Which aqueous solution is expected to have a ph less than 7 at 25c?

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which aqueous solution is expected to have a ph less than 7 at 25c?

Answer:

  1. Normal rain is usually a bit acidic due to dissolved carbon dioxide. Carbon dioxide dissolving in water makes the solution slightly acidic:

CO2(aq) + H2O(l) <==> H+(aq) + HCO3-(aq) Ka = 4.3 x 10-7
Acid rain, on the other hand, is a major pollution problem in many parts of this planet. Acid rain contains a fair amount of strong acids such as H2SO4 and HNO3. It results largely from the burning or combustion of fossil fuels (coal, oil and natural gas) that contain sulfur. Sulfur dioxide is formed which can further react with oxygen in the atmosphere forming sulfur trioxide. Sulfur trioxide dissolves in water to form sulfuric acid:

S(in fossil fuel) + O2 --> SO2(g)
SO2(g) + (1/2)O2(g) --> SO3(g)
SO3(g) + H2O(rain) --> H2SO4(aq)
The net result is to increase the acidity of the rain, which damages trees, aquatic life and corrodes away stones and metal surfaces.

(a) The pH in acid rain can drop to 3 or even a bit lower in heavily polluted areas. Calculate the concentrations of [H+] and [OH-] at pH 3.30 at 25oC.

pH = -log[H+]

[H+] = 10-pH = 10-3.30 = 5.0 x 10-4 M

in aqueous solution at 25oC, [H+][OH-] = 1.0 x 10-14

[OH-] = (1.0 x 10-14)/ (5.0 x 10-4) = 2.0 x 10-11 M

(b) When sulfur dioxide (SO2) dissolves in water, sulfurous acid (H2SO3, Ka1=1.7 x 10-2, Ka2=6.4 x 10-8) is formed. Sulfurous acid can donate a proton to a water molecule. Write a balanced chemical equation for this reaction, and identify the stronger Bronsted-Lowry acid and base in the equation. (Consider only the first proton of H2SO3)

H2SO3(aq) + H2O(l) <==> HSO3-(aq) + H3O+(aq)

H3O+ is the stronger acid (as indicated by Ka1 < 1) and HSO3- is the stronger base.

© Ignore the further ionization (since Ka2<<Ka1) of HSO3-, and calculate the pH of a solution in which the initial concentration of H2SO3 is 4.0 x 10-4M. (Since the concentration is less than 0.01 M and Ka1 is greater than 0.0001, you need to use the quadratic formula)

Ka1 = [H+][HSO3-]/[H2SO3]

[H+] [HSO3-] [H2SO3]
initial 10-7 0 4.0 x 10-4M
shift +x +x -x
equilibrium +x (10-7 is negligible) +x 4.0 x 10-4M - x
1.7 x 10-2 = x2/(4.0 x 10-4M - x)

x2 + (1.7 x 10-2)x - 6.8 x 10-6 = 0

Using the quadratic formula, we get

x = 0.00039 = [H+] so pH = 3.4

(d) Now suppose that all the dissolved SO2 from part © has been oxidized initially to SO3, so that the initial concentration of H2SO4 is 4.0 x 10-4M. Calculate the pH in this case. (H2SO4, Ka1 is very large, assume complete ionization, Ka2 = 1.2 x 10-2) Since the initial concentration is low and Ka2 > 10-4, you should use the quadratic formula.

H2SO4(aq) + H2O(l) ==> HSO4-(aq) + H3O+(aq) (assume complete)

HSO4-(aq) + H2O(l) <==> SO42-(aq) + H3O+(aq)

1.2 x 10-2 = [H+][SO42-]/[HSO4-]

[H+] [SO42-] [HSO4-]
initial 4.0 x 10-4 0 4.0 x 10-4M
shift +x +x -x
equilibrium 4.0 x 10-4 + x +x 4.0 x 10-4M - x
1.2 x 10-2 = (4.0 x 10-4 + x) x/ (4.0 x 10-4M - x)

x2 + (1.2 x 10-2) x - (4.8 x 10-6) = 0

Using the quadratic formula, we get

x = 0.00039

[H+] = 4.0 x 10-4 + x = 0.00079, so pH =3.1

(e) Nature also has its own way of fighting acid rain. Lakes have a natural buffering capacity, especially in regions where there is limestone which gives rise to dissolved calcium carbonate. Write an equation for the reaction of a small amount of acid rain containing sulfuric acid falling into a lake containing carbonate (CO32-) ions. Discuss how the lake will resist further pH changes. What happens if a large excess of acid rain is deposited in the lake?

CO32- is the conjugate base of the weak acid HCO3- and, thus, can react with the strong acid H3O+ in the sulfuric acid solution according to

CO32-(aq) + H3O+(aq) <==> HCO3-(aq) + H2O(l)

with K = 1/Ka2 of H2CO3 (thus, large K)

Likewise, HSO4- will also react with CO32-:

CO32-(aq) + HSO4-(aq) <==> SO42-(aq) + HCO3-(aq)

with K = (Ka2 of H2SO4)/(Ka2 of H2CO3) (thus, large K)

Thus, a HCO3-/CO32- buffer will be produced.

An excess of acid rain will consume all the CO32- and HCO3-, converting all to H2CO3 (no more buffer)

  1. The active component of aspirin is acetylsalicylic acid, HC9H7O4, which has a Ka of 3.0 x 10-4.
    (a) Calculate the pH of a solution made by dissolving 0.500 g (500 mg) of acetylsalicylic acid in water and diluting to 50.0 mL.
    (b) Repeat the calculation, assuming that the same mass of acetylsalicylic acid is dissolved and diluted to 1000.0 mL of solution.
    © Is the pH lower or higher in the more dilute solution? Which solution has a higher fraction of its acetylsalicylic acid ionized?
    (a) Ka=3.0 x 10-4 = [H+][A-]/[HA] where A = C9H7O4

Molar mass of acetylsalicylic acid = 180.16 g/mole

0.500 g ==> 2.78 x 10-3 mol

[HA]initial = 2.78 x 10-3 mol / 0.0500 L = 5.55 x 10-2

[H+] [A-] [HA]
initial 0 0 5.55 x 10-2M
shift +x +x -x
equilibrium + x +x 5.55 x 10-2M - x
3.0 x 10-4 = x2/5.55 x 10-2

x = [H+] = 4.1 x 10-3 M, so pH = 2.4

Using the quadratic equation:

3.0 x 10-4 = x2/(5.55 x 10-2 - x)

x = 3.9 x 10-3 M, so pH = 2.4 (no difference)

(b) same as (a) except for different [HA]initial

[H+] [A-] [HA]
initial 0 0 2.78 x 10-3M
shift +x +x -x
equilibrium + x +x 2.78 x 10-3M - x
We need to use the quadratic equation this time:

3.0 x 10-4 = x2/(2.78 x 10-3M - x)

x= 7.8 x 10-4 = [H+] so pH = 3.1

The pH is higher in the more dilute (lower concentration of acid) solution.

Calculating the %ionized:
For first solution: %ionized = 0.0041/0.0555 = 7.4 %
second solution: %ionized = 0.00078/0.00278 = 28 %

  1. Na+, being the conjugate acid of the strong base NaOH, is not acidic. Cl-, being the conjugate base of the strong acid HCl, is not basic. An aqueous solution containing only NaCl has a pH of 7 (neutral) at 25oC, as expected. An aqueous solution of NH4CH3COO (ammonium acetate) is almost exactly neutral. Explain.

NH4CH3COO is a strong electrolyte, dissociating completely into NH4+ and CH3COO-. These ions being salts of a weak base and a weak acid, respectively, will be acidic (NH4+) and basic (CH3COO-). Thus, we need to take into account:

NH4+(aq) + H2O(l) <==> NH3(aq) + H3O+(aq) Ka = Kw/(1.8 x 10-5)

and

CH3COO-(aq) + H2O(l) <==> CH3COOH (aq) + OH-(aq) Kb=Kw/(1.8 x 10-5)

Since the K’s are similar, the amount of [H+] and [OH-] will be similar and, thus, leads to a neutral solution.

In aqueous solutions of NH4CN, the NH4+(aq) ion is weakly acidic and the CN-(aq) ion) is weakly basic. From the Kb of NH3 and the Ka of HCN, calculate the equilibrium constant for the following reaction:

NH4+(aq) + CN-(aq) <==> NH3(aq) + HCN(aq)
Would a solution containing NH4CN be acidic, basic or neutral? Explain.

NH4+(aq) + H2O(l) <==> NH3(aq) + H3O+(aq) Ka = Kw/(1.8 x 10-5)

CN-(aq) + H2O(l) <==> HCN(aq) + OH-(aq) Kb = Kw/(4.9 x 10-10)

H3O+(aq) + OH-(aq) <==> H2O(l) + H2O(l) K=1/Kw

Adding the equations and multiplying the K’s
NH4+(aq) + CN-(aq) <==> NH3(aq) + HCN(aq) K = Kw/(4.9 x 10-10) (1.8 x 10-5) = 1.1

Since CN- is stronger as a base than NH4+ as an acid, the solution will be basic.

  1. Papaverine hydrochloride (papHCl), a salt of a weak base and a strong acid, is a drug used as a muscle relaxant. It is a weak acid. At 25o C, a 0.205 M solution of papHCl has a pH of 3.31. Compute the Ka of papHCl.
    [H+] = 10-3.31 = 4.90 x 10-4 M

Thus,

Ka = (4.90 x 10-4)2/(0.205 - 4.90 x 10-4) = 1.2 x 10-6

  1. Comment on the following statement:

Weak acids become strong when they are dilute.

Weak acids obey the following equilibrium:

HA(aq) <==> H+(aq) + A-(aq)

On the other hand, a strong acid dissociates completely.

Dilution involves concentration, it is not related to the strength of an acid. The strength of an acid is measured purely by how much (or what fraction) of it ionizes. Only a small fraction of a weak acid ionizes in solution. However, as the initial concentration of the weak acid is decreased, the fraction that dissociates becomes increasingly closer to 100% (remember the weak acid handout). The reason why this occurs is kinetics. As the solution becomes more dilute, it becomes more increasingly difficult for H+ to see A-, which is required to reform the acid HA. Thus, it is simply a manifestation of Le Chatelier’s principle: Bigger volume will favor the side of the reaction that has more particles.

  1. “Tris” is short for tris(hydroxymethyl)aminomethane. This weak base is widely used in biochemical research for the preparation of buffers. It offers low toxicity and a pKb of 5.92 making it convenient for the control of pH in clinical applications. A buffer is made by mixing 0.050 mol of tris with 0.025 mol of HCl in a volume of 2.00 L. Calculate the pH of the solution.
    Using Henderson-Hasselbach equation

pH = pKa + log [base]/[acid]

[acid] = 0.025 mol/2.00 L

[base] = (0.050-0.025) mol / 2.00 L

Thus, [acid]=[base]

pH = pKa = (14-5.92) = 8.08

  1. Suppose you were designing a buffer system for imitation blood, and you wanted the buffer to maintain the blood at the realistic pH of 7.40. All other things being equal, which buffer system would be preferable: H2CO3/HCO3- or H2PO4-/HPO42-? Explain.
    Buffers most effectively resist a change in pH in either direction when the concentrations of acidic and basic species are about the same (HA/A- ~ 1.0). Thus, one normally chooses a buffer that has a pKa close to the desired pH. H2CO3 has a pKa of 6.4 while H2PO4- has a pKa of 7.2. Thus, H2PO4- would be more suitable.

  2. Decide if each of the following substances should be classified as a Lewis acid or a Lewis base:

(a) BCl3 (b) H2N-NH2 © CN- (d) Fe3+

BCl3 is a Lewis acid, B does not have an octet.

H2N-NH2 is a Lewis base, the 2 N’s have a lone pair.

CN- is a Lewis base.

Fe3+ is a Lewis acid.

  1. Liquid ammonia, like water, autoionizes. Write the chemical equation for this reaction. What is the conjugate base of NH3 in this reaction? The conjugate acid?

NH3(l) + NH3(l) <==> NH4+(in ammonia) + NH2-(in ammonia)

NH2- is the conjugate base while NH4+ is the conjugate acid.

In a far away galaxy, ammonia has become the universal solvent and acid-base reactions are performed with ammonia as solvent. In this galaxy, students are also required to take a general chemistry course. Unfortunately, the students have great difficulty remembering which acids are strong in their galaxy. Explain. (On the other hand, the students need not memorize a list of strong bases)

In our galaxy, strong acids are those which completely transfer a proton to a water molecule. Thus, any acid stronger than H3O+ will be considered strong. With ammonia now as solvent, an acid need to be stronger only with respect to NH4+ (which is much weaker than H3O+) to be considered strong. Thus, in this far away galaxy, acids like acetic, formic, HF, even H2CO3 will be considered strong. Gone are the good old days of just memorizing seven strong acids.

On the other hand, OH- is weaker than NH2-. Thus, there is even fewer strong bases with ammonia as solvent.

  1. Acetic acid is a weak Bronsted acid. It is interesting to compare the strength of this acid with a related series of acids where the H atoms of the methyl (CH3) group in acetic acid are replaced by Cl.

Acid Ka
CH3COOH 1.8 x 10-5
ClCH2COOH 1.4 x 10-3
Cl2CHCOOH 3.3 x 10-2
Cl3CCOOH 2.0 x 10-1
Is there a trend? Explain the trend that you observed.

As the H’s are replaced by Cl atoms, the acid becomes stronger. The strength of an acid depends on the stability of its conjugate base. In the case of acetic acid, the conjugate base is CH3COO-. The conjugate base is carrying one electron in excess. Thus, its stability will depend on how well the ion can handle this negative charge. Replacing H with Cl, a much bigger atom and also more electronegative than H, should help in dissipating the negative charge. Thus, as the number of Cl is increased, the acid becomes stronger.