Chapter 15 Chapter Review Answers
1. Water is a very weak electric conductor, because when water self-ionizes, very few
ions are produced. Few ions mean little electricity can be carried.
2. The brackets represent the chemical is dissolved in aqueous solution and its
concentration in moles per liter.
3. Water pH
a. [H3O+] of water at 25 °C = 1 × 10−7 M
b. It is not true at other temperatures, because as water warms, its ionization
increases causing a slight increase in [H+].
4. More pH & Concentrations
a. The [H+] is always greater than the [OH−].
b. The [H3O+] of acidic solutions is greater than 1 × 10−7 M. (pH less than 7)
5. pH
a. pH is the negative logarithm of the hydrogen ion (hydronium ion)
concentration of the solution.
b. pH = −log [H3O+]
or −log [H+]
c. The common log of a number is the value that when an exponent of 10 will
equal the number. Ex. The common log of 1,000 is 3, because 103 = 1,000.
6. Acidic, Basic, or Neutral?
a. Neutral
b. Basic
c. Neutral
d. Acidic
e. Neutral
f. Acidic
g. Basic
7. g, b, c, e, f, d, a, i, h.
8. Calculate concentrations:
a. 0.030 M HCl
[H+] = 0.030 M
[OH−] = 3.3 × 10−13 M
b. 1.0 × 10−4 M NaOH
[H+] = 1.0 × 10−10 M
[OH−] = 1.0 × 10−4 M
−3
+
−3
c. 5.0 × 10 M HNO3
[H ] = 5.0 × 10 M
[OH−] = 2.0 × 10−12 M
d. 0.010 M Ca(OH)2
[H+] = 5.0 × 10−13 M
[OH−] = 0.020 M
9. Find pH from [H+]
a. 2.0
b. 3.0
c. 5.0
d. 4.0
10. Find pH again, but from [OH−]
a. 8.0
b. 5.0
c. 12
d. 7.0
11. Find pH again from [OH−]
a. 12
b. 11
c. 10
12. Find pH again from [H+]
a. 4.7
b. 6.3
c. 2.4
13. Find [H+] from pH
a. 1.0 × 10−3 M
b. 1.00 × 10−7 M
c. 1.00 × 10−11 M
d. 1.0 × 10−5 M
14. Find [OH−] from pH
a. 1.00 × 10−7 M
b. 1.000 × 10−3 M
c. 1.00 × 10−10 M
d. 1.00 × 10−8 M
15. Find [H+] from pH again
a. 5.89 × 10−5 M
b. 2.24 × 10−8 M
c. 3.31 × 10−10 M
16. Nitric Acid Solution with pH = 2.70
a. [H3O+] = 0.00200 M
or
2.00 × 10−3 M
−
−12
b. [OH ] = 5.01 × 10
M
c. 0.011 moles HNO3 in 5.5 L
d. 0.69 g HNO3
e. 0.70 mL of the concentrated HNO3
17. The transition interval of an indicator is the pH range over which an indicator
changes color.
18. A pH indicator is a weak acid or weak base that has a particular color when in
ionized form and a different color when non-ionized. Increasing the H+
concentration forces the indicator to show its color when acidic. Increasing the OH−
concentration forces the indicator to show its color when alkaline.
19. Equivalence & End Points
a. You can determine the equivalence point of a titration experiment or the pH
of the solution without an indicator by using a pH meter.
b. As the solution approaches the end point in a titration, the pH change
increases dramatically.
20. More Titration Information:
a. The end point is the moment in a titration when the indicator changes color.
b. An indicator is used to show when the equivalence point has been reached
because at that point the indicator will change color.
c. The color transition of the indicator will occur in the range of pH for the
equivalence point of the acid-base reaction.
21. The 4 possible acid-base reactions:
a. Strong Acid—Strong Base:
i. Approximate pH at the end point: 7.0
ii. Suitable indicator for the reaction: Bromthymol Blue, Methyl Red, &
Phenolphthalein
b. Strong Acid—Weak Base:
i. Approximate pH at the end point: Below 7.0
ii. Suitable indicator for the reaction: Bromphenol Blue & Methyl Orange
c. Weak Acid—Strong Base:
i. Approximate pH at the end point: Above 7.0
ii. Suitable indicator for the reaction: Phenolphthalein & Phenol Red
d. Weak Acid—Weak Base:
i. Approximate pH at the end point: Either Above 7.0 or Below 7.0
ii. Suitable indicator for the reaction: No Satisfactory indicator for any
reaction, must use one suitable for the exact reaction.
22. Sketch of a graph of a Strong-Acid and Weak-Base Titration:
12
9
pH
6
Equivalence Point is at a pH less than 7.0
3
0
0
25
50
75
100
Volume of Base added (mL)
23. The unknown solution has a pH of approximately 8. Phenolphthalein is colorless
from pH 0 to 8, and Phenol Red is Red from pH just under 8 to 14
24. Mole equivalents for Neutralization:
a. 1.0 mol NaOH will neutralize 1.0 mol HCl
b. 0.75 mol HNO3 will neutralize 0.75 mol KOH
c. 0.10 mol Ba(OH)2 will neutralize 0.20 mol HF
d. 0.90 mol H2SO4 will neutralize 0.90 mol Mg(OH)2
25. Balanced Equation: 2 KOH + H2SO4  2 H2O + K2SO4
15.0 𝑚𝐿
1𝐿
0.0250 𝑚𝑜𝑙 𝐻2 𝑆𝑂4
2 𝑚𝑜𝑙 𝐾𝑂𝐻
1
1000 𝑚𝐿
×
×
×
×
×
= 0.0750 M 𝐾𝑂𝐻
1
1000 𝑚𝐿
1𝐿
1 𝑚𝑜𝑙 𝐻2 𝑆𝑂4
10.0 𝑚𝐿
1𝐿
26. Balanced Equation:
Ba(OH)2
+
2 HNO3

2 H2O
+
Ba(NO3)2
12.5 𝑚𝐿
1𝐿
0.0175 𝑚𝑜𝑙 𝐵𝑎(𝑂𝐻)2
2 𝑚𝑜𝑙 𝐻𝑁𝑂3
1
1000 𝑚𝐿
×
×
×
×
×
= 0.0302 M 𝐻𝑁𝑂3
1
1000 𝑚𝐿
1𝐿
1 𝑚𝑜𝑙 𝐵𝑎(𝑂𝐻)2
14.5 𝑚𝐿
1𝐿
27. Finding Concentrations:
a. The [OH−] for a 4.0 × 10−4 M solution of Ca(OH)2 would be 8.0 × 10−4 M,
because every Ca(OH)2 will produce 2 OH− ions in solution.
b. The [H+] would be 1.2 × 10−11 M.
28. Find pH from the solution concentrations:
a. 7.0
b. 3.0
c. 12
d. 5.0
29. A solution that has a pH of 6.0 will have a [H3O+] of 1.0 × 10−6 M.
30. A 5.0 × 10−5 M solution of Ba(OH)2 will have a pH of 10.
31. Calculate pH & Concentration:
a. pH = 10.
b. [H3O+] = 0.00316 M
32. Calculate Concentrations:
a. [OH−] = 1.08 × 10−4 M
b. [H+] = 9.26 × 10−11 M
33. Calculate More Concentrations:
a. [H3O+] = 1.26 × 10−9 M
b. [OH−] = 7.94 ×10−10 M
34. pH = 4.8
35. 0.011 M HCl
36. 9.0 × 10−4 M Ca(OH)2
37. 0.030 M HNO3
38. The acid chemical is diprotic as there are two slope changes on the graph. The
acidic solution is from a weak acid, as the pH originates near 2, and the first
hydrogen removed at a pH less than 7.0, and the second hydrogen at a pH of near
9.0. The titration curve ends at a pH of 13, which is the pH of a 0.10 M NaOH
solution.
39. pH of Blood
a. CO2 & HCO3− are the chemicals that control [H+] in the blood.
b. Excess CO2 in the blood results in acidosis, which triggers increased
breathing to remove the CO2 from the blood.
c. Hyperventilation is rapid breathing for an extended period of time. This
causes a drop of CO2 in the blood and the pH becomes too high as H+ ions
are released by HCO3− ions to compensate for the loss of CO2. This is why
people that hyperventilate are told to breath in and out into a paper bag.
This forces more CO2 to remain in the blood and maintain the proper blood
pH.