CH 117 Spring 2015 Worksheet 15
1.
Draw and label the titration curve for a weak acid titrated with a strong base.
2.
Draw and label the titration curve for a weak base titrated with a strong acid.
3.
There are four types of titration problems (listed below) we will do with weak acids/weak bases in this chapter. Describe how to set up each of the following calculations to solve for pH. ON ALL OF THESE CALCULATIONS, DO NOT FORGET TO FACTOR
IN THE FINAL, TOTAL VOLUME WHEN YOU HAVE FINISHED WITH YOUR
FIRST ICE CHART. a). Before the equivalence point – Calculate the number of moles of acid and base present in the titration. Set up a limiting reagent ICE chart and get rid of all of the
LR. You should be left with a buffer solution

use Henderson-Hasselbach to calculate the pH. b). At the half equivalence point
– pH = pK a c). At the equivalence point
– Calculate the number of moles of acid and base present.
Set up a limiting reagent ICE chart and get rid of all of the LR. You will be left with only conjugate acid or conjugate base in solution. Set up a second, regular ICE chart by dissolving the conjugate acid or base in water. Solve for x in the ICE chart, which will give you either the [H
3
O + ] or [OH ]

use these concentrations to calculate pH. d). After the equivalence point – Calculate the number of moles of acid or base present. Set up a limiting reagent ICE chart and get rid of all of the LR. You should

CH 117 Spring 2015 Worksheet 15 be left with either strong base + conjugate base or strong acid + conjugate acid. Use only the strong concentration to calculate the pH or pOH.
4.
Find the pH at each of the following points in the titration of 25 mL of 0.3 HF with 0.3 M
NaOH.
Identify type of titration – weak acid + strong base
K a
for HF is 7.2 x 10 -4 a). After adding 10 mL of base
Identify where you are in the titration – before the equivalence point, so we expect a buffer calculation.
Calculate moles of acid and base.
Moles of HF - .025 L x 0.3 M = .0075 moles HF
Moles of NaOH - .01 L x 0.3 M = .003 moles NaOH

should be the limiting reagent since there is less of it available
Set up a limiting reagent ICE chart. Reaction should reflect a strong base (OH ) reacting with the weak acid.
Initial
HF OH -
.0075 moles .003 moles

F -
0 moles
H
2
O n/a
Change -.003 moles -.003 moles
Equilibrium .0045 moles 0 moles
+.003 moles n/a
.003 moles n/a
Left with a buffer solution – about equal amounts of conjugate base and weak acid.
First calculate the concentrations using the total volume, then plug into Henderson-
Hasselbach.
[HF] = .0045 moles / (.025 +.01 L) = .129 M
[F ] = .003 moles / (.025 + .01 L) = .086 M pH = pK a
+ log([base]/[acid]) pH = -log(7.2 x 10 -4 ) + log(.086/.129) = 5.37 b). After adding 12.5 mL of base
Identify where you are in the titration – probably at the midpoint (half-equivalence point), but go through the calculations anyways to verify the assumption.
Calculate moles of acid and base.
Moles of HF = .0075 moles
Moles of NaOH - .0125 L x .3 M = .00375 moles
HF OH 
Initial .0075 moles .00375 moles
F -
0 moles
H
2
O n/a
Change -.00375 moles
-.00375 moles
+.00375 moles n/a
Equilibrium .00375 moles
0 moles .00375 moles n/a
We are at the midpoint since there are exactly equal amounts of weak acid and conjugate base left in solution.

CH 117 Spring 2015 Worksheet 15 pH = pK a

pH = -log(7.2 x 10 -4 ) = 3.14 c). After adding 25 mL of base
Equivalence point or end point calculation – moles of acid = moles of base
Calculate moles of acid and base.
Moles of HF = .0075 moles HF
Moles of NaOH = .025 L x 0.3 M = .0075 moles NaOH
Set up limiting reagent ICE chart. Both acid and base are limiting, so get rid of both entirely.
Initial
HF OH -
.0075 moles .0075 moles

F -
0 moles
H
2
O n/a
Change -.0075 moles
Equilibrium 0 moles
-.0075 moles
0 moles
+.0075 moles n/a
.0075 moles n/a
You are left with only conjugate base in solution or the anion of the salt created through the titration, so you need to set up another, regular ICE chart dissolving this salt in water. (Reaction looks suspiciously like the reverse of the first ICE chart)
Work through this like a normal ICE chart, using x’s in your change row and solving for the value of x, etc.
Before you can put any numbers in this ICE chart, you MUST calculate the concentration of anion using the total volume. In other words, since this is a normal
ICE chart (not a limiting reagent one), YOU MUST WORK IT IN
CONCENTRATION.
[F ] = .0075 moles / (.05 L) = .15 M
F H
2
O
Initial
Change
.15 M
-x n/a n/a

HF
0 M
+x
OH
0 M
+x
-
Equilibrium .15 – x n/a x x
Now solve for x using the K expression. Notice since this is a BASIC salt solution, we will need to use a K b
rather than a K a
.
K b
= K w
/K a
= (1.0 x 10 -14 )/(7.2 x 10 -4 ) = 1.39 x 10 -11
1.39 x 10 -11 = x 2 / (.15 – x)

Use the 5% rule to disregard the x in the denominator since .15 is so much less than the K value.
1.39 x 10 -11 = x 2 /.15

x 2 = 1.39 x 10 -11 * (.15) = 2.09 x 10 -12 x = 1.44 x 10 -6 M x corresponds to the value of [OH ]

pOH = -log(1.44 x 10 -6 ) = 5.84 pH = 14 – 5.84 = 8.16 d). After adding 26 mL of base
After the equivalence point
Calculate the moles of acid and base.
Moles of HF = .0075 moles HF
Moles of NaOH = .026 L x (.3 M) = .0078 moles NaOH
Set up limiting reagent ICE chart, HF is the limiting reagent in this case

CH 117 Spring 2015 Worksheet 15
Initial
Change
HF
.0075 moles
-.0075 moles
Equilibrium 0 moles
OH -
.0078 moles
-.0075 moles
.0003 moles

F -
0 moles
+.0075 moles
.0075 moles
H
2
O n/a n/a n/a
You are left with a solution that has both strong base and weak conjugate base

ONLY use the strong base to calculate pH, since the strong base will overpower any weak base that is present.
[OH ] = .0003 moles / (.025 + .026 L) = .006 M pOH = -log(.006 M) = 2.23 pH = 14 – 2.23 = 11.77
5.
Calculate the molar solubility and the solubility in grams/liter of MgF
2
. The K sp
of MgF
2 is 6.4 x 10
-9
.
Use an ICE chart to calculate molar solubility. Initially you have none of the ions in solution. Remember that solids play no part in the equilibrium calculations!
Initial
Change
MgF n/a n/a
2
(s)

Mg
0
+x
2+ 2 F
0
-
+2x
Equilibrium n/a x 2x
Set up the equilibrium expression to solve for x, which is the same as the molar solubility of the entire salt.
K sp
= 6.4 x 10 -9 = x * (2x) 2 = 4x 3

x = .0012 M
To calculate the grams/L solubility, simply manipulate the units.
.0012 mol/L * molar mass of MgF
2
(62.3 g/mol) = .0748 grams/L