Practice Problems Acids & Bases

Answer

HI is a strong acid, so will ionize completely.

In Arrhenius theory, the acid dissociates into a hydrogen ion and the anion. For HI, then:

HI(aq) → H⁺(aq) + I^–(aq)

In Brønsted-Lowry theory, the acid is a hydrogen ion donor and another species is required to act as the hydrogen ion acceptor. In aqueous solution, the hydrogen ion acceptor is water, so:

HI(aq) + H₂O(l) → H₃O⁺(aq) + I^–(aq)

Answer

(a) HOClO

HOClO(aq) + H₂O(l) → ← H₃O⁺(aq) + OClO^–(aq)

(b) CH₃CH₂COOH

CH₃CH₂COOH(aq) + H₂O(l) → ← H₃O⁺(aq) + CH₃CH₂COO^–(aq)

(c) HCN

HCN(aq) + H₂O(l) → ← H₃O⁺(aq) + CN^–(aq)

(d) C₆H₅OH

C₆H₅OH(aq) + H₂O(l) → ← H₃O⁺(aq) + C₆H₅O^–(aq)

Answer

(a) The bond energy of the H-X bond is directly related to the acid strength. If the H-X bond is weak, then the bond is easily broken and gives a stronger acid. If the bond is strong, then it is difficult to break the bond, that is difficult to donate H⁺, so is a weaker acid. Anions with large ionic radii are the conjugate base of strong acids. If an anion has a large ionic radius, then the HX bond formed will be relatively long. Long bonds give small bond energies, which leads to strong acids. Small anionic radii lead to short and strong bonds, which is the description of a weak acid.

(b) The central atom (if it is more electronegative than H) and terminal oxygen atoms draw electron density away from the hydrogen atoms; this makes the H atoms more H⁺ - like, thus a stronger acid.

Answer

(a) HOClO₂(aq) + H₂O(l) → ← H₃O⁺(aq) + OClO₂^–(aq)

Acid: HOClO₂

Base: H₂O

Conjugate Acid: H₃O⁺

Conjugate Base: OClO₂^–

(b) HSeO₄^–(aq) + NH₃(aq) → ← NH₄⁺(aq) + SeO₄^2–(aq)

Acid: HSeO₄^–

Base: NH₃

Conjugate Acid: NH₄⁺

Conjugate Base: SeO₄^2–

(c) HCO₃^–(aq) + OH^–(aq) → ← CO₃^2–(aq) + H₂O(l)

Acid: HCO₃^–

Base: OH^–

Conjugate Acid: CO₃^2–

Conjugate Base: H₂O

(d) C₅H₅NH⁺(aq) + H₂O(l) → ← C₅H₅N(aq) + H₃O⁺(aq)

Acid: C₅H₅NH⁺

Base: H₂O

Conjugate Acid: H₃O⁺

Conjugate Base: C₅H₅N

Answer

The acid strength will change depending upon the relative electronegativity experienced by the ionizable hydrogen atom. All of these examples are substituted by Cl atoms, which have a high electronegativity. The general trend will be determined by first, the number of Cl atoms (more Cl atoms will give more total electroneagtivity), and second, by the location of the Cl atom relative to the ionizable H atom (a closer Cl atom withdraws more electron density from the H atom than a more remote Cl atom). Using these principals:

2,4,6-Trichlorophenol has three electronegative chlorine atoms so is the most acidic. 2,4-Dichlorophenol has two Cl atoms, so is next. 3-Chlorophenol and 4-Chlorophenol both have only a single Cl atom but in 4-chlorophenol the Cl atom is farther from the H atom, so is the least acidic. Phenol is less acidic than any of these because it has no electronegative Cl atoms.

Net: (least acidic) phenol < 4-chlorophenol < 3-chlorophenol < 2,4-dichlorophenol < 2,4,5-trichlorophenol

Answer

Assume 25 °C in all cases. Write out the reactions in order to find the [H₃O⁺] or [OH^–] concentration for each acid or base. Then use the definitions pH = –log[H₃O⁺] or pOH = –log[OH^–] and the relation pH + pOH = 14.00 (only true at 25 °C). Note that these are all strong acids or bases.

(a) 0.039 M HCl

HCl(aq) + H₂O(l) → H₃O⁺(aq) + Cl^–(aq)

[H₃O⁺] = [HCl] = 0.039

pH = –log[0.039] = 1.41

(b) 0.070 M KOH

KOH(aq) → K⁺(aq) + OH^–(aq)

[OH^–] = [KOH] = 0.070

pOH = –log[0.070] = 1.15

pH = 14.00 – pOH = 14.00 – 1.15 = 12.85

(c) 0.65 M HBr

HBr(aq) + H₂O(l) → H₃O⁺(aq) + Br^–(aq)

[H₃O⁺] = [HBr] = 0.65

pH = –log[0.65] = 0.19

(d) 2.5×10^–4 M Ca(OH)₂

Ca(OH)₂(aq) → Ca²⁺(aq) + 2 OH^–(aq)

[OH^–] = 2×[Ca(OH)₂] = 2×2.5×10^–4 = 5.0×10^–4

pOH = –log[5.0×10^–4] = 3.30

pH = 14.00 – pOH = 14.00 – 3.30 = 10.70

Answer

(a) 0.073 M LiOH

LiOH(aq) → Li⁺(aq) + OH^–(aq) (strong base)

0.073 M LiOH = 0.073 M OH^–

pOH = –log[OH^–] = –log[0.073] = 1.14

(b) 1.75 M NaOH

NaOH(aq) → Na⁺(aq) + OH^–(aq) (strong base)

1.75 M NaOH = 1.75 M OH^–

pOH = –log[OH^–] = –log[1.75] = 0.243

(c) 0.045 M Ba(OH)₂

Ba(OH)₂(aq) → Ba²⁺(aq) + 2 OH^–(aq) (strong base)

0.045 M Ba(OH)₂ = 2(0.045) M OH^– = 0.090 M OH^–

pOH = –log[OH^–] = –log[0.090] = 1.046

(d) 9.1×10^–2 M HClO₄

HClO₄(aq) + H₂O(l) → H₃O⁺(aq) + ClO₄^–(aq) (strong acid)

9.1×10^–2 M HClO₄ = 9.1×10^–2 M H₃O⁺

pH = –log[H₃O⁺] = –log[9.1×10^–2] = 1.041

pOH = 14.00 – pH (assuming 25 °C) = 14.00 – 1.041 = 12.96

Alternatively: [OH^–] = K_w/[ H₃O⁺] = 1.0×10^–14/9.1×10^–2 = 1.1×10^–13

pOH = –log[OH^–] = –log[1.1×10^–13] = 12.96

Answer

Given that pH = –log[H₃O⁺], the desired concentration of hydronium ion is

[H₃O⁺] = 10^–pH = 10^–3.60 = 2.5×10^–4 M.

Since HCl is a strong acid, i.e., HCl(aq) + H₂O(l) → H₃O⁺(aq) + Cl^–(aq)

[H₃O⁺] = [HCl] = 0.100 M in the given solution, so this needs to be diluted to obtain the desired concentration.

Recalling M_conc × V_conc = M_dil × V_dil, the amount of 0.100 M HCl required is

M_conc = 0.100 M

M_dil = 2.5×10^–4 M

V_dil = 2.00 L

so V_conc = (2.5×10^–4 M × 2.00 L)/(0.100 M) = 0.0050 L = 5.0 mL.

Thus to obtain the pH = 3.60 solution, add 5.0 mL of the 0.100 M HCl solution to a 2.00 L volumetric flask and add water to the mark.

Answer

The first hydrogen atom in sulfuric acid is 100% ionized but the second is not. However, if the second hydrogen atom also were 100% ionized, then [H₃O⁺] = 2×0.00048 = 0.00096 M giving a pH = –log[H₃O⁺] = –log[0.00096] = 3.02. Thus, the pH for the sulfuric acid solution must be greater than 3.02 so the vinegar solution has the lower pH.

Answer

Write the balanced equation, write the mass action expression for K_a, find K_a from pK_a, set up a table of concentrations, and solve from there.

HN₃(aq) + H₂O(l) → ← H₃O⁺(aq) + N₃^–(aq)

Initial C_init 0 0

Change – x +x +x

Equilibrium C_init – x x x

In this case we do not know the initial concentration so we need to find x from some other information: the pH.

x = [H₃O⁺] = 10^–pH = 10^–3.10 = 7.9×10^–4 M

Now we can plug into the mass action expression, knowing the value of x:

so the initial concentration of HN₃ = 0.033 M

Answer

Find the molarity of the aspirin sample, use the balanced reaction, write the mass action expression for K_a, set up a table of concentrations, and solve.

o–C₆H₄(OCOCH₃)COOH = C₉H₈O₄ = 9(12.011) + 8(1.008) + 4(15.999) = 180.159 g/mol

initial molarity = (1.00 g/180.159 g/mol)/0.300 L = 0.0185 M

o–C₆H₄(OCOCH₃)COOH(aq) + H₂O(l) → ← H₃O⁺(aq) + o–C₆H₄(OCOCH₃)COO^–(aq)

Initial 0.0185 0 0

Change – x +x +x

Equilibrium 0.0185 – x x x

Since K_a is unknown, x must be found from other information, in this case the pH.

x = [H₃O⁺]_e = 10^–pH = 10^–2.62 = 2.4×10^–3 M = [C₉H₇O₄^–]_e

[C₉H₈O₄]_e = 0.0185 – 2.4×10^–3 = 0.0161 M

These concentrations can be used to find K_a

Answer

Treat this as any equilibrium problem: write the balanced chemical reaction, write the mass action expression for K_b, set up a table of concentrations, and then solve for the unknown variables using the given information. Assume 25 °C.

C₁₈H₂₁NO₃(aq) + H₂O(l) → ← OH^–(aq) + [C₁₈H₂₁NHO₃]⁺(aq)

Initial [B]_ini 0 0

Change – x +x +x

Equilibrium [B]_ini – x x x

[B]_ini is found from the concentration information given:

molar mass of codeine = 18(12.01) + 21(1.01) + 1(14.01) + 3(16.00) = 299.4 g/mol

so the molar concentration of the saturated solution is:

[B]_ini = 0.0278 M

The pH gives the hydronium ion concentration:

[H₃O⁺] = 10^–pH = 10^–9.8 = 2×10^–10 M

The hydroxide ion concentration is found from the hydronium ion concentration by:

[OH^–] = K_w/[ H₃O⁺] = 1.0×10^–14/2×10^–10 = 5×10^–5 M

Then,

x = [OH^–] = 5×10^–5 M

[B]_ini – x = 0.0278 – 5×10^–5 = 0.0278 M

so K_b is found from:

Answer

Write the reaction, write the mass action expression and find K_a, set up a table of concentrations, plug the equilibrium concentrations into the mass action expression, and then solve.

CCl₃COOH(aq) + H₂O(l) → ← H₃O⁺(aq) + CCl₃COO^–(aq)

Initial 0.105 0 0

Change – x +x +x

Equilibrium 0.105 – x x x

[HA]_init/K_a = 0.105/0.30 = 0.35 < 100 so no approximation

Using the quadratic equation gives

x = –0.38 or x = 0.082; only the positive value makes sense so

x = [H₃O⁺] = 0.082 M

pH = –log[H₃O⁺] = –log[0.082] = 1.09

Answer

Treat this as any equilibrium problem: write the balanced chemical reaction, write the mass action expression for K_b, set up a table of concentrations, and then solve for the unknown variables using the given information. Assume 25 °C.

C₅H₁₁N(aq) + H₂O(l) → ← OH^–(aq) + C₅H₁₁NH⁺(aq)

Initial 0.00250 0 0

Change – x +x +x

Equilibrium 0.00250 – x x x

Approximate? 0.00250/1.3×10^–3 = 2 < 100 No!

Solve this using the quadratic equation to give:

x = 1.3×10^–3, –2.6×10^–3

Only the positive root makes sense, so [OH^–] = x = 1.3×10^–3 M

pOH = –log[OH^–] = –log(1.3×10^–3) = 2.89

pH = 14.00 – pOH = 14.00 – 2.89 = 11.11

Answer

This is a polyprotic weak acid. The first ionization is treated exactly like every other weak acid:

H₃PO₄(aq) + H₂O(l) → ← H₃O⁺(aq) + H₂PO₄^–(aq)

Initial 0.15 0 0

Change – x +x +x

Equilibrium 0.15 – x x x

Approximate? 0.15/7.1×10^–3 = 21, NO!

Using the quadratic equation gives

x = 0.029 or x = –0.036

Only the positive value makes sense, so [H₃O⁺] = [H₂PO₄^–] = 0.029 M

[H₃PO₄] = 0.15 – 0.029 = 0.12 M

The second and third ionizations do not contribute significantly to these concentrations.

To find the hydrogen phosphate concentration, look at the second ionization step:

H₂PO₄^–(aq) + H₂O(l) → ← H₃O⁺(aq) + HPO₄^2–(aq)

Initial 0.029 0.029 0

Change – x +x +x

Equilibrium 0.029 – x 0.029 + x x

Approximate? 0.029/6.3×10^–8 = 4.6×10⁵, YES!

x = [HPO₄^2–] = 6.3×10^–8 M

To find the phosphate concentration, look at the third ionization step:

HPO₄^2–(aq) + H₂O(l) → ← H₃O⁺(aq) + PO₄^3–(aq)

Initial 6.3×10^–8 0.029 0

Change – x +x +x

Equilibrium 6.3×10^–8 – x 0.029 + x x

Approximate? 6.3×10^–8/4.2×10^–13 = 1.5×10⁵, YES!

x = [PO₄^3–] = 9.1×10^–19 M

Then:

(a) pH = –log[H₃O⁺] = –log[0.029] = 1.54

(b) [H₃PO₄] = 0.12 M

(c) [H₂PO₄^–] = 0.029 M

(d) [HPO₄^2–] = K_a2 = 6.3×10^–8 M

(e) [PO₄^3–] = 9.1×10^–19 M

Answer

(a) RbClO₄(aq)

RbClO₄(aq) → Rb⁺(aq) + ClO₄^–(aq)

Rb⁺ is the cation of a strong base (RbOH) so no cation hydrolysis occurs.

ClO₄^– is the anion of a strong acid (HClO₄) so no anion hydrolysis occurs.

Since no hydrolysis occurs, the solution will be neutral.

(b) CH₃CH₂NH₃Br(aq)

CH₃CH₂NH₃Br(aq) → CH₃CH₂NH₃⁺(aq) + Br^–(aq)

CH₃CH₂NH₃⁺ is the cation of a weak base, CH₃CH₂NH₂, hydrolysis occurs:

CH₃CH₂NH₃⁺(aq) + H₂O(l) → ← H₃O⁺(aq) + CH₃CH₂NH₂(aq)

Br^– is the anion of a strong acid (HBr) so no anion hydrolysis occurs:

Since hydronium ion is formed by the cation hydrolysis, the solution will be acidic.

(c) HCOONH₄(aq)

HCOONH₄(aq) → NH₄⁺(aq) + HCOO^–(aq)

NH₄⁺(aq) + H₂O(l) → ← H₃O⁺(aq) + NH₃(aq)

HCOO^– is the anion of a weak acid, HCOOH, so anion hydrolysis occurs:

HCOO^–(aq) + H₂O(l) → ← HCOOH(aq) + OH^–(aq)

The equilibrium constant for the cation hydrolysis is 1.0×10^–14/1.8×10^–5 = 5.6×10^–10.

The equilibrium constant for the anion hydrolysis is 1.0×10^–14/1.8×10^–4 = 5.6×10^–11.

The equilibrium reaction with the larger equilibrium constant will be favored, so the cation hydrolysis dominates, so that more hydronium ion will be formed than hydroxide ion and the solution will be acidic.

Answer

(a) CH₃CH₂COOK(aq)

CH₃CH₂COOK(aq) → K⁺(aq) + CH₃CH₂COO^–(aq)

K⁺ is the cation of a strong base (RbOH) so no cation hydrolysis occurs.

CH₃CH₂COO^– is the anion of a weak acid (CH₃CH₂COOH) so anion hydrolysis occurs.

CH₃CH₂COO^–(aq) + H₂O(l) → ← OH^–(aq) + CH₃CH₂COOH(aq)

Since hydroxide ion is formed by the anion hydrolysis, the solution will be basic.

(b) Mg(NO₃)₂(aq)

Mg(NO₃)₂(aq) → Mg²⁺(aq) + 2 NO₃^–(aq)

Mg²⁺ is the cation of a strong base (Mg(OH)₂) so no cation hydrolysis occurs.

NO₃^– is the anion of a strong acid (HNO₃) so no anion hydrolysis occurs.

Since no hydrolysis occurs, the solution will be neutral.

(c) NH₄CN(aq)

NH₄CN(aq) → NH₄⁺(aq) + CN^–(aq)

NH₄⁺(aq) + H₂O(l) → ← H₃O⁺(aq) + NH₃(aq)

CN^– is the anion of a weak acid, HCN, so anion hydrolysis occurs:

CN^–(aq) + H₂O(l) → ← HCN(aq) + OH^–(aq)

The equilibrium constant for the cation hydrolysis is 1.0×10^–14/1.8×10^–5 = 5.6×10^–10.

The equilibrium constant for the anion hydrolysis is 1.0×10^–14/6.2×10^–10 = 1.6×10^–5.

The equilibrium reaction with the larger equilibrium constant will be favored, so the anion hydrolysis dominates, so that less hydronium ion will be formed than hydroxide ion and the solution will be basic.

Answer

Hydrolysis occurs by the reaction of water with the conjugate species of weak acids or weak bases.

(a) NaNO₃(aq) → Na⁺(aq) + NO₃^–(aq)

(Sodium salts and nitrates are freely soluble in water, so the reaction goes to completion.)

Na⁺ is the conjugate acid of NaOH, a strong base, so does not hydrolyze, i.e.

Na⁺(aq) + H₂O(l) → No Reaction

NO₃^– is the conjugate base of HNO₃, a strong acid, so also does not hydrolyze, i.e.,

NO₃^– (aq) + H₂O(l) → No Reaction

Thus, the solution is neutral, pH ~ 7.

(b) CH₃COOK(aq) → K⁺(aq) + CH₃COO^–(aq)

(Potassium salts and acetates are freely soluble in water, so the reaction goes to completion.)

K⁺ is the conjugate acid of KOH, a strong base, so does not hydrolyze, i.e.

K⁺(aq) + H₂O(l) → No Reaction

CH₃COO^– is the conjugate base of CH₃COOH, a weak acid (acetic acid), so does hydrolyze, i.e.

CH₃COO^–(aq) + H₂O(l) → ← CH₃COOH(aq) + OH^–(aq)

The solution generates hydroxide, so pH > 7.

(c) NH₄I(aq) → NH₄⁺(aq) + I^–(aq)

(Ammonium salts and most halides are freely soluble in water, so the reaction goes to completion.)

NH₄⁺ is the conjugate acid of NH₃, a weak base (ammonia), so does hydrolyze, i.e.

NH₄⁺(aq) + H₂O(l) → ← NH₃(aq) + H₃O⁺(aq)

I^– is the conjugate base of a strong acid, HI, so does not hydrolyze, i.e.

I^–(aq) + H₂O(l) → No Reaction

The solution generates hydronium ion, so pH < 7.

(d) Na₃PO₄(aq) → 3 Na⁺(aq) + PO₄^3– (aq)

(Sodium salts are freely soluble in water, so the reaction goes to completion.)

Na⁺ is the conjugate acid of NaOH, a strong base, so does not hydrolyze, i.e.

Na⁺(aq) + H₂O(l) → No Reaction

PO₄^3– is the conjugate base of HPO₄^2–, a weak acid, so does hydrolyze, i.e.

PO₄^3– (aq) + H₂O(l) → ← HPO₄^2– (aq) + OH^–(aq)

The solution generates hydroxide, so pH > 7.

The only acidic solution is (c) so has the lowest pH.

Answer

(a) NaOCl is a soluble salt, so NaOCl(aq) → Na⁺(aq) + OCl^–(aq)

Na⁺ is the conjugate acid of a strong base, NaOH, so does not hydrolyze, i.e.

Na⁺(aq) + H₂O(l) → No Reaction

OCl^– is the conjugate base of a weak acid, HOCl, so does hydrolyze, i.e.

OCl^–(aq) + H₂O(l) → ← HOCl(aq) + OH^–(aq)

The solution will be basic.

(b) The equilibrium constant is found from the hydrolysis reaction.

OCl^–(aq) + H₂O(l) → ← HOCl(aq) + OH^–(aq)

and has the form of a K_b, which must be found from the K_a of the conjugate acid HOCl:

assuming 25 °C.

(Remember that this relationship is only true for conjugate acid–base pairs.)

(c) Use the hydrolysis reaction that generates hydroxide, write the mass action expression for the equilibrium constant, set up a table of concentrations, use this to find the pOH, and then find the pH.

OCl^–(aq) + H₂O(l) → ← HOCl(aq) + OH^–(aq)

Initial 0.082 0 0

Change – x +x +x

Equilibrium 0.082 – x x x

Approximate? 0.080/3.4×10^–7 = 240000 >> 100, so yes.

x = [OH^–] = 1.6×10^–4 M

pOH = –log[OH^–] = –log[1.6×10^–4] = 3.78

pH = 14.00 – pOH = 14.00 – 3.78 = 10.22

Answer

Write the chemical reactions (the solubility of KOH and the acid–base equilibrium for NH₃), write the mass action expression for K_b (ammonia is a base), set up a table of concentrations, and solve.

KOH(aq) → K⁺(aq) + OH^–(aq)

NH₃(aq) + H₂O(l) → ← NH₄⁺(aq) + OH^–(aq)

Initial 0.15 0 0.015

Change – x +x +x

Equilibrium 0.15 – x x 0.015 + x

Approximate? 0.15/1.8×10^–5 = 8000 > 100, yes; 0.015/1.8×10^–5 = 800 > 100, also yes.

x = [NH₄⁺] = 1.8×10^–4 M

Answer

(a) Write the solubilization reaction:

CH₃COONa(aq) → Na⁺(aq) + CH₃COO^–(aq)

Acetate ion is basic, so will react with any acid present, so

HI(aq) + CH₃COO^–(aq) → CH₃COOH(aq) + I^–(aq)

Thus, all of the HI will be consumed and the solution cannot contain 0.15 M HI.

(b) Write the solubilization reactions:

KNO₂(aq)→ K⁺(aq) + NO₂^–(aq)

KNO₃(aq) → K⁺(aq) + NO₃^–(aq)

Nitrite ion is a base, but there is no acid present in solution so no further reactions occur and the given concentrations are compatible.

(c) Calcium hydroxide is a strong base and nitric acid is a strong acid, so reaction will occur

Ca(OH)₂(aq) + 2 HNO₃(aq) → Ca²⁺(aq) + 2 NO₃^–(aq) + 2 H₂O(l)

The given concentrations are incompatible.

(d) Write the solubilization reaction:

NH₄Cl(aq) → NH₄⁺(aq) + Cl^–(aq)

Ammonium ion is an acid and will react with any base present:

NH₄⁺(aq) + NaOH(aq) → Na⁺(aq) + NH₃(aq) + H₂O(l)

The given concentrations are incompatible.

Answer

Write the dissolution reaction for sodium formate, the acid–base equilibrium for formic acid, write the mass action expression for K_a, set up a table of concentrations, and solve.

HCOONa(aq) → Na⁺(aq) + HCOO^–(aq)

HCOOH(aq) + H₂O(l) → ← H₃O⁺(aq) + HCOO^–(aq)

Initial 0.405 0 0.326

Change – x +x +x

Equilibrium 0.405 – x x 0.326 + x

Approximate? 0.405/1.8×10^–4 = 2000 > 100, yes; 0.362/1.8×10^–4 = 2000 > 100, also yes.

x = [H₃O⁺] = 2.2×10^–4 M

pH = –log[H₃O⁺] = –log[2.2×10^–4] = 3.66

Answer

Lewis acids are electron pair acceptors (often cationic) and Lewis bases are electron pair donors (often anionic or have a free lone pair of electrons to donate).

(a) OH^–(aq) + Al(OH)₃(s) → Al(OH)₄^–(aq)

Lewis Acid: Al(OH)₃

Lewis Base: OH^–

(b) Cu²⁺(aq) + 4 NH₃(aq) → [Cu(NH₃)₄]²⁺(aq)

Lewis Acid: Cu²⁺

Lewis Base: NH₃

(c) CO₂(g) + OH^–(aq) → HCO₃^–(aq)

Lewis Acid: CO₂

Lewis Base: OH^–

Answer

Assume 25 °C so [H₃O⁺][OH^–] = K_w = 1.0×10^–14 and pH + pOH = 14.00.

If [H₃O⁺] = 2×[OH^–], then 1.0×10^–14 = [H₃O⁺][OH^–] = (2×[OH^–])[OH^–] = 2[OH^–]²

or [OH^–] = 7.1×10^–8 M and [H₃O⁺] =1.4×10^–7 M. There is no reason why a solution could not have these concentrations.

If pH = 2×pOH, 14.00 = pH + pOH = 2×pOH + pOH = 3 pOH

or pOH = 4.67 and pH = 9.33. Again, these are perfectly reasonable and there is no reason why such a solution could not be prepared.

However, the two solutions are not the same. In the first case, the pH = –log[H₃O⁺] = –log[1.4×10^–7] = 6.85, which is different from the second solution having pH = 9.33.

Answer

Both water and HCl contribute significant amounts of hydronium ion, so both reactions must be considered. This problem, then, is a common–ion problem where HCl provides the common ion to the water autoionization equilibrium. So, write the two reactions, write the mass action expression for K_w, set up a table of concentrations, and solve. Assume 25 °C.

HCl(aq) + H₂O(l) → H₃O⁺(aq) + Cl^–(aq)

H₂O(l) + H₂O(l) → ← H₃O⁺(aq) + OH^–(aq)

K_w = [H₃O⁺]_e[OH^–]_e = 1.0×10^–14

Initial   1.0×10^–8 0

Change   +x +x

Equilibrium   1.0×10^–8 + x x

1.0×10^–14 = [1.0×10^–8 + x][x]

1.0×10^–14 = 1.0×10^–8x + x²

x² + 1.0×10^–8x – 1.0×10^–14 = 0

Solve with the quadratic equation to give

x = 9.5×10^–8, –1.1×10^–7

Only the positive value makes sense chemically, so

[H₃O⁺] = 1.0×10^–8 + x = 1.0×10^–8 + 9.5×10^–8 = 1.05×10^–7 M

pH = –log[H₃O⁺] = –log[1.05×10^–7] = 6.979 = 6.98 (despite the sig fig rules, pH values beyond 2 decimal places are nearly always meaningless)

Answer

First, determine the molar concentration of CO₂ in saturated water using Henry's law.

Change the volume solubility of CO₂ in water to a molar solubility using the ideal gas law:

One of the assumptions in the problem is that all of the dissolved CO₂ reacts with water to form carbonic acid, i.e.

CO₂(aq) + H₂O(l) → H₂CO₃(aq)

Now the problem becomes a typical weak acid equilibrium:

H₂CO₃(l) + H₂O(l) → ← H₃O⁺(aq) + HCO₃^–(aq)

Initial 1.1×10^–5 0 0

Change   +x +x

Equilibrium 1.1×10^–5 – x x x

Approximate? 1.1×10^–5/4.4×10^–7 = 25 <100, so no.

Using the quadratic equation gives

x = 2.0×10^–6, –2.4×10^–6

Only the positive value makes sense, so

x = [H₃O⁺] = 2.0×10^–6 M

pH = – log[H₃O⁺] = – log[2.0×10^–6] = 5.70