Chapter 9
Applying Your Knowledge- Even Numbered
2.
Neutralization is the chemical reaction that occurs between acids and bases. Acids
provide protons while Arrhenius bases provide OH1-(aq) and they combine to form
H2O(l).
4.
a. A substance that gives up a proton, H+, such as HCl(aq) or HNO3(aq).
b. A substance that can accept a proton like OH-. H+(aq) + OH-(aq) --> H2O
H O H
H
. It is a water molecule with
c. The hydronium ion has the formula H3O1+,
an extra proton bonded to one of the unshared pairs on the oxygen.
d. The hydroxide ion has the formula OH-. It is typically released by bases such as
potassium hydroxide, KOH.
6.
Acidic solutions at 25oC have a pH below 7. This is often expressed as a range from0 to 7.
The reality is that the pH can be less than 0 (a negative number) if [H3O1+] is greater than
1 Molar. [H3O1+] molarities > 1M are common for strong acids and for more concentrated
solutions of acid. Usually pH is not even used for these; the molarity is just stated instead.
8.
a. Vinegar, acidic
c. Aspirin, acidic
10.
a.
b.
c.
d.
e.
b. Citrus fruits, acidic
d. Black coffee, acidic
sour taste, acid
bitter taste, base
slippery feeling, base
change color of red litmus to blue, base
change color of blue litmus to red, acid
12.
The pH is a measure of the concentration of hydronium ions in mols /liter.
Mathematically the definition is, pH = - log[H3O1+].
14.
An antacid is a base used to neutralize excess hydrochloric acid, HCl(aq), in the stomach.
16.
a. The total balanced equation between hydrobromic acid and calcium hydroxide is
CaBr2(aq) + 2 H2O(l)
2 HBr(aq)
+
Ca(OH)2(aq)
Strong acid
Soluble salt
Strong base
Nonelectrolyte molecule
b. The balanced total molecular reaction between the strong acid nitric acid and the weak
base aluminum hydroxide is shown here.
3 H2O(l) + Al(NO3)3(aq)
3 HNO3(aq) + Al(OH)3(s)
18.
A solution with a pH of 2 has an hydronium ion concentration of 10-pH = 10-2 = 0.01,
while a solution with pH = 10 has an H3O1+(aq) ion concentration of 10-pH = 10-10. The
solution with pH = 2 is more acidic than the solution with pH = 10 by a factor of 108.
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20.
a.
b.
c.
d.
Cherries, pH = 3.2 is acidic
Crackers, pH = 8.5 is basic
Bananas, pH = 4.6 is acidic
Drinking water, pH = 8.0 is basic
22.
Soap solution, pH = 10, is more acidic (i.e. less basic) than household ammonia, pH = 11.
Soap solution has a pH = 10 and an [H3O1+] = 10-pH = 10-10.
Household ammonia has a pH = 11 and an [H3O1+] = 10-pH = 10-11.
The relative or comparative acidity of soap to household ammonia can be figured by
dividing the soap solution H3O1+ concentration by the household ammonia H3O1+
concentration.
[H 3O + ] = 10 -10-11moles H
Household ammonia
[H 3 O + ] 10 moles H
Soap solution
3O
1+ / liter
1+
/ liter
3O
= 10–10 -(-11 ) = 10–10+11 =10
Soap solution is 10 times more acidic than household ammonia.
24.
A mixture of NaHCO3 and H2CO3 will act a buffer because the acidic HCO31- can provide
H1+ to neutralize added base. The CO32- will react with added acid or H1+ to produce
HCO31-. This action will stabilize the H1+ concentration and pH when either acid or base
is added.
26.
Baking soda is used to neutralize acid burns because the baking soda is a weak base. It
will neutralize the acid, but not create caustic burns from any excess.
Baking soda has the formula NaHCO3. The reaction is as follows.
H2O(l) + CO2(g) + NaCl(aq)
NaHCO3(aq) + HCl(aq)
28.
The grams of HBr in a liter of 2.0 M HBr is figured using the relation;
mass solute = molarity x volume of solution x molar mass
80.9 g
2.0 mole HBr 1.0 L
g HBr =
x
x
= 162. g HBr = 160 g HBr
1
1 liter
1 mol HBr
rounded to 2 significant digits.
30.
The mols of KOH in 125 mL of 0.500 M KOH is figured using the relation;
mols of solute = molarity x volume of solution
0.500 mole KOH
125 mL 1 liter
x
x
= 0.0625 mols KOH
mols of KOH =
1 liter
1000 mL
1
rounded to 3 significant digits.
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32.
The calculation can be done in one set up:
1.5 mole NaCl
1.0 L 58.44 g NaCl
×
×
= 87.66 g NaCl = 88. g NaCl
g NaCl =
1 liter
1
1 mole
The 87.66 g NaCl is rounded off to 88. g because the 1.5 M NaCl and the 1.0 L have two
significant digits each.
34.
The pH for a solution is determined from the definition pH = - log[H3O1+]. A simple
graphic relationship between the hydronium ion concentration and pH is shown here.
The top row gives the concentration [H3O1+] and the second line gives the matching pH.
This is useful only for [H3O1+] = 1 x 10whole number.
[H3O1+] 100 10-1 10-2 10-3 10-4 10-5 10-6 10-7 10-8 10-9 10-10 10-11 10-12 10-13 10-14
pH
0 1
2
3
4
5
6
7
8
9 10
11 12 13 14
a. The pH for a 1 x 10-2 M HCl solution can be figured two ways.
pH = -log[H3O1+]; pH = -log 1 x 10-2 ; pH = 2
The graphical relationship shown above can be used to get the same result.
b. The pH for a 1 x 10-3 M HNO3 solution can be figured two ways.
The [H3O1+] = 0.001; pH = -log [H3O1+]; pH = -log 1 x 10-3 ; pH = 3
The graphical relationship shown above can be used to get the same result.
36.
The molarity of the hydronium ion = [H3O1+]= 10-pH
a. pH = 1 ; [H3O1+] = 10-pH; [H3O1+] = 10-1 = 1 x 10-1 or 0.1
b. pH = 0 ; [H3O1+] = 10-pH; [H3O1+] = 10-0 = 1 x 10-0 or 1
c. pH = 5 ; [H3O1+] = 10-pH;
d. pH = 3 ; [H3O1+] = 10-pH;
38.
[H3O1+] = 10-5 = 1 x 10-5 or 0.00001
[H3O1+] = 10-3 = 1 x 10-3 or 0.001
HBr(aq) + LiOH(aq) Æ H2O + LiBr(aq)
The mol ratio is: 1 mol HBr = 1 mol LiOH
For a neutral solution, the mols of LiOH and mols of HBr must be equal.
The set up to figure the volume in milliliters is
1 mol HBr
0.045 mols LiOH
1L
1000 mL
x
x
x
= mL HBr solution = 90. mL
1 mol LiOH 0.50 mol HBr
1
1L
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40.
When acidic and basic solutions are mixed, a neutralization reaction occurs. First write
the balanced equation for the neutralization. HI(aq) + NaOH(aq)Æ H2O + NaI(aq)
Find the number of mols of solute in each given amount of solution.
The mols of NaOH in 100 mL of 1.5 M NaOH is figured using the relation;
mols of base = molarity base x volume of solution
1.5 mole NaOH
100 mL
1 liter
x
x
mols of NaOH =
= 0.15 mols NaOH
1 liter
1
1000 mL
rounded to 2 significant digits.
mols of acid = molarity of acid x volume of solution
0.50 mole HI 31 mL
1 liter
x
x
mols of HI =
= 0.016 mols HI
1 liter
1
1000 mL
rounded to 2 significant digits.
The #mol of the base, NaOH, exceeds the #mols of acid, HI. Since the mol ratio in the
balanced equation is 1:1, 0.016 mol HI will react with 0.016 mol NaOH. There will be 0.13
mols NaOH left over and the final solution will be basic.
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