AcidBase Reactions

1
Acid-Base Reactions

• Do the stoichiometry problem first. Acid-base
  reactions are complete, and the limiting reagent
  is completely consumed.
• Stoichiometry problems are worked in moles, not
  molarity.
• No. of mmol (Molarity) x (volume in mL)
• The result tells what is present in the solution
  before any weak-acid or weak-base equilibria take
  place.

2

• Convert the quantities to molarity by dividing
  by the total solution volume.
• Identify the type of pH problem from the species
  present after the acid-base reaction has reached
  completion.
• a) strong acid or strong base only
• b) weak acid or weak base only
• c) salt of WA or salt of WB only
• d) buffer solution

3
Examples of Acid-Base Reaction Problems

• Consider our earlier buffer solution 1.00 L of
  soln that was 0.10 M in HC2H3O2 and 0.20 M in
  NaC2H3O2. The pH of this solution is 5.05.
• Now bubble 0.010 mol of HCl gas into the
  solution. Assuming that no volume change occurs,
  what is the new pH?
• First identify the nature of each species and
  any possible acid-base reactions that can occur.

4

• HC2H3O2 is a weak acid. NaC2H3O2 is the salt of
  a weak acid, i.e., a weak base. HCl is a strong
  acid.
• Find the number of moles of all substances
  initially present
• (1.0 L)(0.10 M) 0.10 mol HC2H3O2
• (1.0 L)(0.20 M) 0.20 mol NaC2H3O2
• The acid HCl will react with the base NaC2H3O2
  until the limiting reagent is used up.
• Now, work the stoichiometry problem.

5

• HCl NaC2H3O2 ? HC2H3O2
  NaCl
• I. 0.010 mol 0.20 mol
  0.10 mol 0
• C. 0.010 0.010
  0.010 0.010
• F. 0 0.19
  0.11 0.010
• Recall that the NaCl is neutral and plays no
  further role.
• Now, which of the four types of pH problems does
  this represent?
• We have NaC2H3O2, the salt of a weak acid, and
  HC2H3O2, a weak acid. The HCl is gone.
• Thus, this is a buffer solution.

6

• Convert the quantities to molarities.
• 0.19 mol NaC2H3O2 / 1.0 L 0.19 M
• 0.11 mol HC2H3O2 / 1.0 L 0.11 M
• The salt dissociates first
• NaC2H3O2 ? Na C2H3O2
• 0 0.19
  0.19 M
• The weak acid is in equilibrium with its ions
• HC2H3O2 ? H C2H3O2
• I. 0.11 0
  0.19
• C. x x
  x
• E. 0.11x x
  0.19x

7

• x H 1.04 x 105 M
• pH 4.98
• Or use the Henderson-Hasselbalch equation
• pH 4.98

8

• Now take the same original buffer solution and
  instead add 0.010 mole of solid NaOH. Assuming
  that no volume change occurs, what is the new pH?
• This time, the strong base NaOH reacts with the
  weak acid HC2H3O2.
• NaOH HC2H3O2 ? NaC2H3O2 H2O
• I. 0.010 mol 0.10 mol
  0.20 mol
• C. 0.010 0.010
  0.010
• F. 0 0.09 mol
  0.21 mol

9

• 0.09 mol HC2H3O2 / 1.0 L 0.09 M
• 0.21 mol NaC2H3O2 / 1.0 L 0.21 M
• This is a buffer solution (weak acid its salt).
• pH 5.11
• If we had simply placed 0.010 mol of HCl in a
  liter of water, the pH would be 2.00. If we had
  placed 0.010 mol of NaOH in a liter of water, the
  pH would be 12.00 (verify this for yourself!).

10
Mixing of Acid and Base Solutions

• Ex. Calculate the pH of the solution resulting
  when 200 mL of 0.20 M NH3 is mixed with 100 mL of
  0.10 M HCl.
• Find the number of moles of each substance.
• (200 mL)(0.20 mmol/mL) 40 mmol NH3
• (100 mL)(0.10 mmol/mL) 10 mmol HCl
• 2) Identify the acid-base reaction and do the
  stoichiometry problem (Find the amounts of
  products formed).

11

• NH3 HCl ? NH4 Cl
• Initial 40 mmol 10 mmol
  0 0
• Change 10 10
  10 10
• Final 30 mmol 0
  10 mmol 10 mmol
• Identify the type of pH problem from the species
  present after the reaction.
• a) Strong acid or strong base alone
• b) Weak acid or weak base alone
• c) Salt of WA or salt of WB alone
• d) Buffer solution
• NH3 NH4Cl weak base its salt
• buffer solution

12

• Equilibrium problems must be worked in molarity.
  The total mixed volume must be taken into
  account.
• 30 mmol NH3 / 300 mL total 0.10 M
• 10 mmol NH4Cl / 300 mL total 0.033 M
• The salt is completely dissociated
• NH4Cl ? NH4 Cl
• 0 0.033 M
• The weak base is in equilibrium

13

• NH3 H2O ? NH4 OH
• Initial 0.10 M
  0.033 M 0
• Change x
  x x
• Equil 0.10x
  0.033x x
• X OH 5.4 x 105 M
• pOH 4.27
• pH 9.73

14

• Using the Henderson-Hasselbalch equation
• The Ka of NH4 Kw / Kb 5.6 x 1010
• The pKa 9.25
• pH 9.25 log(Conj Base/Conj Acid)
• 9.25 log(0.10 / 0.033)
• 9.73