wolpa/ap chemistry/cdo chapter 18 acid-base equilibria

WOLPA/AP CHEMISTRY/CDO

Chapter 18

Acid-Base Equilibria


Acid – Base Reactions

• Strong acid + strong base:

• HNO3 + Ca(OH)2

• H+(aq) + OH-(aq) ----------> H2O(l)


Acid-Base Reactions

• Weak acid + strong base:

• HF + KOH

• HF(aq) + OH-(aq) <----> F-(aq) + H2O(l)


Acid-Base Reactions

• Strong acid + weak base:

• HClO4 + NH3

• H+(aq) + NH3(aq) <----------> NH4+(aq)

• H2SO4 + Na2CO3

• H+(aq) + CO32-(aq) <------> HCO3

-(aq)


The Common Ion Effect

The common ion effect is when an ion common to the ionization of an acid is present in the solution in an amount greater than that produced by the acid ionization. The presence of this ion, according to LeChatelier's principle, limits the extent to which the acid will ionize and thus affects the pH of the solution.


Example

• Determine the pH of a solution of 0.25 M acetic acid. Ka = 1.8 x 10-5


Example

• Determine the pH of a solution of 0.25 M acetic acid in a solution of 0.10 M sodium acetate.


Example

Determine the pH of a solution prepared by mixing 50.0 mL of 0.100 M HOCl with 50.0 mL of 0.100 M NaOCl (Ka = 3.5x108).


Buffers

• Acid-base buffers confer resistance to a change in the pH of a solution when hydrogen ions (protons) or hydroxide ions are added or removed. An acid-base buffer typically consists of a weak acid, and its conjugate base.

• Prepared by adding both the weak acid HB and the salt of its conjugate base B- to water.


General Expressions for Buffer Solutions

• Assume equilibrium is established, therefore

• Ka = [H3O+] [B-] / [HB] and

• [H3O+] = Ka [HB] /[B-]

or• Kb = [HB] [OH-] / [B-] and • [OH-] = Kb [B-] /[HB]


Example

• Determine the pH of a solution of 0.10 M acetic acid in a solution of 0.10 M sodium acetate.


Effect of H3O+ or OH- on Buffer System

• A buffer system contains one species (HB) that will react with added hydroxide ions and another species (B-) that will react with added hydronium ions.

• Both reactions will go virtually to completion hence, the added hydronium or hydroxide ions are consumed and the effect on the overall pH is negligible.


Buffer Calculations

• Determine the concentrations of HB and B- after the addition of H3O+ or OH-


Addition of an Acid

• 0.10 M HC2H3O2 with 0.10 M NaC2H3O2 add 50.0 mL 0.10 M HCl


Addition of a Base

• 0.10 M HC2H3O2 with 0.10 M NaC2H3O2 add 50.0 ml 0.10 M NaOH


The Henderson-Hasselbalch Equation

• pH = pKa + log [conjugate base]/[acid]

• This equation clearly shows that the pH of the solution of a weak acid and its conjugate base is controlled primarily by the strength of the acid.

• Can also be written

• pOH = pKb + log [conjugate acid]/[base]


Henderson-Hasselbalch Equation

Allows us to predict pH when HB/B mixed.• When [B] /[HB] = 1 (i.e. [HB]=[B]), pH = pKa


Example

Calculate pH of solution containing 0.040M Na2HPO4 and 0.080M KH2PO4. pKa2=7.20.


Example

Determine the ratio of the concentration of the conjugate acid to concentration of the conjugate base for a weak acid in which the pH was 5.45 and pKa was 5.75.


Example

Determine the pH of a solution consisting of 0.100 M NH3 and 0.150 M NH4Cl.


Preparation of a Buffer Solution

Selection of weak acid or weak base

• (pKa ~ pH)


Titrations of Weak Acids and Strong Bases

Typical net ionic equation

• HC2H3O2 + OH- ------> C2H3O2- +

H2O

• K = 1/Kb(C2H3O2-) = 1/5.6 x 10-10 =

1.8 x 109

• The reaction goes essentially to completion


pH Changes

• -pH starts off at about 2.4, the pH of 1 M HC2H3O2

• -Region, centered around the half-way point of the titration where pH changes very slowly. In this region there are appreciable amounts of two different species: unreacted HC2H3O2and C2H3O2

- ions. Hence, we have a buffer system• -Equivalence point - we have a solution of

sodium acetate. This solution is basic because C2H3O2

- is a weak base. pH at equivalence is > 7.00


pH Calculations

• Write a net ionic equation to determine the extent of the reaction.

• Calculate the initial pH from the Ka or Kb.

• Calculate pH at midpoint using the buffer relation.

• Calculate pH at equivalence point using Ka of conjugate acid or Kb of conjugate base.


pH Titration Curves

• Titration curve: plot of pH of the solution as a function of the volume of base (acid) added to an acid (base).

Titration of 0.100 M HA with 0.100 M NaOH

0

2

4

6

8

10

12

14

0 10 20 30 40

Volume Base Added, mL

pH

WA

SA


• Sharp rise in curve is equivalence point.

• pH at equivalence point is 7.0 for SA but higher for WA.

• Equivalence point can be used to determine the concentration of the titrant.


Example

The equivalence point for 15.00 mL of an acid occurred when 25.00 mL of 0.075 M NaOH was added. What was the molarity of the acid?


SA–SB Titrations• Base removes some acid and pH increases.• Let nb = moles of base added

na,r = moles of acid remaining

na,r = na nb = MaVa MbVb

• Moles of hydronium ion same as moles of acid remaining.

nH3O+ = na,r;

• Valid until very close to equivalence point.


• Equivalence point(EP): pH = 7.00

• Beyond EP: pH due only to base added (i.e. excess base). Use total volume.


Example

• Determine pH of 10.0 mL of 0.100M HCl after addition of 5.00, 10.0 and 15.0mL of 0.100M NaOH.


Titration of SB with SA• Acid removes some of the base and pH is changed by amount

of base removed.Let na = moles of acid added nb,r = moles of base remaining

nb,r = MbVb MaVa

• Moles of hydroxide ion same as moles of base remaining.

• nOH = nb,r;

– Valid until EP.

• EP: pH = 7.00• Beyond EP: pH due only to excess acid. Use total volume.


Example

• Determine pH of 10.0 mL of 0.100M NaOH after addition of 5.00, 10.0 and 15.0mL of 0.100M HCl.


WA with SB Titration• As above base removes some of the acid and pH is

changed by amount of acid removed. Let nb = moles of base added

nHA = moles of acid remainingnHA = MHAVHA MbVb

nA = nb = MbVb

• Up to equivalence point moles of hydronium ions must be determined from equilibrium expression.

• Equivalence point: pH = pH of salt of WA• Beyond Equivalence point: Use amount of excess

base to determine pH.


Example

• Determine pH of 10.0 mL of 0.100M HA after addition of 5.00, 10.0 and 15.0mL of 0.100M NaOH. Ka = 1.75x105.


WB–SA Titrations• Acid removes some of the base and decreases the pH.

Let na = moles of acid added nb,r = moles of base remaining

nb,r = CbVb CaVa nBH+ = na = CaVa

• Moles of hydroxide ions must be determined from equilibrium expression. Valid until EP.

• EP: pH = pH of salt of weak base.• Beyond EP: pH due only to presence of acid added

after endpoint (i.e. excess acid) as seen for strong base. Volume correction needed as above (total volume).


Example

Determine pH of 10.0 mL of 0.100M B after addition of 5.00, 10.0 and 15.0mL of 0.100M HCl. Kb = 1.75x105.


Acid-Base Indicators

• An acid-base indicator (HIn) is usually an organic dye that is itself a weak acid governed by an equilibrium constant.

• The acid form has one color and the base form has another. In principle, the color of the indicator changes when [H3O+] = Ka of the indicator.(because HIn = In- at this point)


Choice of Indicator

1. Strong acid-Weak base

• Solution at equivalence point is weakly acidic. Choose indicator which turns color below pH 7. ex methyl red (pH 5)


Choice of Indicator

2. Weak acid - Strong base

• Solution at equivalence point is weakly basic, Choose indicator which turns color above pH 7. ex. phenolphthalein (pH 9)


Choice of Indicator

• Strong acid - Strong base

• Solution at equivalence point is neutral. However, pH changes so rapidly near the end point that any indicator that changes color between 5 and 9 will work.

wolpa/ap chemistry/cdo chapter 18 acid-base equilibria

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