CH 117 Spring 2015
Worksheet 14
1. What is the difference between the equivalence point and the end point of a titration?
The equivalence point is a theoretical, calculated value where there are exactly
equal amounts of acid and base present in the titration solution. The end point is the
experimental point at which the amount of acid and base should be the same,
detected in the lab by indicators. If you have done the titration correctly and
selected an appropriate indicator, the equivalence point and end point will be the
same.
2. Draw and label the titration curve for the titration of a strong acid with a strong base.
3. Draw and label the titration curve for the titration of a strong base with a strong acid.
CH 117 Spring 2015
Worksheet 14
4. What is the pH at the point in a titration at which 20.00 mL of 1.0 M KOH has been
added to 25.00 mL of 1.0 M HBr?
First identify the type of titration – this is a strong acid titrated with a strong base,
so the first curve we drew.
Strong acid/strong base titrations are nothing more than limiting reagent problems,
so we’ll use a limiting reagent ICE table to visualize what’s happening. We really
only care about the reactants side for strong acid/strong base titrations since the salt
and the water that are produced will both HAVE NO EFFECT on pH. Titration
limiting reagent ICE charts must be worked in units of moles, so make sure to
multiply volume by molarity for each chemical to get started.
Moles of HBr = .025 L x 1.0 M = .025 moles HBr
Moles of KOH = .020 L x 1.0 M = .02 moles KOH
Since there is less of the KOH to begin with, that will be our limiting reagent.
Remember to use the ICE chart to get rid of all of the limiting reagent and
subtract/add from the other reagents as appropriate.
Initial
Change
Equilibrium
(Final)
HBr
.025 moles
-.02 moles
.005 moles

KOH
.02 moles
-.02 moles
0 moles
KBr
n/a
n/a
n/a
H2 O
n/a
n/a
n/a
So, our final solution has .005 moles of HBr remaining. Since HBr is a strong acid,
we can calculate pH simply by taking the –log of the concentration of acid left.
Since we have moles, we must divide by the FINAL TOTAL VOLUME to calculate
molarity of acid remaining. This is an important step in titration problems. Do not
forget that you are working in moles, so it is necessary to calculate a concentration
before finding the pH or pOH.
[HBr] = [H3O+] = .005 moles / (.025 + .02 L) = .111 M
pH = -log[H3O+] = -log(.111) = 0.95
5. What is the pH at the point in a titration at which 55.00 mL of 1.0 M KOH has been
added to 25.00 mL of 1.0 M HBr?
Identify type of titration – strong acid + strong base
The process involved is identical to the process in #4 with different numbers.
Moles of HBr = .025 L x 1.0 M = .025 moles HBr
Moles of KOH = .055 L x 1.0 M = .055 moles of KOH
HBr is our limiting reagent now.
Initial
Change
Equilibrium
HBr
.025 moles
-.025 moles
0 moles
KOH
.055 moles
-.025 moles
.03 moles

KBr
n/a
n/a
n/a
H2 O
n/a
n/a
n/a
CH 117 Spring 2015
Worksheet 14
Final solution has .03 moles of KOH  [KOH] = .03 moles/ (.025 L + .055 L) = .375
M
[KOH] = [OH-} = .375 M
pOH = -log(.375) = .43
pH = 14 - .43 = 13.6