Question 1

What is the pH of an aqueous solution at 25.0 °C in which [H⁺] is 0.00250 M?

Accepted Answer

The pH of a solution is calculated using the formula pH = -log[H⁺]. For [H⁺] = 0.00250 M, pH = -log(0.00250) = 2.60.

Question 2

What is the pH of an aqueous solution at 25.0 °C that contains 3.98 × 10^-⁹M hydroxide ion?

Accepted Answer

The answer of What is the pH of an aqueous...

Question 3

An aqueous solution contains 0.100 M NaOH at 25.0 °C. The pH of the solution is _ .

Accepted Answer

The pH of a solution is calculated using the formula pH = 14 - pOH. Since NaOH is a strong base, it dissociates completely in water, giving a pOH of -log(0.100) = 1. Therefore, the pH is 14 - 1 = 13.0.

Question 4

The acid- dissociation constants of phosphoric acid (H₃PO₄) are K_a1= 7.5 × 10^-³, K_a2= 6.2 × 10^-⁸, and K_a3= 4.2 × 10^-¹³at 25.0 °C. What is the pH of a 2.5 M aqueous solution of phosphoric acid?

Accepted Answer

The answer of The acid- dissociation constants of phosphoric acid...

Question 5

The acid- dissociation constant for chlorous acid, HClO₂, at 25.0 °C is 1.0 × 10^-². Calculate the concentration of H⁺if the initial concentration of acid is 0.10 M.

Accepted Answer

The concentration of H⁺ can be calculated using the formula for the acid dissociation constant: Ka = [H⁺][ClO₂⁻]/[HClO₂]. Assuming x is the concentration of H⁺, Ka = x²/(0.10 - x). Given Ka = 1.0 × 10⁻², solving for x gives approximately 2.7 × 10⁻² M.

Question 6

An aqueous solution contains 0.050 M of methylamine. The concentration of hydroxide ion in this solution is _ M. K_bfor methylamine is 4.4 × 10^-⁴.

Accepted Answer

The concentration of hydroxide ion can be calculated using the formula for Kb: [OH-] = √(Kb × [methylamine]) = √(4.4 × 10^-4 × 0.050) ≈ 4.5 × 10^-3 M.

Question 7

HZ is a weak acid. An aqueous solution of HZ is prepared by dissolving 0.020 mol of HZ in sufficient water to yield 1.0 L of solution. The pH of the solution was 4.93 at 25.0 °C. The K_aof HZ is .

Accepted Answer

The answer of HZ is a weak acid. An aqueous...

Question 8

Calculate the concentration (in M) of hydronium ions in a solution at 25.0 °C with a pOH of 4.223.

Accepted Answer

The pH can be calculated using pH + pOH = 14. Therefore, pH = 14 - 4.223 = 9.777. The concentration of hydronium ions is then 10^-pH = 10^-9.777 = 1.67 × 10^-10 M.

Question 9

The K_aof hydrazoic acid (HN₃) is 1.9 × 10^-⁵at 25.0 °C. What is the pH of a 0.35 M aqueous solution of HN₃?

Accepted Answer

To find the pH, use the formula: pH = -log[H⁺]. First, set up the expression for Kₐ: Kₐ = [H⁺][N₃⁻]/[HN₃]. Assume [H⁺] = x, then Kₐ = x²/(0.35 - x) ≈ x²/0.35. Solve for x: x = √(Kₐ × 0.35) = √(1.9 × 10⁻⁵ × 0.35) ≈ 2.6 × 10⁻³. Therefore, pH = -log(2.6 × 10⁻³) ≈ 2.6.

Question 10

What is the concentration (in M) of hydronium ions in a solution at 25.0 °C with pH = 4.282?

Accepted Answer

The concentration of hydronium ions [H₃O⁺] can be found using the formula [H₃O⁺] = 10^(-pH). For pH = 4.282, [H₃O⁺] = 10^(-4.282) = 5.22 × 10⁻⁵ M.

Question 11

The acid- dissociation constant, K_a, for gallic acid is 4.57 × 10^-³. What is the base- dissociation constant, K_b, for the gallate ion?

Accepted Answer

The relationship between K_a and K_b is given by K_w = K_a × K_b, where K_w is the ion-product constant of water (1.0 × 10^-14 at 25°C). Solving for K_b gives K_b = K_w / K_a = (1.0 × 10^-14) / (4.57 × 10^-3) = 2.19 × 10^-12.

Question 12

What is the pOH of a 0.0150 M solution of barium hydroxide?

Accepted Answer

The answer of What is the pOH of a 0.0150...

Question 13

What is the pH of a 0.0150 M aqueous solution of barium hydroxide?

Accepted Answer

Barium hydroxide, Ba(OH)₂, is a strong base that dissociates completely in water to produce two hydroxide ions (OH⁻) per formula unit. Therefore, a 0.0150 M solution of Ba(OH)₂ produces 0.0300 M OH⁻. The pOH is calculated as -log(0.0300) = 1.52. The pH is then 14 - 1.52 = 12.48, which rounds to 12.5.

Question 14

K_afor HF is 7.0 × 10^-⁴. K_bfor the fluoride ion is _ _.

Accepted Answer

The relationship between K_a and K_b for a conjugate acid-base pair is K_w = K_a × K_b, where K_w is the ion-product constant of water (1.0 × 10^-14). Solving for K_b gives K_b = K_w / K_a = 1.0 × 10^-14 / 7.0 × 10^-4 = 1.4 × 10^-11.

Question 15

The acid- dissociation constants of sulfurous acid (H₂SO₃) are K_a1= 1.7 × 10^-²and K_a2= 6.4 × 10^-⁸at 25.0 °C. Calculate the pH of a 0.163 M aqueous solution of sulfurous acid.

Accepted Answer

The correct answer is A because the pH of a 0.163 M aqueous solution of sulfurous acid can be calculated using the dissociation constants and the ICE table. The pH is found to be 1.4.

Question 16

The conjugate base of HSO₄^-is

Accepted Answer

The conjugate base of HSO₄⁻ is formed by removing a proton (H⁺), resulting in SO₄²⁻.

Question 17

The K_aof hypochlorous acid (HClO) is 3.00 × 10^-⁸at 25.0 °C. Calculate the pH of a 0.0385 M hypochlorous acid solution.

Accepted Answer

The equation for the dissociation of hypochlorous acid is:

HClO (aq) ⇌ H+ (aq) + ClO– (aq)

The acid dissociation constant, Ka, is given by:

Ka = [H+][ClO–]/[HClO]

At equilibrium, the concentrations of H+ and ClO– will be equal, so we can write:

Ka = [H+]2/[HClO]

Rearranging this equation, we get:

[H+] = sqrt(Ka*[HClO])

The value of Ka for HClO is given by the expression:

Ka = [H+][ClO−]/[HClO] = 3.00 × 10−8

[HClO] is given as 0.0385 M. Substituting these values into the equation for [H+], we get:

[H+] = sqrt(3.00 × 10−8 * 0.0385) = 1.43 × 10−4

The pH is defined as the negative logarithm (base 10) of the [H+]:

pH = -log[H+] = -log(1.43 × 10−4) = 4.47

Therefore, the best choice is E, with a pH value of 4.47.

Question 18

The pH of a 0.10 M solution of a weak base is 9.82. What is the K_bfor this base?

Accepted Answer

The answer of The pH of a 0.10 M solution...

Question 19

K_bfor NH₃is 1.8 × 10^-⁵. What is the pH of a 0.35 M aqueous solution of NH₄Cl at 25.0 °C?

Accepted Answer

NH₄Cl is an acidic salt, and its pH can be calculated using the Kb of NH₃ to find the Ka of NH₄⁺. The pH of the solution is approximately 4.9.

Question 20

What is the conjugate base of OH^-?

Accepted Answer

The conjugate base of OH⁻ is formed by removing a proton (H⁺) from it, resulting in O²⁻.

Quiz 16: Acid-Base Equilibria

What is the pH of an aqueous solution at 25.0 °C in which [H+] is 0.00250 M?

What is the pH of an aqueous solution at 25.0 °C that contains 3.98 × 10- 9 M hydroxide ion?

An aqueous solution contains 0.100 M NaOH at 25.0 °C. The pH of the solution is _ .

The acid- dissociation constants of phosphoric acid (H3PO4) are Ka1 = 7.5 × 10- 3, Ka2 = 6.2 × 10- 8, and Ka3 = 4.2 × 10- 13 at 25.0 °C. What is the pH of a 2.5 M aqueous solution of phosphoric acid?

The acid- dissociation constant for chlorous acid, HClO2, at 25.0 °C is 1.0 × 10- 2. Calculate the concentration of H+ if the initial concentration of acid is 0.10 M.

An aqueous solution contains 0.050 M of methylamine. The concentration of hydroxide ion in this solution is _ M. Kb for methylamine is 4.4 × 10- 4.

HZ is a weak acid. An aqueous solution of HZ is prepared by dissolving 0.020 mol of HZ in sufficient water to yield 1.0 L of solution. The pH of the solution was 4.93 at 25.0 °C. The Ka of HZ is .

Calculate the concentration (in M) of hydronium ions in a solution at 25.0 °C with a pOH of 4.223.

The Ka of hydrazoic acid (HN3) is 1.9 × 10- 5 at 25.0 °C. What is the pH of a 0.35 M aqueous solution of HN3?

What is the concentration (in M) of hydronium ions in a solution at 25.0 °C with pH = 4.282?

The acid- dissociation constant, Ka, for gallic acid is 4.57 × 10- 3. What is the base- dissociation constant, Kb, for the gallate ion?