6.E: Acid-Base Equilibrium (Exercises)
6.1: Brønsted-Lowry Acids and Bases
Write equations that show NH 3 as both a conjugate acid and a conjugate base.
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One example for NH 3 as a conjugate acid: \(\ce{NH2-}(aq) + \ce{H3O+}(aq) ⟶ \ce{NH3}(aq) + \ce{H2O}(l)\); as a conjugate base: \(\ce{NH4+}(aq)+\ce{OH-}(aq)⟶\ce{NH3}(aq)+\ce{H2O}(l)\)
Show by suitable reaction equations that each of the following species can act as a Brønsted-Lowry acid:
- HCl (aq)
- CH 3 CO 2 H (aq)
- \(\ce{NH4+}\) (aq)
- \(\ce{HSO4-}\) (aq)
- Answer
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a.\(\ce{HCl}(aq) + \ce{H2O}(l) ⟶ \ce{H3O+}(aq) + \ce{Cl-}(aq)\)
b. \(\ce{CH3CO2H}(aq) + \ce{H2O}(l) ⟶ \ce{CH3CO2-}(aq) + \ce{H3O+}(aq)\)
c. \(\ce{NH4+}(aq) + \ce{H2O}(l) ⟶ \ce{NH3}(aq) + \ce{H3O+}(aq)\)
d. \(\ce{HSO4-}(aq) + \ce{H2O}(l) ⟶ \ce{H3O+}(aq) + \ce{SO42-}(aq)\)
Show by suitable reaction equations that each of the following species can act as a Brønsted-Lowry base:
- H 2 O (l)
- NH 3 (aq)
- S 2− (aq)
- \(\ce{H2PO4-}\) (aq)
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a. \(\ce{H2O}(l) + \ce{H2O}(l) ⟶ \ce{H3O+}(aq) + \ce{OH-}(aq)\)
b.\(\ce{NH3}(aq) + \ce{H2O}(l) ⟶ \ce{NH4+}(aq) + \ce{OH-}(aq)\)
c. \(\ce{S2-}(aq) + \ce{H2O}(l) ⟶ \ce{HS-}(aq) + \ce{OH-}(aq)\)
d. \(\ce{H2PO4-}(aq) + \ce{H2O}(l) ⟶ \ce{H3PO4}(aq) + \ce{OH-}(aq)\)
Decide if each of the following are a conjugate acid-base pair:
a. H 2 O (l), H 3 O + (aq)
b. OH - (aq), H 3 O + (aq)
c. NH 3 (aq), OH - (aq)
d. H 2 CO 3 (aq), HCO 3 - (aq)
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a. Yes, H 2 O (l) is the base and H 3 O + is the conjugate acid.
b. No.
c. No.
d. Yes, H 2 CO 3 s the acid and HCO 3 - is the conjugate base.
What is the conjugate acid of each of the following? What is the conjugate base of each?
- \(\ce{OH-}\)
- H 2 O
- \(\ce{HCO3-}\)
- NH 3
- \(\ce{HSO4-}\)
- H 2 O 2
- HS −
- \(\ce{H5N2+}\)
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H 2 O, O 2− ; H 3 O + , \(\ce{OH^-}\) ; H 2 CO 3 , \(\ce{CO3^2-}\); \(\ce{NH4+}\), \(\ce{NH2-}\); H 2 SO 4 , \(\ce{SO4^2-}\); \(\ce{H3O2+}\), \(\ce{HO2-}\); H 2 S; S 2− ; \(\ce{H6N2^2+}\), H 4 N 2
Identify and label the Brønsted-Lowry acid, its conjugate base, the Brønsted-Lowry base, and its conjugate acid in each of the following equations:
- \(\ce{HNO3}(aq) + \ce{H2O}(l) ⟶ \ce{H3O+}(aq) + \ce{NO3-}(aq)\)
- \(\ce{CN-}(aq) + \ce{H2O}(l) ⟶ \ce{HCN}(aq) + \ce{OH-}(aq)\)
- \(\ce{H2SO4}(aq) + \ce{Cl-}(aq) ⟶ \ce{HCl}(aq) + \ce{HSO4-}(aq)\)
- \(\ce{HSO4-}(aq) + \ce{OH-}(aq) ⟶ \ce{SO42-}(aq) + \ce{H2O}(l)\)
- \(\ce{O2-}(aq) + \ce{H2O}(l) ⟶ \ce{2OH-}(aq)\)
- \(\ce{H2S}(aq) + \ce{NH2-}(aq) ⟶ \ce{HS-}(aq) + \ce{NH3}(aq)\)
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The labels are Brønsted-Lowry acid = BA; its conjugate base = CB; Brønsted-Lowry base = BB; its conjugate acid = CA.
a. HNO 3 (BA), H 2 O(BB), H 3 O + (CA), \(\ce{NO3- (CB)}\)
b. CN − (BB), H 2 O(BA), HCN(CA), \(\ce{OH^-}\) (CB)
c. H 2 SO 4 (BA), Cl − (BB), HCl(CA), \(\ce{HSO4- (CB)}\)
d. \(\ce{HSO4- (BA)}\), OH - (BB), \(\ce{SO4^2- (CB)}\), H 2 O(CA)
e. O 2− (BB), H 2 O(BA) \(\ce{OH^-}\) (CB and CA)
f. H 2 S(BA), \(\ce{NH2- (BB)}\), HS − (CB), NH 3 (CA)
What is the [H 3 O + ] if an aqueous solution has an [OH - ] concentration of 1.65 x 10 -5 at 100 o C? K w at 100 o C is 5.6 × 10 −13
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3.39 x 10 -8 M
What are amphiprotic species? Illustrate with suitable equations.
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Amphiprotic species may either gain or lose a proton in a chemical reaction, thus acting as a base or an acid. An example is H 2 O.
As an acid: \(\ce{H2O}(l) + \ce{NH3}(aq) \rightleftharpoons \ce{NH4+}(aq) + \ce{OH-}(aq)\). As a base: \(\ce{H2O}(l) + \ce{HCl}(aq) \rightleftharpoons \ce{H3O+}(aq) + \ce{Cl-}(aq)\)
State which of the following species are amphiprotic and write chemical equations illustrating the amphiprotic character of these species.
- NH 3
- \(\ce{HPO4-}\)
- Br −
- \(\ce{NH4+}\)
- \(\ce{ASO4^3-}\)
- Answer
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amphiprotic: \(\ce{NH3 + H3O+ ⟶ NH4OH + H2O}\), \(\ce{NH3 + OCH3- ⟶ NH2- + CH3OH}\); \(\ce{HPO4^2- + OH- ⟶ PO4^3- + H2O}\), \(\ce{HPO4^2- + HClO4 ⟶ H2PO4- + ClO4-}\); not amphiprotic: Br − ; \(\ce{NH4+}\); \(\ce{AsO4^3-}\)
6.2: pH and pOH
Explain why a sample of pure water at 40 °C is neutral even though [H 3 O + ] = 1.7 × 10 −7 M . K w is 2.9 × 10 −14 at 40 °C.
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In a neutral solution [H 3 O + ] = [OH − ]. At 40 °C,
[H 3 O + ] = [OH − ] = (2.910 −14 ) 1/2 = 1.7 × 10 −7 .
The ionization constant for water ( K w ) is 2.90 × 10 −14 at 40 °C. Calculate [H 3 O + ], [OH − ], pH, and pOH for pure water at 40 °C.
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x = 1.70 × 10 −7 M = [H 3 O + ] = [OH − ]
pH = -log (1.70 × 10 −7 ) = −(−6.769) = 6.769
pOH = pH = 6.769
Calculate the pH and the pOH of each of the following solutions at 25 °C for which the substances ionize completely:
- 0.000259 M HClO 4
- 0.21 M NaOH
- 0.000071 M Ba(OH) 2
- 2.5 M KOH
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pH = 3.587; pOH = 10.413; pH = 0.68; pOH = 13.32; pOH = 3.85; pH = 10.15; pH = −0.40; pOH = 14.4
What are the hydronium and hydroxide ion concentrations in a solution whose pH is 6.52?
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[H 3 O + ] = 3.0 × 10 −7 M ; [OH − ] = 3.3 × 10 −8 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?
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[OH − ] = 3.1 × 10 −12 M
6.3: Relative Strengths of Acids and Bases
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Arrange these acids in order of increasing strength.
- acid A: pKa=1.52
- acid B: pKa=6.93
- acid C: pKa=3.86
Given solutions with the same initial concentration of each acid, which would have the highest percent ionization?
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Acids in order of increasing strength: acidB<acidC<acidA.
Given the same initial concentration of each acid, the highest percent of ionization is acid A because it is the strongest acid.
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.
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[H 2 O] > [CH 3 CO 2 H] > \(\ce{[H3O+]}\) ≈ \(\ce{[CH3CO2- ]}\) > [OH − ]
Gastric juice, the digestive fluid produced in the stomach, contains hydrochloric acid, HCl. Milk of Magnesia, a suspension of solid Mg(OH) 2 in an aqueous medium, is sometimes used to neutralize excess stomach acid. Write a complete balanced equation for the neutralization reaction, and identify the conjugate acid-base pairs.
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\(\underset{\large\ce{BB}}{\ce{Mg(OH)2}(s)}+\underset{\large\ce{BA}}{\ce{HCl}(aq)}⟶\underset{\large\ce{CB}}{\ce{Mg^2+}(aq)}+\underset{\large\ce{CA}}{\ce{2Cl-}(aq)}+\underset{\:}{\ce{2H2O}(l)}\)
What is the ionization constant at 25 °C for the weak acid \(\ce{CH3NH3+}\), the conjugate acid of the weak base CH 3 NH 2 , K b = 4.4 × 10 −4 .
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\(K_\ce{a}=2.3×10^{−11}\)
Which base, CH 3 NH 2 or (CH 3 ) 2 NH, is the strongest base? Which conjugate acid, \(\ce{(CH3)2NH2+}\) or (CH 3 ) 2 NH, is the strongest acid?
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The strongest base or strongest acid is the one with the larger K b or K a , respectively. In these two examples, they are (CH 3 ) 2 NH and \(\ce{CH3NH3+}\).
What two common assumptions can simplify calculation of equilibrium concentrations in a solution of a weak acid?
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Assume that the change in initial concentration of the acid as the equilibrium is established can be neglected, so this concentration can be assumed constant and equal to the initial value of the total acid concentration.
Assume we can neglect the contribution of water to the equilibrium concentration of H 3 O + .
What is the effect on the concentrations of \(\ce{NO2-}\), HNO 2 , and OH − when the following are added to a solution of KNO 2 in water:
- HCl
- HNO 2
- NaOH
- NaCl
- KNO
The equation for the equilibrium is:
\[\ce{NO2-}(aq)+\ce{H2O}(l)⇌\ce{HNO2}(aq)+\ce{OH-}(aq)\]
Answer
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Adding HCl will add H 3 O + ions, which will then react with the OH − ions, lowering their concentration. The equilibrium will shift to the right, increasing the concentration of HNO 2 , and decreasing the concentration of \(\ce{NO2-}\) ions. Adding HNO 2 increases the concentration of HNO 2 and shifts the equilibrium to the left, increasing the concentration of \(\ce{NO2-}\) ions and decreasing the concentration of OH − ions. Adding NaOH adds OH − ions, which shifts the equilibrium to the left, increasing the concentration of \(\ce{NO2-}\) ions and decreasing the concentrations of HNO 2 . Adding NaCl has no effect on the concentrations of the ions. Adding KNO 2 adds \(\ce{NO2-}\) ions and shifts the equilibrium to the right, increasing the HNO 2 and OH − ion concentrations.
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?
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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.
NH 3 : [OH − ] = 3.1 × 10 −3 M ;
\(\ce{[NH4+]}\) = 3.1 × 10 −3 M ;[NH 3 ] = 0.533 M ;
HNO 2 : \(\ce{[H3O+]}\) = 0.011 M ;
\(\ce{[NO2- ]}\) = 0.0438 M ;[HNO 2 ] = 1.07 M ;
(CH 3 ) 3 N: [(CH 3 ) 3 N] = 0.25 M ;
[(CH 3 ) 3 NH + ] = 4.3 × 10 −3 M ;[OH − ] = 4.3 × 10 −3 M ;
\(\ce{NH4+ : [NH4+]}\) = 0.100 M ;
[NH 3 ] = 7.5 × 10 −6 M ;
[H 3 O + ] = 7.5 × 10 −6 M
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\(K_\ce{b}=1.8×10^{−5};\) \(K_\ce{a}=4.5×10^{−4};\) \(K_\ce{b}=7.4×10^{−5};\) \(K_\ce{a}=5.6×10^{−10}\)
Determine K a for hydrogen sulfate ion, \(\ce{HSO4-}\). In a 0.10- M solution the acid is 29% ionized.
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\(K_\ce{a}=1.2×10^{−2}\)
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. Ionization constants can be found in the text or Resource Tables.
- 0.0092 M HClO, a weak acid
- 0.0784 M C 6 H 5 NH 2 , a weak base
- 0.0810 M HCN, a weak acid
- 0.11 M (CH 3 ) 3 N, a weak base
- 0.120 M \(\ce{Fe(H2O)6^2+}\) a weak acid, K a = 1.6 × 10 −7
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a. \(\ce{\dfrac{[H3O+][ClO- ]}{[HClO]}}=\dfrac{(x)(x)}{(0.0092−x)}≈\dfrac{(x)(x)}{0.0092}=3.5×10^{−8}\)
Solving for x gives 1.79 × 10 −5 M . This value is less than 5% of 0.0092, so the assumption that it can be neglected is valid. Thus, the concentrations of solute species at equilibrium are:
[H 3 O + ] = [ClO] = 1.8 × 10 −5 M [HClO] = 0.00092 M [OH − ] = 5.6 × 10 −10 M ;
b. \(\ce{\dfrac{[C6H5NH3+][OH- ]}{[C6H5NH2]}}=\dfrac{(x)(x)}{(0.0784−x)}≈\dfrac{(x)(x)}{0.0784}=4.6×10^{−10}\)
Solving for x gives 6.01 × 10 −6 M .
This value is less than 5% of 0.0784, so the assumption that it can be neglected is valid. Thus, the concentrations of solute species at equilibrium are: \(\ce{[CH3CO2- ]}\) = [OH − ] = 6.0 × 10 −6 M [C 6 H 5 NH 2 ] = 0.00784 [H 3 O + ] = 1.7× 10 −9 M
c. \(\ce{\dfrac{[H3O+][CN- ]}{[HCN]}}=\dfrac{(x)(x)}{(0.0810−x)}≈\dfrac{(x)(x)}{0.0810}=4×10^{−10}\) Solving for x gives 5.69 × 10 −6 M . This value is less than 5% of 0.0810, so the assumption that it can be neglected is valid. Thus, the concentrations of solute species at equilibrium are: [H 3 O + ] = [CN − ] = 5.7 × 10 −6 M [HCN] = 0.0810 M [OH − ] = 1.8 × 10 −9 M
d. \(\ce{\dfrac{[(CH3)3NH+][OH- ]}{[(CH3)3N]}}=\dfrac{(x)(x)}{(0.11−x)}≈\dfrac{(x)(x)}{0.11}=7.4×10^{−5}\) Solving for x gives 2.85 × 10 −3 M . This value is less than 5% of 0.11, so the assumption that it can be neglected is valid. Thus, the concentrations of solute species at equilibrium are: [(CH 3 ) 3 NH + ] = [OH − ] = 2.9 × 10 −3 M [(CH 3 ) 3 N] = 0.11 M [H 3 O + ] = 3.5 × 10 −12 M
e. \(\ce{\dfrac{[Fe(H2O)5(OH)+][H3O+]}{[Fe(H2O)6^2+]}}=\dfrac{(x)(x)}{(0.120−x)}≈\dfrac{(x)(x)}{0.120}=1.6×10^{−7}\) Solving for x gives 1.39 × 10 −4 M . This value is less than 5% of 0.120, so the assumption that it can be neglected is valid. Thus, the concentrations of solute species at equilibrium are: [Fe(H 2 O) 5 (OH) + ] = [H 3 O + ] = 1.4 × 10 −4 M \(\ce{[Fe(H2O)6^2+]}\) = 0.120 M [OH − ] = 7.2 × 10 −11 M
White vinegar is a 5.0% by mass solution of acetic acid in water. If the density of white vinegar is 1.007 g/cm 3 , what is the pH?
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pH = 2.41
The pH of a 0.15- M solution of \(\ce{HSO4-}\) is 1.43. Determine K a for \(\ce{HSO4-}\) from these data.
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\(K_\ce{a}=1.2×10^{−2}\)
The pH of a solution of household ammonia, a 0.950 M solution of NH 3 , is 11.612. Determine K b for NH 3 from these data.
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\(K_\ce{b}=1.77×10^{−5}\)
6.4: Diprotic & Polyprotic Acids
Which of the following concentrations would be practically equal in a calculation of the equilibrium concentrations in a 0.134- M solution of H 2 CO 3 , a diprotic acid: [H 3 O + ],[H 3 O + ], [OH − ], [H 2 CO 3 ], [HCO 3 − ],[HCO 3 − ], [CO 3 2 − ]?[CO 3 2 − ]? No calculations are needed to answer this question.
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[H 3 O + ] and [HCO 3 − ] are practically equal.
Nicotine, C 10 H 14 N 2 , is a base that will accept two protons ( K 1 = 7 × 10 −7 , K 2 = 1.4 × 10 −11 ). What is the concentration of each species present in a 0.050- M solution of nicotine?
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[C 10 H 14 N 2 ] = 0.049 M
[C 10 H 14 N 2 H + ] = 1.9 × 10 −4 M \(\ce{[C10H14N2H2^2+]}\) = 1.4 × 10 −11 M [OH − ] = 1.9 × 10 −4 M [H 3 O + ] = 5.3 × 10 −11 M
The ion HTe − is an amphiprotic species; it can act as either an acid or a base.
K a2 for the reation of HTe − with H 2 O is 1.5 x 10 -11
What is K b for the reaction in which HTe − functions as a base in water is 4.3 x 10 -12
Demonstrate whether or not the second ionization of H 2 Te can be neglected in the calculation of [HTe − ] in a 0.10 M solution of H 2 Te.
- Answer
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\[\frac{\left[\mathrm{Te}^{2-}\right]\left[\mathrm{H}_3 \mathrm{O}^{+}\right]}{\left[\mathrm{HTe}^{-}\right]}=\frac{(x)(0.0141+x)}{(0.0141-x)} \approx \frac{(x)(0.0141)}{0.0141}=1.5 \times 10^{-11} \notag\]
Solving for x gives 1.5 × 10 −11 M . Therefore, compared with 0.014 M , this value is negligible (1.1 ×× 10 −7 %).
6.5 Hydrolysis of Salt Solutions
Determine whether aqueous solutions of the following salts are acidic, basic, or neutral:
- FeCl 3
- K 2 CO 3
- NH 4 Br
- KClO 4
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a. acidic
b. basic
c. acidic
d. neutral
- Why are aqueous solutions of salts such as CaCl 2 neutral? Why is an aqueous solution of NaNH 2 basic?
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Aqueous solutions of salts such as CaCl 2 are neutral because it is created from hydrochloric acid (a strong acid) and calcium hydroxide (a strong base). An aqueous solution of NaNH 2 is basic because it can deprotonate alkynes, alcohols, and a host of other functional groups with acidic protons such as esters and ketones.
Predict whether the aqueous solutions of the following are acidic, basic, or neutral.
- Li 3 N
- NaH
- KBr
- C 2 H 5 NH 3 Cl
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a. Li 3 N is a base because the lone pair on the nitrogen can accept a proton.
b. NaH is a base because the hydrogen has a negative charge with a lone pai that can accept a proton.
c. KBr is neutral because it is formed from HBr (a strong acid) and KOH (a strong base) so no reaction.
d. C 2 H 5 NH 3 Cl is acidic because it can donate a proton.
(make sure you can write out reactions to explain each answer!)
6.5 Lewis Acids & Bases
Identify the nature of each of the following as either a Lewis Acid or a Lewis Base:
a. NH 3
b. H 2 O
c. SO 2
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a. NH 3 is a lewis base because nitrogen has a lone pair of electrons to "donate."
b. H 2 O is a lewis base because oxygen has two lone pairs of electrons to "donate."
c. SO 2 is a lewis acid because sulfur has an unfilled octet and thus is able to accept a pair of electrons.