SCH4U
Chapter 8 Practice Test
1) According to the Brønsted Lowry definition of an acid-base reaction...
a) what would the products of the following reaction have to be? HA + B —>A- + HB+
b) why is the following reaction not an acid-base reaction? CaCl2(aq) + 2NaHSO4(aq) —> Ca(HSO4)2(s) + 2NaCl(aq)
The Hydrogen ion is still attached to the SO42- in the products as it was in the reactants (H+ not transferred).
2) Identify the base, conjugate base, acid, and conjugate acid in the following reaction.
H2PO4- + HS- —> H2S + HPO42acid
base
conj A. conj B.
3) Which of the following are...
i) acids?
ii) bases?
iii) neutral?
iv) amphoteric?
b) HPO42c) SO32a) HNO3
acid
amph.
base
d) Clneut.
e) NH3
amph.
f) Al3+
acid
g) Na+
neut.
4) If the Ka of an acid HA is 3x10-5, what is the Kb of the ion A-? Kb = 10-14/3x10-5 = 3.3x10-10
5) H2PO4 has 3 ionizable Hydrogens with the following K values: Ka1=7.1x10-3, Ka2=6.3x10-8, Ka3=4.2x10-13
a) Which one would be used to determine the pH of a solution of phosphoric acid? Ka1 since it is the largest.
b) List the concentrations of the 4 forms of this molecule in solution in order of decreasing concentration.
Since even Ka1 is much less than one, the order would be [H3PO4]>[H2PO4-]>[HPO42-]>[PO43-]
6) Which of the following salts will produce the most basic solution?
b) Na2HPO4
c) Na3PO4
a) NaH2PO4
The answer is c) since Ka3 is the smallest, Kb3 (10-14/Ka3) will be the greatest of the three Kb values.
7) Calculate a Kb value to determine if NaH2PO4 will be an acid or a base.
H2PO4- can undergo either of the following two reactions:
H2PO4- + H2O —> HPO42- + H3O+ (Ka2 = 6.3x10-8)
-14
-14
-3
-12
(Kb1 = 10 /Ka1 = 10 /7.1x10 = 1.4x10 )
H2PO4- + H2O —> H3PO4 + OH- (Kb1 = 1.4x10-12)
Since Ka2>Kb1, the acid reaction will predominate and the salt solution will be acidic.
8) a) What would be the pH of a 2.5x10-5 M NaOH(aq) if this substance is a strong base?
[OH-] = [NaOH] x 100% x 1 = 2.5x10-5 (100% since it is strong, 1 since it has only 1 OH- in the formula)
[H+] = 10-14/[OH-] = 10-14/2.5x10-5 = 4x10-10
pH = -log[H+] = -log(4x10-10) = 9.4
b) What would be the pH of a 7.5x10-4 M HC2H3O2(aq) is this monoprotic acid has a percent ionization of 1.3%?
[H+] = [HC2H3O2] x 0.013 x 1 =9.75x10-6 (0.013 since 1.3% strength, 1 since it is monoprotic)
pH = -log[H+] = -log(9.75x10-6) = 5
9) The Ka of an indicator HIn is 4x10-5. At what pH will this indicator change colour?
changes colour when [HA]=[A-], therefore Ka = [H+]([A-]/[HA]) = [H+]
pH of colour change = -log(Ka) = 4.4
10) What is a primary standard and why are these substances necessary for accurate titrations?
A primary standard is an acid or base with a large molar mass, that is stable in pure form as a solid, and absorbs a
precisely predictable amount of H20 and CO2 from the atmosphere. As such, it is possible to measure out a very
precise number of moles in pure form. Solutions of primary standards are used to titrate the strong acids or bases
used as titrants in the titration of unknown solutions. This is necessary because none of the good titrants satisfy all
of the criteria for measuring out a precise number of moles in pure form.
11) Calculate the pH of a 1.5x10-3 M solution of HA if the Ka of this acid is 6x10-5.
This is an equilibrium problem.
Ka = [H+][A-]/[HA]
HA
since [A-]i = 0 Q=0 thus the equil will shift to prod.
—>
H+
+
A-
I
1.5x10-3
0
0
C
-x
+x
+x
E
1.5x10-3 - x
x
x
Ka = [H+][A-]/[HA] = (x)(x)/(1.5x10-3 -x) = x2/1.5x10-3
x = (Ka x 1.5x10-3)0.5 = (6x10-5 x 1.5x10-3)0.5 = 3x10-4
E
1.5x10-5 - 3x10-4
so, our assumption
was not really valid,
but it is pretty close.
3x10-4
3x10-4
pH = -log[H+] = -log(3x10-4) = 3.5
12) Sketch titration curves for the following titrations.
a) Show where and if buffering is occuring on each graph.
b) Be sure to show the appropriate difference in the slope of the line for this part of the titration curve between
strong samples and weak samples.
c) Identify the equivalence point on each curve.
d) If the indicator changes colour at a pH of 7, show the relative location of the end point compared to the
equivalence point on each graph.
13) Hypochlorous acid (HClO; Ka=2.9x10-8) is a weak acid.
a) What ratio of [HClO] and [NaClO] would you use to create a buffer that would hold a pH of 7.9.
desired [H+] = 10-pH = 10-7.9 = 1.26x10-8
Ka = [H+]([A-]/[HA])
([A-]/[HA]) = Ka/[H+] = 2.9x10-8 / 1.26x10-8 = 2.3
Therefore, we need to have a ratio of ([A-]/[HA]) in our solution of 2.3
Since this ratio is between 0.1 and 10, the buffer will be effective.
If we use a solution where [HClO]=0.1M, then will need to add enough NaClO to make a concentration of 0.23M.
b) For a system like this to be an effective buffer, what are the limits on the ratio of these two ions in solution?
The ratio must be between 0.1 and 10 for the buffer to be effective.
c) What do we mean by the capacity of a buffer solution?
It is the number of moles of acid or base that can be added to the system before the ratio of the concentrations of
the acid and its conjugate are pushed outside of the range of 0.1 to 10.
For example, if the ratio [A-]/[HA] = 5 and [HA]=0.1 M, then [A-] must equal 0.5 M.
If 0.045 M of a strong base is added to the solution, it will react with the HA to make A- ions. This will increase [A] to 0.545 M and reduce [HA] to 0.055 M. This will make [A-]/[HA] = 9.9 If any more base was added to the
system, the ratio would exceed 10 and the buffer would no longer be effective. Thus, the capacity of this system is
0.045 mols of base per litre of buffer.
However, if 0.045 M of acid is added to the buffer solution, it would react with the A-, reducing [A-] to 0.455 M and
increasing [HA] to 0.145. [A-]/[HA] would be 3.14, still well within the limiting range. Since the buffer starts out
with more A- than HA, it has a much greater capacity to buffer against the addition of a strong acid than a strong
base.
d) Why are buffers important to biological systems?
Amino acids are weak acids. Thus, proteins will be altered by the pH of their environment. For enzymes and other
proteins to work effectively, they must be kept at an appropriate pH. Since many activities (blood [CO2], acidic
foods, etc.) can alter the acidity of bodily fluids, buffers must be present to counteract these effects.
14) The Ka of HOCN (Hydrogen cyanate) is 3.5x10-4
a) Calculate the pH of a 0.10 M solution of HOCN.
This is an equilibrium problem.
Ka = [H+][A-]/[HA]
HA
since [A-]i = 0 Q=0 thus the equil will shift to prod.
—>
H+
+
A-
I
0.1
0
0
C
-x
+x
+x
E
0.1 - x
x
x
Ka = [H+][A-]/[HA] = (x)(x)/(0.1 -x) = x2/0.1
x = (Ka x 0.1)0.5 = (3.4x10-4 x 0.1)0.5 = 0.0058
E
0.1-0.0058
0.0058
0.0058
pH = -log[H+] = -log(0.0058) = 2.2
The pH will be 2.2
b) Calculate the percent ionization for the acid in this solution.
%I = 100% x [H+]eq / [HA]i = 100% x 0.0058 / 0.1 = 5.8%
c) Would you expect the percent ionization to be the same at different concentrations?
No, since the [H+]eq is not linearly related to [HA]i
d) Calculate the resulting pH when 50.0 mL of 0.10 M HOCN is titrated with 30.0 mL of 0.10 M NaOH.
This is a titration problem, which means that it is an equilibrium problem.
First, we need to calculate the initial concentrations of HA and A- in the solution, after the titrant is added, but
before the HOCN equilibrium is established (a mathematical short cut that eleminates the need to solve a
quadratic equation)
Since NaOH is a strong base, it will react quantitatively with HOCN
initial moles of HOCN =
CV = 0.1 x 0.05 L = 0.005 moles
initial moles of NaOH =
CV = 0.1 x 0.03 L = 0.003 moles
total volume after mixing =
0.05 L + 0.03 L = 0.08 L
HOCN
+
NaOH
—>
OCN-
+
initial
0.005 mol
0.003 mol (limiting reactant)
0.000 mol
final
0.002 mol
0.000 mol
0.003 mol
conc.
= 0.002 mol / 0.08 L
= 0.025 M
0.000 M
= 0.003 mol / 0.08 L
= 0.0375 M
We can now set up our equilibrium calculations (Ka expr., balanced equation, and ICE table)
Ka = [H+][A-]/[HA]
HA
I
—>
0.025
H+
+
10-7
A0.0375
Q = 10-7(0.0375)/0.025 = 1.5x10-7 << Ka, thus the equilib. will shift to the prod.
C
-x
+x
+x
E
0.025 - x
10-7 + x
0.0375+x
Since Ka = 3.5x10-4, x should be much smaller than either 0.025 or 0.0375 and
much larger than 10-7, so we can simplify our equilibrium conditions to:
E
0.025
x
0.0375
Ka = [H+][OCN-]/[HOCN] = (x)(0.0375)/(0.025) = 1.5x
x = Ka/1.5= 3.5x10-4 /1.5 = 2.3x10-4
E
0.02477
2.3x10-4
0.03773
pH = -log[H+] = -log(2.3x10-4) = 3.6
The pH will be 3.6
15) The Ka of a weak acid is 4.5x10-9. The molecular acid and conjugate base have different colours.
a) Could this acid be used as an acid/base indicator?
Yes.
b) If your answer to a) is yes, at what pH would the indicator change colour?
Ka = [H+]([A-]/[HA])
If [A-]>[HA] then the A- colour predominates, and the HA colour predominates if [A-]>[HA]
Thus, the colour change occurs when [A-]/[HA] = 1, at which point Ka = [H+].
Thus, the colour change occurs when pH = pKa = -log(4.5x10-9) = 8.3
Na+ + H2O
c) Would an indicator that changed colour at a moderately basic pH be useful in the titration of:
To be an effective indicator, the end point (pH at which the colour change occurs) must be close to the equivalence
point (a small difference in volume of titrant added to reach end point (Vt) compared to equivalence point).
i) a weak base? No, these have equivalence points that are acidic in pH
ii) a weak acid? Yes, these have equivalence points that are basic in pH
iii) a strong base or strong acid? Yes, although the pH of the equivalence point will be at pH=7, the
steepness of the pH curve at the equivalence point is such that the difference in Vt to reach equivalence point rather
than end point is very small.
d) Will an acid/base indicator act as a buffer around the pH of its colour change? Yes.
e) If the answer to d) is yes, do we not need to worry about the effects of this in a titration experiment?
No, because so little indicator chemical is added to the titration system, its capacity as a buffer is too small to be
significant.