh(aq)  H + (aq) + Ph - (aq) (colorless) (pink)  Color change will be sharp, occurring with the addition of a single drop of titrant

pH Color Less than 8 colorless Between 8 & 9 light pink Greater than 9 pink

Finally done with Unit 10!