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Chapter 18. Acid-Base Equilibria. Acid – Base Reactions. Strong acid + strong base: HNO 3 + Ca(OH) 2 H + (aq) + OH - (aq) ----------> H 2 O(l). Acid-Base Reactions. Weak acid + strong base: HF + KOH HF(aq) + OH - (aq) <----> F - (aq) + H 2 O(l). Acid-Base Reactions.

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chapter 18

Chapter 18

Acid-Base Equilibria

WOLPA/AP CHEMISTRY/CDO

acid base reactions
Acid – Base Reactions
  • Strong acid + strong base:
  • HNO3 + Ca(OH)2
  • H+(aq) + OH-(aq) ----------> H2O(l)

WOLPA/AP CHEMISTRY/CDO

acid base reactions1
Acid-Base Reactions
  • Weak acid + strong base:
  • HF + KOH
  • HF(aq) + OH-(aq) <----> F-(aq) + H2O(l)

WOLPA/AP CHEMISTRY/CDO

acid base reactions2
Acid-Base Reactions
  • Strong acid + weak base:
  • HClO4 + NH3
  • H+(aq) + NH3(aq) <----------> NH4+(aq)
  • H2SO4 + Na2CO3
  • H+(aq) + CO32-(aq) <------> HCO3-(aq)

WOLPA/AP CHEMISTRY/CDO

the common ion effect
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.

WOLPA/AP CHEMISTRY/CDO

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

WOLPA/AP CHEMISTRY/CDO

example1
Example
  • Determine the pH of a solution of 0.25 M acetic acid in a solution of 0.10 M sodium acetate.

WOLPA/AP CHEMISTRY/CDO

example2
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 = 3.5x108).

WOLPA/AP CHEMISTRY/CDO

buffers
Buffers
  • Acid-base buffersconfer 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.

WOLPA/AP CHEMISTRY/CDO

general expressions for buffer solutions
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]

WOLPA/AP CHEMISTRY/CDO

example3
Example
  • Determine the pH of a solution of 0.10 M acetic acid in a solution of 0.10 M sodium acetate.

WOLPA/AP CHEMISTRY/CDO

effect of h 3 o or oh on buffer system
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.

WOLPA/AP CHEMISTRY/CDO

buffer calculations
Buffer Calculations
  • Determine the concentrations of HB and B- after the addition of H3O+ or OH-

WOLPA/AP CHEMISTRY/CDO

addition of an acid
Addition of an Acid
  • 0.10 M HC2H3O2 with 0.10 M NaC2H3O2 add 50.0 mL 0.10 M HCl

WOLPA/AP CHEMISTRY/CDO

addition of a base
Addition of a Base
  • 0.10 M HC2H3O2 with 0.10 M NaC2H3O2 add 50.0 ml 0.10 M NaOH

WOLPA/AP CHEMISTRY/CDO

the henderson hasselbalch equation
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]

WOLPA/AP CHEMISTRY/CDO

henderson hasselbalch equation
Henderson-Hasselbalch Equation

Allows us to predict pH when HB/B mixed.

  • When [B] /[HB] = 1 (i.e. [HB]=[B]), pH = pKa

WOLPA/AP CHEMISTRY/CDO

example4
Example

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

WOLPA/AP CHEMISTRY/CDO

example5
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.

WOLPA/AP CHEMISTRY/CDO

example6
Example

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

WOLPA/AP CHEMISTRY/CDO

preparation of a buffer solution
Preparation of a Buffer Solution

Selection of weak acid or weak base

  • (pKa ~ pH)

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titrations of weak acids and strong bases
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

WOLPA/AP CHEMISTRY/CDO

ph changes
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

WOLPA/AP CHEMISTRY/CDO

ph calculations
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.

WOLPA/AP CHEMISTRY/CDO

ph titration curves
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).

WOLPA/AP CHEMISTRY/CDO

slide26
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.

WOLPA/AP CHEMISTRY/CDO

example7
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?

WOLPA/AP CHEMISTRY/CDO

sa sb titrations
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.

WOLPA/AP CHEMISTRY/CDO

slide29
Equivalence point(EP): pH = 7.00
  • Beyond EP: pH due only to base added (i.e. excess base). Use total volume.

WOLPA/AP CHEMISTRY/CDO

example8
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.

WOLPA/AP CHEMISTRY/CDO

titration of sb with sa
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.

WOLPA/AP CHEMISTRY/CDO

example9
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.

WOLPA/AP CHEMISTRY/CDO

wa with sb titration
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 remaining

nHA = 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.

WOLPA/AP CHEMISTRY/CDO

example10
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 = 1.75x105.

WOLPA/AP CHEMISTRY/CDO

wb sa titrations
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).

WOLPA/AP CHEMISTRY/CDO

example11
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 = 1.75x105.

WOLPA/AP CHEMISTRY/CDO

acid base indicators
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)

WOLPA/AP CHEMISTRY/CDO

choice of indicator
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)

WOLPA/AP CHEMISTRY/CDO

choice of indicator1
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)

WOLPA/AP CHEMISTRY/CDO

choice of indicator2
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