Chemistry 160
Suggested Problems Tuesday 8 Jan-Monday 14 Jan
If you are reading this before Monday night, please check back at that point. I will have the final
suggested problems list up by 7pm Monday night – UNLESS THERE IS A POWER OUTAGE…
including material covered in lecture that day.
15.4 If you mix equal concentrations of reactants and products, which os the following reactions proceed
to the right and which proceed to the left?
a.
b.
HF(aq) + NO3-(aq) HNO3(aq) + F-(aq)
NH4+(aq) + CO32-(aq) HCO3-(aq) + NH3(aq)
15.8 Calculate the pH of each of the following solutions:
a. A sample of seawater that has an OH- concentration of 1.58 * 10-6 M
b. A sample of acid rain that has an H3O+ concentration of 6.0 * 10-5 M
15.10 Calculate the pH of
a. 0.050 M HClO4
c. 4.0 M KOH
15.11 Calculate the pH of a solution prepared by dissolving 0.25 g of BaO in enough water to
make 0.500 L of solution.
15.14 Acetic acid (CH3CO2H) is the solute that gives vinegar its characteristic odor and sour
taste. Calculate the pH and the concentrations of all species present (H3O+, CH3CO2-, CH3CO2H,
and OH-) in:
a. 1.00 M CH3CO2H
b. 0.0100 M CH3CO2H
15.45 For each of the following reactions, identify the Bronsted-Lowry acids and base and the
conjugate acid-base pairs.
a. CN-(aq) + H2O(l)   OH-(aq) + HCN(aq)
b. H2PO4-(aq) + H2O(l)   H3O+(aq) + HPO42-(aq)
c. HPO42- (aq) + H2O(l)   OH-(aq) + H2PO4-(aq)
d. NH4+(aq) + NO2-(aq)   HNO2(aq) + NH3(aq)
15.49 If you mix equal concentrations of reactants and products, which of the following reactions
proceed to the right and which proceed to the left? Use the data in Table 15.1.
a. HSO4-(aq) + CH3CO2-(aq)   CH3CO2H(aq) + SO42=(aq)
b. HNO2(aq) + NO3-(aq)   HNO3(aq) + NO2-(aq)
c. HCO3-(aq) + F-(aq)   HF(aq) + CO32-(aq)
d. NH4+(aq) + CN-(aq)   HCN(aq) + NH3(aq)
15.51 For each of the following solutions, calculate [OH-] from [H3O+], or [H3O+] from [OH-].
Classify the solution as acidic, basic, or neutral.
a. [H3O+] = 2.5 * 10-4 M
b. [H3O+] = 2.0 M
c. [OH-] = 5.6 * 10-9 M
15.53 What is the pH to the correct number of significant figures for solutions with the following
concentrations of H3O+ or OH-?
a. [OH-] = 7.6 * 10-3 M
e. [H3O+] = 2.18 * 10-10 M
15.55 What is the H3O+ concentration to the correct number of significant figures for solutions
with the following pH values?
a. 9.0
b. 7.00
e. 2.63
15.107 Given the following pH values for some common foods, calculate [H3O+] and [OH-].
a. Dill pickles, 3.2
b. Eggs, 7.8
c. Apples, 3.1
d. Milk, 6.4
15.112 At 0oC, the density of liquid water is 0.9998 g/mL and the value of Kw is 1.14 * 10-15.
What fraction of the molecules in liquid water are dissociated at 0oC? What is the percent
dissociation at 0oC? What is the pH of a neutral solution at 0oC?
15.4 In each case, consider which is the stronger acid – that will react/dissociate more, and will push the
reaction equilibrium to the other side. The stronger acid will not want to hold onto its proton, the HA
form will not exist, and therefore, the conjugate base of the acid will be more present.
a.
HNO3- is stronger than HF. Therefore, the equilibrium will lie to the left,
because HNO3- will be dissociated.
b.
NH4+ is a stronger acid than NH3. Therefore, the equilibrium will lie to the
right.
15.8 a. First calculate H3O+ concentration
[H3O+] = Kw/[OH-] = 1.0 * 10-14/1.58 * 10-6 = 6.3*10-9
Then calculate pH = -log[H+] = -log(6.3 * 10-9) = 8.20
b. Already know [H3O+] = 6.0 * 10-5 M
pH = -log (6.0 * 10-5) = 4.22
15.10 Assume both are strong acids/strong bases considering it’s in that section of the book…
therefore, both will fully dissociate.
a. [H3O+] = initial acid concentration = 0.050 M
pH = -log(0.050) = 1.30
c. [OH=] = 4.0 M
[H3O+] = Kw/[OH-] = 1.0 * 10-14/4.0 = 2.5 * 10-15 M
pH = -log(2.5 * 10-15) = 12.30
15.11 First, determine the molarity of the solution. BaO molecular weight = 153.33 amu.
As in Worked example 15.8 in the book, we’ll need to consider the reaction of BaO with
water to determine how much OH- is present in the solution.
BaO(s) + H2O(l)   Ba(OH)2(aq)
So, for every mol of Ba(OH)2 produced, two mol OH- are made.
0.25 g BaO (1 mol BaO/153.33 g BaO)(1 mol Ba(OH)2/1 mol BaO)(2 mol OH-/1 mol
Ba(OH)2) = 3.26 * 10-3 mol OHThis is dissolved in 0.500 L, so molarity =
3.26 * 10-3 mol/0.500 L = 6.52 * 10-3 M OH[H3O+] = Kw/[OH-] = 1 * 10-14/6.52 * 10-3 = 1.53 * 10-12 M
pH = 11.81
15.14 a. species present initially: CH3CO2H (acid) and H2O (acid or base)
Of the two potential reactions, acetic acid + water has the higher Ka value:
CH3CO2H + H2O   CH3CO2- + H3O+ Ka = 1.8 * 10-5
initial(M)
change(M)
equil(M)
CH3CO2H + H2O   CH3CO2- + H3O+
1.00
0
~0
-x
+x
+x
1.00 - x
x
x
Ka = [CH3CO2-][H3O+]/[CH3CO2H] = x2/1.00 – x ~ x2/1.00
x = 4.2 * 10=3 ** Since x is less than 5% of our original concentration, x is negligible
enough to not include in the above calculation.
[H3O+] = [CH3CO2-] = x = 4.2 * 10-3
[CH3CO2H] = 1.00 – x = 1.00 – 4.2 * 10-3 = 1.00 M
[OH-] = Kw/[H3O+] = 1.0 * 10-14/4.2 * 10-3 = 2.4 * 10-12 M
b. Same species initially present, with the same principal reaction.
initial(M)
change(M)
equil.(M)
CH3CO2H + H2O   CH3CO2- + H3O+
0.0100
0
~0
-x
+x
+x
0.0100 – x
x
x
Ka = x2/0.0100 – x ~ x2/0.0100
x = 4.24 * 10-4
This is 4.2% of the original concentration… which is dangerously close to the 5%
boundary. So why don’t I use this as an opportunity to practice the quadratic equation
setup?
Place in quadratic equation format, starting from
1.8 * 10-5 = x2/0.0100 – x
1.8 * 10-7 – 1.8 * 10-5x = x2
2
-5
-7
x + 1.8 * 10 x – 1.8 * 10 = 0
x = -(1.8 * 10-5) +- [(1.8 * 10-5)2 – 4(-1.8 * 10-7)]1/2 / 2(1)
= (-1.8 * 10-5) +- (8.5 * 10-4) / 2
The two possible solutions for x are 4.2 * 10-4 and -4.3 * 10-4, but you can’t have a
negative concentration… so the answer would have to be 4.2 * 10-4.
[H3O+] = [CH3CO2-] = x = 4.2 * 10-4
[CH3CO2H] = 0.0100 – x = 0.0096 M
[OH-] = Kw/[H3O+] = 2.4 * 10-11 M
15.45 a. CN- is base, HCN is the conjugate acid
H2O is the acid, OH- is the conjugate base
b. H2PO4 is the acid, HPO42- is the conjugate base
H2O is the base, H3O+ is the conjugate acid
c. HPO42- is the base, H2PO4 is the conjugate acid
H2O is the acid, OH- is the conjugate base
d. NH4+ is the acid, NH3 is the conjugate base
NO2- is the base, HNO2 is the conjugate acid
15.49 a. HSO4- is a stronger acid than CH3CO2H
Therefore, the reaction will proceed to the right
b. HNO3 is stronger than HNO2
Therefore, the reaction will proceed to the left
c. HF is stronger than HCO3Therefore, the reaction will proceed to the left
d. NH4+ is stronger than HCN
Therefore, the reaction will proceed to the right
15.51 Assuming 25oC…
a. [OH-] = Kw/[H3O+]
= 1 * 10-14/2.5 * 10-4
= 4 * 10-11 = [OH-]
[H3O+] > [OH-] therefore acidic
b. [OH-] = 1 * 10-14/2.0
=5 * 10-15
Solution is very acidic!
c. [H3O+] = Kw/[OH-]
=1 * 10-14/5.6 * 10-9
= 1.8 * 10-6 = [H3O=]
Since [H3O+] > [OH-], this solution is also acidic
15.53 a. First convert to [H3O+}
= Kw/[OH-] = 1 * 10-14/7.6 * 10-3
= 1.32 * 10-12
pH = -log[H3O+] = -log (1.32 * 10-12) = 11.88
**Sig figs: the original [OH-] had two sig figs. In a logarithm, the sig figs are only
those numbers to the right of the decimal point; all numbers to the left are related to the
exponent, so they aren’t counted as sig figs. That is why the number is reported as it is.
e. This one doesn’t have to be converted first.
pH = -log(2.18 * 10-10) = 9.662
Again, three sig figs to the right of the decimal place to account for the three sig figs in
the initial concentration.
15.55 a. pH = -log[H3O+]
[H3O+] = 10-pH = 10-9.0 = 1 * 10-9 M
b. [H3O+] = 10-pH = 10-7.00 = 1.0 * 10-7 M
e. [H3O+] = 10-pH = 10-2.63 = 2.3 * 10-3 M
15.107 a. pH = 3.2 [H3O+] = 10-pH = 10-3.2 = 6 * 10-4 M
[OH-] = Kw/[H3O+] = 1 * 10-14/6 * 10-4 = 2 * 10-11
b. [H3O+] = 10-7.8 = 2 * 10-8
[OH-] = 1 * 10-14/2 * 10-8 = 5 * 10-7
c. [H3O+] = 10-3.1 = 8 * 10-4
[OH-] = 1 * 10-14/8 * 10-4 = 1 * 10-11
d. [H3O+] = 10-6.4 = 4 * 10-7
[OH-] = 3 * 10-8
Interesting to see which of our foods are acidic/basic… or not!
15.112 Given density at 0oC, what is the concentration of H2O in terms of molarity?
Density = 0.9998 g/mL  convert this into mol/L (molarity)
[H2O] = [0.9998 g (1 mol/18.02 g)/0.001 L] = 55.48 M
Then, we will calculate the concentration of [H3O+] at the same temperature.
Kw = [H3O+][OH-] Assuming a neutral solution, [H3O+] = [OH-]
so you can say [H3O+] = (Kw)1/2
= (1.14 * 10-15)1/2
= 3.38 * 10-8 M
And then calculate pH from here… pH = -log(3.38 * 10-8) = 7.472
You can determine the amount of dissociation by comparing the [H3O+ ] (which is
dissociated H2O) to the initial [H2O].
= 3.38 * 10-8 M / 55.48 M = 6.09 * 10-10 M
This is the fraction dissociated. Take this number, and multiply by 100% to determine
percent dissociated.
6.09 * 10-10 M * 100% = 6.09 * 10-8 % dissociated. In other words… not very
dissociated, just like water at 25oC.
I’ll write up some more weak acid/weak base equilibrium problems before the first exam – too
time consuming, and won’t get it done before the end of tonight! At least you have a few to
practice with.