Chem 1B
Dr. Abel
1
WS 6 Acid- Base Worksheet (pH) KEY
1. Calculate the pH and pOH of a solution that is prepared by dissolving 0.35 g of NaOH in
enough water to make 500. mL of solution.
0.35 g NaOH
= 8.75043 x 10 -3 moles
39.998 g/mole
8.75043 x 10 -3 moles
= 1.75009 x 10 -2 M = [OH- ]
0.500 L
pOH = - log[OH- ] = 1.76
pH = 14.00 - pOH = 12.24
2. Make the following conversions to fill in the table:
€
pH
2.15
[H3O+]
7.1x10-3
[OH-]
1.4x10-12
11.85
7.57
2.7 x 10-8
3.7 x 10-7
6.43
pOH
3. What is the pH of the solution obtained when 125 mL of 0.606 M NaOH is diluted to 15.0 L
with water?
4. A 0.250 mol sample of a generic acid HA is dissolved in enough water to form 655 mL of
solution. If the pH of this solution is 3.44, what is the Ka value of HA?
HA (aq) + H2O (l) H3O+ (aq) + A- (aq)
[HA] initially = 0.250 mol/0.655 L = 0.38168 M
pH gives us [H3O+]eq = 10-3.44 = 3.6308 x 10-4
Concentration
Initial
Change
Equilibrium
Ka =
€
HA
0.38168
-x
0.38168 -x
H3O+
0
+x
x = 3.6308 x 10-4
[H 3O + ][A- ] = (3.6308x10-4 )(3.6308x10-4 ) = 3.5 x 10-7
[HA]
0.38080
A0
+x
3.6308 x 10-4
Chem 1B
Dr. Abel
2
check assumption!
Ignoring [H3O+] from water: 1.0 x 10-7 / 3.6 x 10-4 x 100 = 0.028%<5% OK
5. Dimethylamine, (CH3)2NH, a key intermediate in detergent manufacture, has a Kb of
5.9 x 10-4. What is the pH of 1.5 M (CH3)2NH? Make sure you start out with a balanced
equation for reaction of dissolving this base in water.
(CH3)2NH (aq) + H2O (l)
Concentration
Initial
Change
Equilibrium
(CH3)2NH2+ (aq) + OH-(aq)
(CH3)2NH
1.5
-x
1.5 – x ~ 1.5
OH0
+x
x
(CH3)2NH2+
0
+x
x
Here, 5.9 x 10-4 = x2 / 1.5
so x2 = 0.000885
thus x = 2.9749 x 10-2 = [OH-]
check assumptions!
Ignoring change in [HA]: 3.0 x 10-2 / 1.5 x 100 = 2.0%<5% OK
Ignoring [H3O+] from water: 1.0 x 10-7 / 3.6 x 10-4 x 100 = 0.028%<5% OK
From this can easily get to [H3O+] = 3.4 x 10-13 so pH = 12.47, basic as
you’d expect from a solution of a weak base. Or easier, pOH = 1.53 so pH
= 12.47.
6. A 0.10 M solution of nicotinic acid (HC6H4NO2) is prepared at 25°C. What are the
equilibrium concentrations of each substance? What is the pH?? Ka = 1.4 x 10-5.
HC 6 H 4 NO 2 (aq) + H 2 O(l) ⇔ H 3O+ (aq) + C 6 H 4 NO 2 (aq)
I
0.10M
-0.00 M
0.00 M
C
-x
-+x
+x
E 0.10 - x
-x
x
x2
x = 1.1832x10 −3 M
0.10
[HC 6 H 4 NO 2 ] = 0.10M
Ka =
[H 3O+ ] = [C 6 H 4 NO 2 ] = 1.1832x10 −3 M
pH = − log(1.1832x10 −3 M ) = 2.93
0.0012
Check assumptions :
x100 = 1.2% < 5% OK!
0.10
1.0x10 -7
x100 = 0.0083% < 5% OK!
0.0012
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Chem 1B
Dr. Abel
3
7. Consider a solution of the salt (NH4)3PO4 in water
(NH4)3PO4 → 3 NH4+ + PO4-3
a) Which ion or ions can and will affect the final pH of this solution?
NH4+ and PO4-3 will both affect pH, as both are conjugates of weak species, so both
have pH affecting properties.
b) What do you expect the pH of this solution to be, acidic, basic or neutral? Why do you
predict this? (You should be able to figure this out – write out reactions of the ions with
water, what results??)
Cite appropriate constants to support your answer and write the net ionic equation showing
the resulting reaction of the ion(s) with water and indicate how it relates to your answer.
NH4+ + H2O ⇄ H3O+ + NH3
Ka = 5.6 x 10-10
PO4-3+ H2O ⇄ OH- + HPO4-2 Kb = 2.8 x 10-2 (Ka for HPO42- given in book,
converted to Kb)
Since phosphate has the larger equilibrium constant, this reaction will
dominate, so the pH will be basic
8. Pick an appropriate cation to combine with the anion HCO3- to make a salt that will
produce a basic solution. Explain your choice.
HCO3- + H2O
H2CO3 + OHK = Kb = Kw/Ka1 = 1.0 x 10-14/4.3 x 10-7 = 2.3 x 10-8
HCO3- + H2O
CO32- + H3O+
K = Ka = Ka2 = 5.6 x 10-11
Since Kb>Ka this anion gives a basic solution when it reacts with water. Thus if it is
with a cation that does not react with water, than the solution will be basic.
NaHCO3 will work!
9. For each of the following acid-base reactions:
•First identify which is the acid and which the base
•Write the balanced chemical equation showing the reaction of the acid with the base
•Predict whether the resulting solution (i.e., the products) will be acidic, basic or
neutral.
•Explain your choice
a) methylamine (CH3NH2, Kb = 4.4 x 10-4) + HCl
HCH3NH3+ + Cl-
Cl- won’t react with water (it is the conjugate of a strong acid, so a very weak base), but
HCH3NH2+ will react with water in the following reaction:
HCH3NH2+ + H2O
CH3NH2 + H3O+ So, the solution is acidic
b) HCN (Ka = 6.2 x 10-10) + KOH
KCN + H2O
K won’t react with water (it is the conjugate of a strong base so a very weak acid), but CNwill react with water in the following reaction:
+
Chem 1B
CN- + H2O
Dr. Abel
HCN + OH-
So, the solution is basic
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