10.E: Acid-Base Equilibria - Homework
Brønsted-Lowry Acids and Bases
-
Show by suitable net ionic equations that each of the following species can act as a Brønsted-Lowry acid:
- H 3 O +
- HCl
- NH 3
- CH 3 CO 2 H
- Answer
-
-
\( \mathrm{H}_3 \mathrm{O}^{+}(a q) \longrightarrow \mathrm{H}^{+}(a q)+\mathrm{H}_2 \mathrm{O}(l) \)
-
\( \mathrm{HCl}(\mathrm{aq}) \longrightarrow \mathrm{H}^{+}(a q)+\mathrm{Cl}^{-}(a q) \)
-
\( \mathrm{NH}_3(a q) \longrightarrow \mathrm{H}^{+}(a q)+\mathrm{NH}_2^{-}(a q) \)
-
\( \mathrm{CH}_3 \mathrm{CO}_2 \mathrm{H}(a q) \longrightarrow \mathrm{H}^{+}(a q)+\mathrm{CH}_3 \mathrm{CO}_2^{-}(a q) \)
-
\( \mathrm{NH}_4{ }^{+}(a q) \longrightarrow \mathrm{H}^{+}(a q)+\mathrm{NH}_3(a q) \)
-
\( \mathrm{HSO}_4^{-}(a q) \longrightarrow \mathrm{H}^{+}(a q)+\mathrm{SO}_4^{2-}(a q) \)
-
-
Show by suitable net ionic equations that each of the following species can act as a Brønsted-Lowry base:
- (a) H 2 O
- (b) OH −
- (c) NH 3
- (d) CN −
- Answer
-
-
\( \mathrm{H}_2 \mathrm{O}(l)+\mathrm{H}^{+}(a q) \longrightarrow \mathrm{H}_3 \mathrm{O}^{+}(a q) \)
-
\( \mathrm{OH}^{-}(a q)+\mathrm{H}^{+}(a q) \longrightarrow \mathrm{H}_2 \mathrm{O}(l) \)
-
\( \mathrm{NH}_3(a q)+\mathrm{H}^{+}(a q) \longrightarrow \mathrm{NH}_4^{+}(a q) \)
-
\( \mathrm{CN}^{-}(a q)+\mathrm{H}^{+}(a q) \longrightarrow \mathrm{HCN}(a q) \)
-
-
What is the conjugate acid of each of the following? What is the conjugate base of each?
- OH −
- H 2 O
- HCO 3 -
- NH 3
- Answer
-
- H 2 O
- H 3 O +
- H 2 CO 3
- NH 4 +
-
Identify and label the Brønsted-Lowry acid, its conjugate base, the Brønsted-Lowry base, and its conjugate acid in each of the following equations:
- \( \mathrm{HNO}_3+\mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{H}_3 \mathrm{O}++\mathrm{NO}_3^{-} \)
- \( \mathrm{H}_2 \mathrm{SO}_4+\mathrm{Cl}^{-} \longrightarrow \mathrm{HCl}+\mathrm{HSO}_4^{-}\)
- \( \mathrm{HSO}_4{ }^{-}+\mathrm{OH}^{-} \longrightarrow \mathrm{SO}_4{ }^2-+ \mathrm{H}_2 \mathrm{O}\)
- \( \mathrm{O}^{2-}+\mathrm{H}_2 \mathrm{O} \longrightarrow 2 \mathrm{OH}- \)
- Answer
-
- HNO 3 (BA), H 2 O(BB), H 3 O + (CA), NO 3 - (CB)
- H 2 SO 4 (BA), Cl − (BB), HCl(CA), HSO 4 - (CB)
- HSO 4 - (BA), OH − (BB), SO 4 2 - (CB), H 2 O(CA)
- O 2− (BB), H 2 O(BA) OH − (CB and CA)
-
State which of the following species are amphiprotic and write chemical equations illustrating the amphiprotic character of these species.
- NH 3
- Br −
- Answer
-
- Acid: \(\mathrm{HPO}_4^{2-}+\mathrm{OH}^{-} \longrightarrow \mathrm{PO}_4^{3-}+\mathrm{H}_2 \mathrm{O} \) Base: \( \mathrm{H}_2 \mathrm{O}(a q)+\mathrm{NH}_3(a q) \rightleftharpoons \mathrm{NH}_4^{+}(a q)+\mathrm{OH}^{-}(a q) \)
- Acid: \(\mathrm{HPO}_4{ }^{2-}+\mathrm{OH}^{-} \longrightarrow \mathrm{PO}_4{ }^{3-}+\mathrm{H}_2 \mathrm{O} \) Base: \( \mathrm{HPO}_4{ }^{2-}+\mathrm{HClO}_4 \longrightarrow \mathrm{H}_2 \mathrm{PO}_4^{-}+\mathrm{ClO}_4 \)
- not amphiprotic
pH and pOH
-
Calculate the pH and the pOH of each of the following solutions at 25 °C for which the substances ionize completely:
- (a) 0.000259 M HClO 4
- (b) 0.21 M NaOH
- (c) 0.000071 M Ba(OH) 2
- (d) 2.5 M KOH
- Answer
-
- pH = 3.587; pOH = 10.413
- pOH = 0.68; pH = 13.32;
- pOH = 3.85; pH = 10.15;
- pOH = −0.40; pH = 14.4
-
Calculate the hydronium ion concentration and the hydroxide ion concentration in lime juice from its pH. See
Figure 14.2
for useful information.
- Answer
- \( \left[\mathrm{H}_3 \mathrm{O}^{+}\right]=1 \times 10^{-2} \mathrm{M} ;\left[\mathrm{OH}^{-}\right]=1 \times 10^{-12} \mathrm{M} \)
-
The hydroxide ion concentration in household ammonia is 3.2
10
−3
M
at 25 °C. What is the concentration of hydronium ions in the solution?
- Answer
- \( \left[\mathrm{OH}^{-}\right]=3.1 \times 10^{-12} \mathrm{M} \)
Relative Strengths of Acids and Bases
-
.The odor of vinegar is due to the presence of acetic acid, CH
3
CO
2
H, a weak acid. List, in order of descending concentration, all of the ionic and molecular species present in a 1-
M
aqueous solution of this acid.
- Answer
- \( \left[\mathrm{H}_2 \mathrm{O}\right]>\left[\mathrm{CH}_3 \mathrm{CO}_2 \mathrm{H}\right]>\left[\mathrm{H}_3 \mathrm{O}^{+}\right] \approx\left[\mathrm{CH}_3 \mathrm{CO}_2^{-}\right]>\left[\mathrm{OH}^{-}\right] \)
-
Predict which acid in each of the following pairs is the stronger and explain your reasoning for each.
- H 2 O or HF
- HSO 3 - or HSO 4 -
- NH 3 or H 2 O
- NH 3 or PH 3
- Answer
-
- HF; Increasing acid strength to the right on the periodic table
- HSO 4 - ; increasing oxidation number of central atom
- H 2 O; NH 3 is a base, increasing acid strength to the right on the periodic table
- PH 3 ; acidity increases as you go down on the periodic table
-
Rank the compounds in each of the following groups in order of increasing acidity and explain the order you assign.
- HOCl, HOBr, HOI
- HOCl, HOClO, HOClO 2 , HOClO 3
- Answer
-
- HOI < HOBr < HOCl; in a series of the same form of oxyacids, the acidity increases as the electronegativity of the central atom increases. Cl is more electronegative than Br, and I is the least electronegative of the three.
- HOCl < HOClO < HOClO 2 < HOClO 3 ; in a series of oxyacids of the same central element, the acidity increases as the number of oxygen atoms increases (or as the oxidation state of the central atom increases).
-
The active ingredient formed by aspirin in the body is salicylic acid, C
6
H
4
OH(CO
2
H). The carboxyl group (−CO
2
H) acts as a weak acid. The phenol group (an OH group bonded to an aromatic ring) also acts as an acid but a much weaker acid. List, in order of descending concentration, all of the ionic and molecular species present in a 0.001-
M
aqueous solution of C
6
H
4
OH(CO
2
H).
- Answer
- \( \left[\mathrm{H}_2 \mathrm{O}\right]>\left[\mathrm{C}_6 \mathrm{H}_4 \mathrm{OH}\left(\mathrm{CO}_2 \mathrm{H}\right)\right]>\left[\mathrm{H}^{+}\right] 0>\left[\mathrm{C}_6 \mathrm{H}_4 \mathrm{OH}\left(\mathrm{CO}_2\right)^{-}\right] \gg\left[\mathrm{C}_6 \mathrm{H}_4 \mathrm{O}\left(\mathrm{CO}_2 \mathrm{H}\right)^{-}\right]>\left[\mathrm{OH}^{-}\right] \)
-
Why is the hydronium ion concentration in a solution that is 0.10
M
in HCl and 0.10
M
in HCOOH determined by the concentration of HCl?
- Answer
- This is a case in which the solution contains a mixture of acids of different ionization strengths. In solution, the HCO 2 H exists primarily as HCO 2 H molecules because the ionization of the weak acid is suppressed by the strong acid. Therefore, the HCO 2 H contributes a negligible amount of hydronium ions to the solution. The stronger acid, HCl, is the dominant producer of hydronium ions because it is completely ionized. In such a solution, the stronger acid determines the concentration of hydronium ions, and the ionization of the weaker acid is fixed by the [H 3 O + ] produced by the stronger acid.
-
From the equilibrium concentrations given, calculate
K
a
for each of the weak acids and
K
b
for each of the weak bases.
- ClO − : [OH − ] = 4.0 10 −4 M ; [HClO] = 2.38 10 −4 M ; [ClO − ] = 0.273 M ;
- HNO 2 : = 0.011 M ; = 0.0438 M ; [HNO 2 ] = 1.07 M ;
- = 0.100 M ; [NH 3 ] = 7.5 10 −6 M ; [H 3 O + ] = 7.5 10 −6 M
- Answer
-
- K b = 3.49x10 -7
- K a = 4.50 x 10 -4
- K a = 5.62 x 10 -10
-
Determine
K
a
for hydrogen sulfate ion,
In a 0.10-
M
solution the acid is 29% ionized.
- Answer
- \( K_{\mathrm{a}}=1.2 \times 10^{-2} \)
-
Calculate the concentration of all solute species in each of the following solutions of acids or bases. Assume that the ionization of water can be neglected, and show that the change in the initial concentrations can be neglected.
- 0.0092 M HClO, a weak acid
- Answer
-
-
\( \begin{aligned}
& {\left[\mathrm{H}_3 \mathrm{O}^{+}\right]=\left[\mathrm{ClO}^{-}\right]=1.6 \times 10^{-5} \mathrm{M}} \\
& {\left[\mathrm{HClO}^{-}\right]=0.0092 \mathrm{M}} \\
& {\left[\mathrm{OH}^{-}\right]=6.1 \times 10^{-10} \mathrm{M} ;}
\end{aligned} \)
-
\( \begin{aligned}
-
The pH of a 0.15-
M
solution of
is 1.43. Determine
K
a
for
from these data.
- Answer
- \( K_{\mathrm{a}}=1.2 \times 10^{-2} \)
-
Determine whether aqueous solutions of the following salts are acidic, basic, or neutral:
- K 2 CO 3
- NH 4 Br
- KClO 4
- Answer
-
- basic
- acidic
- neutral
Buffers
-
Explain why the pH does not change significantly when a small amount of an acid or a base is added to a solution that contains equal amounts of the base NH
3
and a salt of its conjugate acid NH
4
Cl.
- Answer
- If you add a small amount of base it reacts with the NH 4 + ions in solution to make water and NH 3 - so the pH does not change. If you add a small amount of acid it reacts with the NH 3 in the solution to make NH 4 + ions and water - so again the pH does not change.
-
What is the effect on the concentration of acetic acid, hydronium ion, and acetate ion when the following are added to an acidic buffer solution of equal concentrations of acetic acid and sodium acetate:
- HCl
- NaCl
- KOH
- CH 3 CO 2 H
- Answer
-
- The added HCl will increase the concentration of H 3 O + slightly, which will react with CH 3 CO 2 - and produce CH 3 CO 2 H in the process. Thus, [CH 3 CO 2 - ] decreases and [CH 3 CO 2 H] increases.
- The added NaCl will have no effect on the concentration of the ions.
- The added KOH will produce OH − ions, which will react with the H 3 O + , thus reducing [H 3 O + ]. Some additional CH 3 CO 2 H will dissociate, producing CH 3 CO 2 - ions in the process. Thus, [CH 3 CO 2 H] decreases slightly and [CH 3 CO 2 - ] increases.
- The added CH 3 CO 2 H will increase its concentration, causing more of it to dissociate and producing more CH 3 CO 2 - and H 3 O + in the process. Thus, [H 3 O + ] increases slightly and [CH 3 CO 2 - ] increases.
-
What is the effect on the concentration of ammonia, hydroxide ion, and ammonium ion when the following are added to a basic buffer solution of equal concentrations of ammonia and ammonium nitrate:
- KI
- NH 3
- HI
- NaOH
- NH 4 Cl
- Answer
-
- no change in the equilibrium, since neither K+ or I- ions are in the equlibrium expression
- Adding NH 3 will shift the equilibrium to the right, reducing the added NH 3 and increasing the NH 4 + and OH - concentrations.
- Adding HI will increase the H 3 O + concentration, which will react with the NH 3 . NH 3 will go down, NH 4 + and OH - will go up.
- Adding NaOH will increase the OH- concentration. The equilibrium will respond with NH 3 going up, NH 4 + and OH - will go down.
- Adding NH 4 Cl will increase the NH 4 + concentration. The equilibrium will respond with NH 3 going up, NH 4 + and OH - will go down.
-
A buffer solution is prepared from equal volumes of 0.200
M
acetic acid and 0.600
M
sodium acetate. Use 1.80
10
−5
as
K
a
for acetic acid.
- What is the pH of the solution?
- Is the solution acidic or basic?
- What is the pH of a solution that results when 3.00 mL of 0.034 M HCl is added to 0.200 L of the original buffer?
- Answer
-
- pH = 5.222
- The solution is acidic.
- pH = 5.220