Question 1

Calculate the concentration of bicarbonate ion,HCO₃^-,in a 0.010 M H₂CO₃ solution that has the stepwise dissociation constants K_a1 = 4.3 × 10^-7 and K_a2 = 5.6 × 10^-11.

Accepted Answer

We can use the following equation to calculate the concentration of bicarbonate ion in the solution: [HCO3-] = [H2CO3]/(1 + K1/[H+] + K1K2/[H+]^2) where [H2CO3] is the concentration of dissolved carbonic acid, K1 and K2 are the stepwise dissociation constants, [H+] is the concentration of hydrogen ions (determined by the pH of the solution). First, we need to determine the concentration of dissolved carbonic acid: [H2CO3] = [CO2]*k where [CO2] is the concentration of carbon dioxide in the solution, and k is the solubility constant for CO2 in water (0.032 M/atm at 25°C). Assume that the solution is open to the atmosphere (i.e. aCO2 = 0.0004 atm). [CO2] = aCO2*PCO2 where PCO2 is the partial pressure of CO2, which is taken to be 0.0004 atm. [CO2] = (0.0004 atm)*(0.032 M/atm) = 1.28 x 10^-5 M [H2CO3] = (1.28 x 10^-5 M)*k = (1.28 x 10^-5 M)*(0.032 M/atm) = 4.1 x 10^-7 M Now we can plug in the values for K1 and K2, as well as the pH of the solution (determined by the dissociation of H2S, HA of H₂): [H+] = sqrt(KaKa/[HS-]) [H+] = sqrt((10^-7)*(10^-13)/0.010) [H+] = 3.16 x 10^-11 M [HCO3-] = [H2CO3]/(1 + K1/[H+] + K1K2/[H+]^2) [HCO3-] = (4.1 x 10^-7 M)/(1 + (4.3 x 10^-7 M)/(3.16 x 10^-11 M) + (4.3 x 10^-7 M)*(5.6 x 10^-11 M)/(3.16 x 10^-11 M)^2) [HCO3-] = 6.6 x 10^-6 M Therefore, the concentration of bicarbonate ion is 6.6 x 10^-6 M, which corresponds to answer choice A.

Question 2

Dihydrogen phosphate H₂PO₄^-,has an acid dissociation constant of 6.2 × 10^-7.What is the conjugate base of H₂PO₄^- and what is its base dissociation constant?

Accepted Answer

The conjugate base of dihydrogen phosphate (H₂PO₄⁻) is hydrogen phosphate (HPO₄²⁻). The base dissociation constant (Kb) can be calculated using the relation Kw = Ka * Kb, where Kw is the ion product of water (1.0 × 10⁻¹⁴ at 25°C). Given Ka = 6.2 × 10⁻⁷, Kb = Kw / Ka = (1.0 × 10⁻¹⁴) / (6.2 × 10⁻⁷) = 1.6 × 10⁻⁸.

Question 3

What is the pH of a 0.10 M H₂Se solution that has the stepwise dissociation constants K_a1 = 1.3 × 10^-4 and K_a2 = 1.0 × 10^-11?

Accepted Answer

The pH of the solution can be approximated using only the first dissociation constant (Ka1 = 1.3 × 10^-4) because the second dissociation constant (Ka2 = 1.0 × 10^-11) is much smaller and contributes negligibly to the overall H+ concentration. For a weak acid HA dissociating as HA ⇌ H+ + A^-, the pH can be estimated using the formula pH = 1/2(pKa - log[HA]), where [HA] is the concentration of the acid. Substituting the given values, pKa1 = -log(1.3 × 10^-4) ≈ 3.89, and [HA] = 0.10 M, we get pH ≈ 1/2(3.89 - log(0.10)) = 1/2(3.89 + 1) = 2.44.

Question 4

What is the relationship between K_a and K_b at 25°C for a conjugate acid base pair?

Accepted Answer

The product of the dissociation constants (Ka for the acid and Kb for the base) of a conjugate acid-base pair is equal to the ion product of water (Kw), which is 1 × 10^-14 at 25°C.

Question 5

Calculate the pH of a 0.20 M H₂SO₃ solution that has the stepwise dissociation constants K_a1 = 1.5 × 10^-2 and K_a2 = 6.3 × 10^-8.

Accepted Answer

First, we can write out the chemical equation for the dissociation of H₂SOH as follows: H₂SOH ⇌ H+ + ₂SO- The equilibrium constant for this reaction is K_a1 = [H+][₂SO-]/[H₂SOH], which we can rearrange to obtain an expression for [H+] in terms of the concentration of H₂SOH: [H+] = K_a1[H₂SOH]/[₂SO-] At an initial concentration of 0.20 M, the concentration of H₂SOH is also 0.20 M, so we can substitute these values into the equation above to obtain: [H+] = (1.5 × 10^-2)(0.20 M)/(0.20 M) = 1.5 × 10^-2 Next, we need to consider the dissociation of ₂SO-. Writing out the chemical equation as before, we have: ₂SO- ⇌ H+ + ₃SO2- The equilibrium constant for this reaction is K_a2 = [H+][₃SO2-]/[₂SO-], which we can rearrange to obtain an expression for [H+] in terms of the concentration of ₂SO-: [H+] = K_a2[₂SO-]/[₃SO2-] Using the law of mass action, we know that the concentration of ₃SO2- is equal to the concentration of H+ that was produced in the first step (since one mole of H+ is generated for every mole of ₂SO- that reacts), so we can substitute 1.5 × 10^-2 for [H+] in the equation above to obtain: [H+] = (6.3 × 10^-8)(0.20 M)/(1.5 × 10^-2) = 1.32 Therefore, the pH of the solution is 1.32.

Question 6

How many grams of pyridine are there in 100 mL of an aqueous solution that has a pH of 9.00? The K_b for pyridine is 1.9 × 10^-9 and the equation of interest is C₅H₅N(aq)+ H₂O(l)⇌ C₅H₅NH⁺(aq)+ OH^-(aq).

Accepted Answer

First, we need to use the given equation and the Ka value to find the concentration of pyridine in the solution.

Ka = [C5H5NHS+][OH-]/[C5H5N]

At pH 9, [OH-] = 1.0 x 10^-5 M

Let x be the concentration of pyridine in M

1.9 x 10^-9 = (x^2)/(x + 1.0 x 10^-5)

Solving for x, we get x = 1.22 x 10^-4 M

Now, we can use the density of water (1 g/mL) and the molar mass of pyridine (79.1 g/mol) to convert the concentration to grams per mL and then multiply by 100 mL to get the total grams of pyridine:

1.22 x 10^-4 M * 79.1 g/mol * 1 mL/1 g = 0.00965 g/mL

0.00965 g/mL * 100 mL = 0.965 g

However, we need to account for the fact that pyridine is not the only component of the solution - it is dissolved in water and there may be some pyridine hydroxide present. To do this, we need to use the Henderson-Hasselbalch equation:

pH = pKa + log([C5H5NHS+]/[C5H5N])

At pH 9, we can assume that [C5H5N] is much larger than [C5H5NHS+], so we can simplify:

9.00 = 8.72 + log([C5H5NHS+]/[C5H5N])

0.28 = log([C5H5NHS+]/[C5H5N])

[C5H5NHS+]/[C5H5N] = 1.93

This means that for every 1.93 moles of pyridine, there is 1 mole of pyridine hydroxide. So we can adjust our concentration of pyridine accordingly:

1.22 x 10^-4 M / 2.93 = 4.15 x 10^-5 M

And now we can calculate the grams of pyridine:

4.15 x 10^-5 M * 79.1 g/mol * 1 mL/1 g = 0.00329 g/mL

0.00329 g/mL * 100 mL = 0.329 g

Therefore, the answer is closest to B) 0.42 g.

Question 7

Which of the following can be classified as a weak base?

Accepted Answer

Both CH3NH2 (methylamine) and NH2OH (hydroxylamine) can act as weak bases because they can accept protons (H+ ions) due to the presence of a lone pair of electrons on the nitrogen atom.

Question 8

What is the selenide ion concentration [Se^2-] for a 0.100 M H₂Se solution that has the stepwise dissociation constants of K_a1 = 1.3 × 10^-4 and K_a2 = 1.0 × 10^-11?

Accepted Answer

The answer of What is the selenide ion concentration [Se^2-]...

Question 9

What is the pH of a 0.100 M NH₃ solution that has K_b = 1.8 × 10^-5? The equation for the dissociation of NH₃ is NH₃(aq)+ H₂O(l)⇌ NH₄⁺(aq)+ OH^-(aq).

Accepted Answer

The dissociation of NH4+ in water is NH4+ + H2O ⇌ NH3 + H3O+. The given $K_b$ for NH3 is $1.8 \times 10^{-5}$. To find the pH, we first calculate the $pOH$ using the formula $pOH = \frac{1}{2}(pK_w - \log K_b - \log [NH_4^+])$, where $pK_w = 14$ at 25°C. Substituting the given values, $pOH = \frac{1}{2}(14 - \log(1.8 \times 10^{-5}) - \log(0.100))$. This calculation gives a $pOH$ around 2.87. Since $pH + pOH = 14$, the $pH = 14 - pOH = 14 - 2.87 = 11.13$.

Question 10

What is the pH of a 0.30 M pyridine solution that has a K_b = 1.9 × 10^-9? The equation for the dissociation of pyridine is C₅H₅N(aq)+ H₂O(l)⇌ C₅H₅NH⁺(aq)+ OH^-(aq).

Accepted Answer

First, we need to find the equilibrium constant (Kw) for the dissociation of pyridine:
Kw = [C5H5NH+][OH-]/[C5H5N]
We can write an expression for the equilibrium concentrations in terms of the initial pyridine concentration (0.30 M) and the extent of dissociation (α):
[C5H5NH+] = α × 0.30 M
[OH-] = α × Kw/PNH+ = α × Kw/(0.30-α) M 
where PNH+ is the partial pressure of pyridine.

Substituting these expressions into the Kw equation and simplifying, we get:
Kw = α^2 × (0.30-α) / (1-α) = 1.9 × 10^-9

Solving this quadratic equation for α gives us:
α^2 - 0.30α + 1.9 × 10^-9 = 0
Using the quadratic formula, we get:
α = 0.00000103 or α = 0.29999897

Since α is much smaller than 1, we can assume that the concentration of pyridine is essentially unchanged by the dissociation, and therefore [C5H5N] ≈ 0.30 M.

The concentration of [OH-] is given by:
[OH-] = α × Kw/(0.30-α) M = 0.00000103 × 1.9 × 10^-9 / (0.30-0.00000103) M ≈ 1.0 × 10^-5 M
The pH is defined as the negative logarithm of the [H+] concentration, which can be calculated using the Kw equation:
Kw = [H+][OH-] = (1.0 × 10^-5) × [H+]
[H+] = Kw/[OH-] = 1.0 × 10^-14 / 1.0 × 10^-5 M = 1.0 × 10^-9 M
pH = -log[H+] ≈ 9.38

Therefore, the correct answer is C.

Question 11

Calculate the pH of a 0.020 M carbonic acid solution,H₂CO₃(aq),that has the stepwise dissociation constants K_a1 = 4.3 × 10^-7 and K_a2 = 5.6 × 10^-11.

Accepted Answer

Given the stepwise dissociation constants for carbonic acid (H2CO3), $K_{a1} = 4.3 \times 10^{-7}$ and $K_{a2} = 5.6 \times 10^{-11}$, we can assume that the first dissociation step contributes significantly more to the acidity of the solution than the second step due to the much higher $K_{a1}$ value. Therefore, we can approximate the pH of the solution by considering only the first dissociation step.The dissociation of carbonic acid can be represented as:$$H_2CO_3 \rightleftharpoons H^+ + HCO_3^-$$Using the acid dissociation constant, $K_{a1}$, we can set up the expression:$$K_{a1} = \frac{[H^+][HCO_3^-]}{[H_2CO_3]}$$Assuming that the concentration of $H^+$ and $HCO_3^-$ produced is $x$ and the initial concentration of $H_2CO_3$ is 0.020 M, the equation simplifies to:$$4.3 \times 10^{-7} = \frac{x^2}{0.020 - x}$$Given that $x$ is very small compared to 0.020 M, we can approximate the denominator to 0.020 M, leading to:$$x^2 = 4.3 \times 10^{-7} \times 0.020$$$$x^2 = 8.6 \times 10^{-9}$$$$x = \sqrt{8.6 \times 10^{-9}}$$$$x \approx 9.27 \times 10^{-5}$$The concentration of $H^+$ ions, $x$, is approximately $9.27 \times 10^{-5}$ M. To find the pH:$$pH = -\log[H^+]$$$$pH = -\log(9.27 \times 10^{-5})$$$$pH \approx 4.03$$Therefore, the pH of the 0.020 M carbonic acid solution is approximately 4.03.

Question 12

What is the second stepwise equilibrium constant expression for phosphoric acid H₃PO₄?

Accepted Answer

The second stepwise equilibrium for phosphoric acid involves the dissociation of the hydrogen phosphate ion (HPO4^2-) into the phosphate ion (PO4^3-) and a hydrogen ion (H+). The correct expression includes the concentrations of the products (H+ and PO4^3-) divided by the concentration of the reactant (HPO4^2-), which matches choice D.

Question 13

Ammonia NH₃,has a base dissociation constant of 1.8 × 10^-5.What is the conjugate acid of ammonia and what is its acid dissociation constant?

Accepted Answer

The answer of Ammonia NH₃,has a base dissociation constant of...

Question 14

Which of the following are weak diprotic acids?

Accepted Answer

Carbonic acid, oxalic acid, and sulfurous acid are all weak diprotic acids, meaning they can donate two protons but do not completely dissociate in solution.

Question 15

Aniline, (C₆H₅NH₂,K_b = 4.3 × 10^-10 at 25°C)is an industrially important amine used in the making of dyes.Determine the pH of an aniline solution made by dissolving 3.90 g of aniline in enough water to make 100 mL of solution.

Accepted Answer

Step 1: Write the equation for the dissociation of aniline.
C6H5NH2(aq) + H2O(l) ↔ C6H5NH3+(aq) + OH-(aq)

Step 2: Write the expression for the equilibrium constant (Kb) for the above reaction.
Kb = [C6H5NH3+][OH-]/[C6H5NH2]

Step 3: Calculate the concentration of aniline in the solution.
Moles of aniline = 3.90 g / 93.13 g/mol = 4.18 × 10^-2 mol
Volume of solution = 100 mL = 0.1 L
Concentration of aniline = moles/volume = 4.18 × 10^-2 mol / 0.1 L = 0.418 M

Step 4: Calculate the concentration of hydroxide ions (OH-).
Kb = [C6H5NH3+][OH-]/[C6H5NH2]
Assume that initially, [OH-] = 0 (since aniline is a weak base and will not completely dissociate into ions).
Kb = [C6H5NH3+][OH-]/[C6H5NH2]
4.3 × 10^-10 = x^2 / (0.418 - x)
Assuming x << 0.418, then (0.418 - x) ≈ 0.418.
x = [OH-] = 6.416 × 10^-6 M

Step 5: Calculate the pOH and pH of the solution.
pOH = -log[OH-] = -log(6.416 × 10^-6) = 5.19
pH = 14 - pOH = 8.81

Step 6: Apply the correction factor for the ionization of water.
Since aniline is a weak base, we need to correct for the ionization of water. We can assume that [H+] = [OH-] = 6.416 × 10^-6 M (at 25°C). Therefore, pKa = 14 - pKb = 14 - 4.87 = 9.13
pH = pKa + log([C6H5NH3+]/[C6H5NH2]) = 9.13 + log(0.017) = 9.13 + (-1.77) = 7.36
But remember that we need to apply the correction factor for the ionization of water.
pH = 7.36 + log([H+]/[OH-])
Since [H+] = [OH-] = 6.416 × 10^-6 M,
pH = 7.36 + log(1) = 7.36

Therefore, the pH of the aniline solution is 7.36. The closest answer choice is B) 9.13, but this is not the correct answer.

Question 16

The percent dissociation of acetic acid changes as the concentration of the acid decreases.A 100-fold decrease in acetic acid concentration results in a ________ fold ________ in the percent dissociation.

Accepted Answer

The percent dissociation of weak acids like acetic acid increases as the concentration decreases. This is because the dissociation of acetic acid in water is an equilibrium process, and according to Le Chatelier's principle, decreasing the concentration of the acid shifts the equilibrium towards more dissociation to increase the concentration of the acid. A 100-fold decrease in concentration significantly increases the percent dissociation, but the relationship is not directly proportional, making a 10-fold increase in percent dissociation a reasonable approximation.

Question 17

Which of the following salts are acidic?

Accepted Answer

The salts CuCl₂ and AlCl₃ contain metal cations that can hydrolyze in water, releasing H+ ions and making the solution acidic. NH₄Cl does not contain a metal cation that can hydrolyze, so it is not acidic.

Question 18

Determine the ammonia concentration of an aqueous solution that has a pH of 11.50.The equation for the dissociation of NH₃ (K_b = 1.8 × 10^-5)is NH₃(aq)+ H₂O(l)⇌ NH₄⁺(aq)+ OH^-(aq).

Accepted Answer

The dissociation of ammonia in water is given by the following equation:

NH3 + H2O ⇌ NH4+ + OH-

The equilibrium constant for this equation, known as the base dissociation constant (Kb), is given by:

Kb = [NH4+][OH-]/[NH3]

Since pH + pOH = 14 and pOH can be calculated from the given pH, we can solve for [OH-]:

pOH = 14 - pH = 2.50
[OH-] = 10^-pOH = 3.16 × 10^-3 M

Using the equation for Kb and the value of [OH-], we can solve for [NH3]:

Kb = [NH4+][OH-]/[NH3]
1.8 × 10^-5 = x^2 / (0.050 - x)

Solving for x gives:

x = [NH3] = 0.55 M

Therefore, the answer is B).

Question 19

Acetic acid CH₃COOH,has an acid dissociation constant of 1.8 × 10^-5.What is the conjugate base of acetic acid and what is its base dissociation constant?

Accepted Answer

The conjugate base of acetic acid (CH3COOH) is acetate ion (CH3COO-), and the base dissociation constant (Kb) can be calculated using the relation Kw = Ka * Kb, where Kw is the ion product of water (1.0 × 10^-14 at 25°C). Given Ka = 1.8 × 10^-5, Kb = Kw / Ka = (1.0 × 10^-14) / (1.8 × 10^-5) = 5.6 × 10^-10.

Question 20

Methylamine CH₃NH₂,has a base dissociation constant of 3.7 × 10^-4.What is the conjugate acid of methylamine and what is its acid dissociation constant?

Accepted Answer

The conjugate acid of methylamine (CH3NH2) is formed by adding a proton (H+), resulting in CH3NH3+. The acid dissociation constant (Ka) for the conjugate acid can be found using the relationship Kw = Ka * Kb, where Kw is the ion product of water (1.0 × 10^-14 at 25°C). Given that the base dissociation constant (Kb) for methylamine is 3.7 × 10^-4, we can calculate Ka as (1.0 × 10^-14) / (3.7 × 10^-4) = 2.7 × 10^-11.

Quiz 14: Aqueous Equilibria: Acids and Bases

Calculate the concentration of bicarbonate ion,HCO₃^-,in a 0.010 M H₂CO₃ solution that has the stepwise dissociation constants K_a1 = 4.3 × 10^-7 and K_a2 = 5.6 × 10^-11.

Dihydrogen phosphate H₂PO₄^-,has an acid dissociation constant of 6.2 × 10^-7.What is the conjugate base of H₂PO₄^- and what is its base dissociation constant?

What is the pH of a 0.10 M H₂Se solution that has the stepwise dissociation constants K_a1 = 1.3 × 10^-4 and K_a2 = 1.0 × 10^-11?

What is the relationship between K_a and K_b at 25°C for a conjugate acid base pair?

Calculate the pH of a 0.20 M H₂SO₃ solution that has the stepwise dissociation constants K_a1 = 1.5 × 10^-2 and K_a2 = 6.3 × 10^-8.

How many grams of pyridine are there in 100 mL of an aqueous solution that has a pH of 9.00? The K_b for pyridine is 1.9 × 10^-9 and the equation of interest is C₅H₅N(aq) + H₂O(l) ⇌ C₅H₅NH⁺(aq) + OH^-(aq) .

Which of the following can be classified as a weak base?

What is the selenide ion concentration [Se^2-] for a 0.100 M H₂Se solution that has the stepwise dissociation constants of K_a1 = 1.3 × 10^-4 and K_a2 = 1.0 × 10^-11?

What is the pH of a 0.100 M NH₃ solution that has K_b = 1.8 × 10^-5? The equation for the dissociation of NH₃ is NH₃(aq) + H₂O(l) ⇌ NH₄⁺(aq) + OH^-(aq) .

What is the pH of a 0.30 M pyridine solution that has a K_b = 1.9 × 10^-9? The equation for the dissociation of pyridine is C₅H₅N(aq) + H₂O(l) ⇌ C₅H₅NH⁺(aq) + OH^-(aq) .

Calculate the pH of a 0.020 M carbonic acid solution,H₂CO₃(aq) ,that has the stepwise dissociation constants K_a1 = 4.3 × 10^-7 and K_a2 = 5.6 × 10^-11.

What is the second stepwise equilibrium constant expression for phosphoric acid H₃PO₄?

Ammonia NH₃,has a base dissociation constant of 1.8 × 10^-5.What is the conjugate acid of ammonia and what is its acid dissociation constant?

Which of the following are weak diprotic acids?

Aniline, (C₆H₅NH₂,K_b = 4.3 × 10^-10 at 25°C) is an industrially important amine used in the making of dyes.Determine the pH of an aniline solution made by dissolving 3.90 g of aniline in enough water to make 100 mL of solution.

The percent dissociation of acetic acid changes as the concentration of the acid decreases.A 100-fold decrease in acetic acid concentration results in a fold in the percent dissociation.

Which of the following salts are acidic?

Determine the ammonia concentration of an aqueous solution that has a pH of 11.50.The equation for the dissociation of NH₃ (K_b = 1.8 × 10^-5) is NH₃(aq) + H₂O(l) ⇌ NH₄⁺(aq) + OH^-(aq) .

Acetic acid CH₃COOH,has an acid dissociation constant of 1.8 × 10^-5.What is the conjugate base of acetic acid and what is its base dissociation constant?

Methylamine CH₃NH₂,has a base dissociation constant of 3.7 × 10^-4.What is the conjugate acid of methylamine and what is its acid dissociation constant?

Quiz 14: Aqueous Equilibria: Acids and Bases

Calculate the concentration of bicarbonate ion,HCO3-,in a 0.010 M H2CO3 solution that has the stepwise dissociation constants Ka1 = 4.3 × 10-7 and Ka2 = 5.6 × 10-11.

Dihydrogen phosphate H2PO4-,has an acid dissociation constant of 6.2 × 10-7.What is the conjugate base of H2PO4- and what is its base dissociation constant?

What is the pH of a 0.10 M H2Se solution that has the stepwise dissociation constants Ka1 = 1.3 × 10-4 and Ka2 = 1.0 × 10-11?

What is the relationship between Ka and Kb at 25°C for a conjugate acid base pair?

Calculate the pH of a 0.20 M H2SO3 solution that has the stepwise dissociation constants Ka1 = 1.5 × 10-2 and Ka2 = 6.3 × 10-8.

How many grams of pyridine are there in 100 mL of an aqueous solution that has a pH of 9.00? The Kb for pyridine is 1.9 × 10-9 and the equation of interest is C5H5N(aq) + H2O(l) ⇌ C5H5NH+(aq) + OH-(aq) .

Which of the following can be classified as a weak base?

What is the selenide ion concentration [Se2-] for a 0.100 M H2Se solution that has the stepwise dissociation constants of Ka1 = 1.3 × 10-4 and Ka2 = 1.0 × 10-11?

What is the pH of a 0.100 M NH3 solution that has Kb = 1.8 × 10-5? The equation for the dissociation of NH3 is NH3(aq) + H2O(l) ⇌ NH4+(aq) + OH-(aq) .

What is the pH of a 0.30 M pyridine solution that has a Kb = 1.9 × 10-9? The equation for the dissociation of pyridine is C5H5N(aq) + H2O(l) ⇌ C5H5NH+(aq) + OH-(aq) .

Calculate the pH of a 0.020 M carbonic acid solution,H2CO3(aq) ,that has the stepwise dissociation constants Ka1 = 4.3 × 10-7 and Ka2 = 5.6 × 10-11.

What is the second stepwise equilibrium constant expression for phosphoric acid H3PO4?

Ammonia NH3,has a base dissociation constant of 1.8 × 10-5.What is the conjugate acid of ammonia and what is its acid dissociation constant?

Which of the following are weak diprotic acids?

Aniline, (C6H5NH2,Kb = 4.3 × 10-10 at 25°C) is an industrially important amine used in the making of dyes.Determine the pH of an aniline solution made by dissolving 3.90 g of aniline in enough water to make 100 mL of solution.

The percent dissociation of acetic acid changes as the concentration of the acid decreases.A 100-fold decrease in acetic acid concentration results in a ________ fold ________ in the percent dissociation.

Which of the following salts are acidic?

Determine the ammonia concentration of an aqueous solution that has a pH of 11.50.The equation for the dissociation of NH3 (Kb = 1.8 × 10-5) is NH3(aq) + H2O(l) ⇌ NH4+(aq) + OH-(aq) .

Acetic acid CH3COOH,has an acid dissociation constant of 1.8 × 10-5.What is the conjugate base of acetic acid and what is its base dissociation constant?

Methylamine CH3NH2,has a base dissociation constant of 3.7 × 10-4.What is the conjugate acid of methylamine and what is its acid dissociation constant?

Calculate the concentration of bicarbonate ion,HCO₃^-,in a 0.010 M H₂CO₃ solution that has the stepwise dissociation constants K_a1 = 4.3 × 10^-7 and K_a2 = 5.6 × 10^-11.

Dihydrogen phosphate H₂PO₄^-,has an acid dissociation constant of 6.2 × 10^-7.What is the conjugate base of H₂PO₄^- and what is its base dissociation constant?

What is the pH of a 0.10 M H₂Se solution that has the stepwise dissociation constants K_a1 = 1.3 × 10^-4 and K_a2 = 1.0 × 10^-11?

What is the relationship between K_a and K_b at 25°C for a conjugate acid base pair?

Calculate the pH of a 0.20 M H₂SO₃ solution that has the stepwise dissociation constants K_a1 = 1.5 × 10^-2 and K_a2 = 6.3 × 10^-8.

How many grams of pyridine are there in 100 mL of an aqueous solution that has a pH of 9.00? The K_b for pyridine is 1.9 × 10^-9 and the equation of interest is C₅H₅N(aq) + H₂O(l) ⇌ C₅H₅NH⁺(aq) + OH^-(aq) .

What is the selenide ion concentration [Se^2-] for a 0.100 M H₂Se solution that has the stepwise dissociation constants of K_a1 = 1.3 × 10^-4 and K_a2 = 1.0 × 10^-11?

What is the pH of a 0.100 M NH₃ solution that has K_b = 1.8 × 10^-5? The equation for the dissociation of NH₃ is NH₃(aq) + H₂O(l) ⇌ NH₄⁺(aq) + OH^-(aq) .

What is the pH of a 0.30 M pyridine solution that has a K_b = 1.9 × 10^-9? The equation for the dissociation of pyridine is C₅H₅N(aq) + H₂O(l) ⇌ C₅H₅NH⁺(aq) + OH^-(aq) .

Calculate the pH of a 0.020 M carbonic acid solution,H₂CO₃(aq) ,that has the stepwise dissociation constants K_a1 = 4.3 × 10^-7 and K_a2 = 5.6 × 10^-11.

What is the second stepwise equilibrium constant expression for phosphoric acid H₃PO₄?

Ammonia NH₃,has a base dissociation constant of 1.8 × 10^-5.What is the conjugate acid of ammonia and what is its acid dissociation constant?

Aniline, (C₆H₅NH₂,K_b = 4.3 × 10^-10 at 25°C) is an industrially important amine used in the making of dyes.Determine the pH of an aniline solution made by dissolving 3.90 g of aniline in enough water to make 100 mL of solution.

The percent dissociation of acetic acid changes as the concentration of the acid decreases.A 100-fold decrease in acetic acid concentration results in a fold in the percent dissociation.

Determine the ammonia concentration of an aqueous solution that has a pH of 11.50.The equation for the dissociation of NH₃ (K_b = 1.8 × 10^-5) is NH₃(aq) + H₂O(l) ⇌ NH₄⁺(aq) + OH^-(aq) .

Acetic acid CH₃COOH,has an acid dissociation constant of 1.8 × 10^-5.What is the conjugate base of acetic acid and what is its base dissociation constant?

Methylamine CH₃NH₂,has a base dissociation constant of 3.7 × 10^-4.What is the conjugate acid of methylamine and what is its acid dissociation constant?