#5
Guided Study Titrations:
We have already discussed strong acid strong base titration calculations so we will skip to
Weak acid – Strong base titration
Calculate the pH after the follow total volumes of 0.400 M NaOH are added to 50.00 mL of 0.200 M HCOOH
a)0.00 mL
b)5.00 mL
c)12.50 mL
d)24.50 mL
e)25.00 mL
f)25.50 mL
g)40.00 mL
a)0.00 mL of 0.400 M NaOH are added to 50.00 mL of 0.200 M HCOOH
There is only a weak acid solution, 0.200 M HCOOH.
- Equation - dissociation of HCOOH
HCOOH ↔ COOH- + H+
- ICEBOX (but ignore X)
- Use Ka to solve for X
- X = [H+] Solve for pH
- 5% test?
b)5.00 mL of 0.400 M NaOH added to 50.00 mL of 0.200 M HCOOH
- Equation adding strong base to the acid
HCOOH + OH- ↔ COOH- + HOH
You will have a buffered solution containing HCOOH and COOH- so calculate the pH of the buffered solution
Stoichiometry Part
- Solve for moles
mole HCOOH initial =
mole OH- added =
- ICEBOX using moles subtracting the limiting amount of moles from initial Solve for concentrations with new volumes (equil. mol /liters)
[HCOOH] =
[COOH-] =
Equilibrium part or pH = pKa + log [A-]/[HA]
Formic acid is a stronger acid than the formate ion which is a base (comparing K values) Therefore, use the acid dissociation equilibrium
- Solve for X (H+) using Ka = ?
- Solve for pH or use Henderson Hasselbalch
Notice the pH doesn’t rise much in the buffer region of a titration
c)c. 12.50 mL of .400 M NaOH is addedto 50.00 mL of 0.200 M HCOOH
Stoichiometry Part
- Solve for moles
mole HCOOH initial =
mole OH- added =
- ICEBOX using moles subtracting the limiting amount of moles
- Solve for concentrations with new volumes
[HCOOH] =
[COOH-] =
Equilibrium part orpH = pKa + log [A-]/[HA]
- Solve for X (H+) using Ka =
- Solve for pH or use Henderson Hassel balch
WHEN [HA] = [A-] , pKa = pH. THIS IS THE TITRATION MIDPOINT. We are halfway to the equivalence point. This is important because if we know pH, we can find pKa, for the weak acid.
d) 24.50 mL of .400 M NaOH is addedto 50.00 mL of 0.200 M HCOOH
Stoichiometry Part
- Solve for moles
mole HCOOH initial =
mole OH- added =
- ICEBOX using mole
- WE ARE NEAR THE BUFFER CAPACITY
- Solve for concentrations with new volumes
[HCOOH] =
[COOH-] =
Equilibrium part or pH = pKa + log [A-]/[HA]
- Solve for X (H+) using Ka =
- Solve for pH
e)25.00 mL of 0.4000 M NaOH is added to 50.00 mL of 0.200 M HCOOH
Stoichiometry Part
- Solve for moles
mole HCOOH initial =
mole OH- added =
- ICEBOX using mole
WE HAVE REACHED THE EQUIVALENCE POINT. We now have a solution that contains a base that can undergo hydrolysis. LeChatelier will predict a change in the direction of the reaction since there is almost no HCOOH or OH- left.
Equilibrium part
NEW EQUATION!!!!!!!
COOH- + HOH ↔ HCOOH + OH-
Kb = Kw/ Ka
Kb =
- Solve for concentration of [COOH-] with new volume
- Solve for X using Kb
X = [OH-]
- Solve for pOH then pH
- OR use Henderson Hasselbalch pOH = pKb + log [ HA]
[A-]
f)25.50 mL of 0.4000 M NaOHis added to 50.00 mL of 0.200 M HCOOH
Stoichiometry Part
- Solve for moles
mole HCOOH initial =
mole OH- added =
- ICEBOX with mol
We have neutralized all of the weak acid and have 0.20 mol of strong base remaining. The pH will be determined by the excess strong base.
- Calculate new concentration of [OH-] using new volume
- Solve for pOH, then pH
g)40.00 mL of 0.4000 M NaOH is added to 50.00 mL of 0.200 M HCOOH
- Solve for moles
mole HCOOH initial =
mole OH- added =
- ICEBOX with mol
We have neutralized all of the weak acid and have ______mol of strong base remaining. The pH will be determined by the excess strong base.
- Calculate new concentration of [OH-] using new volume
- Solve for pOH, then pH
Weak Base – Strong Acid
These problems use the same strategies, having a buffer region, equivalence point, and post- equivalence point.
Try this one:
Calculate the pH at each of the following points in the titration of 50.00 mL of a 0.01000 M sodium phenolate (NaOC6H5) solution with 1.000 M HCl solution ( Ka for HOC6H5 = 1.05 x 10 -10 )
a) initially b) midpoint c) equivalence point
Kb for OC6H5- = Kw/Ka =
a) initially
- Equation OC6H5- + HOH ↔ HOC6H5 + OH-
- Solve for X = [OH-] using Kb
- Solve for pOH, then pH
b)midpoint
by definition, pH at the titration midpoint equals pKa for the acid. Therefore, pOH = pKb for the base.
- Solve pKb = - logKb = ______
pOH = ______pH = ______
c)equivalence point
All of the base has been converted to acid at the equivalence point. You must know the concentration of the weak acid at the equivalence point. SO the question is “How many mL of HCl were required to reach the equivalence point?”
- solve for moles of OC6H5-initial =
- solve for ml HCL required to just neutralize
- using the total volume 50.00 + 0.5000 = 50.50 mL, find the concentration of the weak acid
[ HOC6H5] =
- use this concentration and Ka to solve for X (X = [H+])
- solve for pH
- Or use Henderson Hasselbalch
Check out the graphs for these titrations
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