Question 1

What is the percent dissociation of a benzoic acid solution with pH = 2.59? The acid dissociation constant for this monoprotic acid is 6.5 × 10^-5.

Accepted Answer

First, we need to calculate the concentration of H+ ions in the solution using the pH:

pH = -log[H+]
2.59 = -log[H+]
[H+] = 2.12 x 10^-3 M

Now, we can use the expression for the acid dissociation constant (Ka) to calculate the percent dissociation (α) of the acid:

Ka = [H+][A-]/[HA] = 6.5 x 10^-5
(assuming that the initial concentration of benzoic acid, [HA], is equal to the concentration of undissociated benzoic acid at equilibrium)

α = [H+]/Ka x 100%
α = (2.12 x 10^-3 M)/(6.5 x 10^-5) x 100%
α = 3.26%

Therefore, the percent dissociation of the benzoic acid solution is approximately 3.5%, which corresponds to answer choice (C).

Question 2

Benzoic acid (C₆H₅CO₂H = HBz)solutions are sometimes used in experiments to determine the molarity of a basic solution of unknown concentration.What is the pH of a 0.100 M solution of benzoic acid if K_a = 6.5 × 10^-5and the equilibrium equation of interest is HBz(aq)+ H₂O(l)⇌ H₃O⁺(aq)+ Bz^-(aq).

Accepted Answer

The correct answer is B because the pH of a 0.100 M solution of benzoic acid can be calculated using the equilibrium constant (Ka) and the equilibrium equation. The Ka value for benzoic acid is 6.5 × 10^-5, and the equilibrium equation is HBz(aq) + H2O(l) ⇌ H3O+(aq) + Bz-(aq). Using the Ka value and the equilibrium equation, we can calculate the concentration of H3O+ ions in the solution, which can then be used to calculate the pH.

Question 3

A tablet containing 500.0 mg of aspirin (acetylsalicylic acid or HC₉H₇O₄)was dissolved in enough water to make 100 mL of solution.Given that K_a = 3.0 × 10^-4 for aspirin,what is the pH of the solution?

Accepted Answer

To find the pH of the aspirin solution, we first calculate the molarity of aspirin in the solution. Since we have 500.0 mg of aspirin dissolved in 100 mL of solution, we convert the mass of aspirin to grams (500.0 mg = 0.500 g) and then to moles (molar mass of aspirin ≈ 180 g/mol), resulting in approximately 0.00278 moles of aspirin. This gives us a molarity of 0.00278 M in a 0.1 L solution. Using the given $K_a = 3.0 \times 10^{-4}$, we can set up the equation for the dissociation of aspirin (HA) in water: $K_a = \frac{[H^+][A^-]}{[HA]}$. Assuming $x$ is the concentration of $H^+$ ions (and also of $A^-$ since they are produced in a 1:1 ratio), we have $3.0 \times 10^{-4} = \frac{x^2}{0.00278}$. Solving for $x$ gives a $H^+$ concentration of approximately $2.9 \times 10^{-3}$ M. The pH is calculated using $-\log[H^+]$, which gives a pH close to 2.54.

Question 4

The pH of 0.150 M CH₃CO₂H,acetic acid,is 2.78.What is the value of K_a for acetic acid?

Accepted Answer

The Ka expression for acetic acid is:

Ka = [H+][C2H3O2-]/[HC2H3O2]

At the half-equivalence point, [H+] = [C2H3O2-] = [A-] and [HA] = [OH-]. Therefore, we can substitute these values into the Ka expression as follows:

Ka = [H+]^2/[HA]

From the given pH of 2.78, we can calculate [H+] as follows:

pH = -log[H+]
2.78 = -log[H+]
[H+] = 1.94 × 10^-3 M

Substituting this value and the initial concentration of acetic acid ([HA] = 0.150 M) into the Ka expression gives:

1.9 × 10^-5 = (1.94 × 10^-3)^2/(0.150 - 1.94 × 10^-3)

Solving for Ka gives:

Ka = 1.9 × 10^-5

Question 5

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

The equilibrium reactions and their corresponding equilibrium constants are: H₂CO3- ⇌ H+ + HCO3-, Ka1 = 4.3 × 10^-7 HCO3- ⇌ H+ + CO32-, Ka2 = 5.6 × 10^-11 Let's assume x is the concentration of HCO3- initially. Then, the concentration of H₂CO3- is (0.010 - x) because the two concentrations add up to the total bicarbonate concentration of 0.010 M. In the first equilibrium, we can write: Ka1 = [H+][HCO3-]/[H₂CO3-] = (x)(0.010 - x) / (0.010 - x) Simplifying yields: 4.3 × 10^-7 = x(0.010 - x) / (0.010 - x) 4.3 × 10^-7(0.010 - x) = x(0.010 - x) 0.000043 - 4.3 × 10^-7x = x^2 - 0.010x x^2 - 0.010x - 4.3 × 10^-7x + 0.000043 = 0 x^2 - 4.3 × 10^-7x - 0.01x + 0.000043 = 0 Solving this quadratic equation yields: x = 6.6 × 10^-5 M Therefore, the answer is (A), 6.6 × 10^-5 M.

Question 6

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 7

What is the strongest monoprotic acid of the following set if all the acids are at 0.100 M concentration?

Accepted Answer

The strength of an acid is determined by its dissociation constant (Ka). The higher the Ka value, the stronger the acid. Hydrofluoric acid (A) has the highest Ka value (3.5 × 10^-4), making it the strongest acid among the options listed.

Question 8

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 9

Vinegar is a 5.0% solution by weight of acetic acid (CH₃CO₂H)in water.Given that K_a = 1.8 × 10^-5 for acetic acid and assuming the density of vinegar to be 1.00 g/cm³,what is the pH of this vinegar solution?

Accepted Answer

First, we need to calculate the molarity of acetic acid in the solution:

Molarity = (5.0 g acetic acid / 60.05 g/mol) / (100 mL vinegar / 1000 mL) = 0.0832 M

Next, we can use the equilibrium expression for acetic acid in water:

CH3COOH + H2O ⇌ H3O+ + CH3COO-

Ka = [H3O+][CH3COO-] / [CH3COOH]

Since we have a weak acid, we can assume that [H3O+] = [CH3COO-]

Therefore, Ka = [H3O+]^2 / [CH3COOH]

Solving for [H3O+]:

[H3O+] = sqrt(Ka*[CH3COOH]) = sqrt(1.8E-5*0.0832) = 1.28E-3 M

Finally, we can calculate the pH:

pH = -log[H3O+] = 2.41

Question 10

What is the hydronium ion concentration of a 0.100 M acetic acid solution with a K_a = 1.8 × 10^-5? The equation for the dissociation of acetic acid is: CH₃CO₂H(aq)+ H₂O(l)⇌ H₃O⁺(aq)+ CH₃CO₂^-(aq).

Accepted Answer

The equilibrium constant (Ka) for the dissociation of acetic acid can be expressed as follows:
Ka = [H3O+][CH3COO-]/[CH3COOH]
We can assume that the concentration of CH3COO- = [H+] and substitute the given values:
1.8 × 10-5 = [H+]2/0.1
[H+]2 = 1.8 × 10-6
[H+] = sqrt(1.8 × 10-6)
[H+] = 1.34 × 10-3 M
Therefore, the hydronium ion concentration of the solution is 1.3 × 10-3 M, which is closest to option C.

Question 11

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 12

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 13

Calculate the pH of a 0.020 M carbonic acid solution,H₂CO₃(aq),that has the stepwise dissociation constants K_a1 = 4.3 × 10^-7and 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 14

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 15

Para-Aminobenzoic acid (PABA),p-H₂NC₆H₄(COOH),is used in some sunscreens and hair conditioning products.Calculate the pH of an aqueous solution with [PABA] = 0.030 M and K_a = 2.2 × 10^-5.

Accepted Answer

PABA is a weak acid, and its dissociation in water can be represented as HA ⇌ H+ + A-. The pH of the solution can be calculated using the formula pH = 1/2(pKa - log[HA]), where pKa = -log(Ka). Given Ka = 2.2 × 10^-5, pKa = -log(2.2 × 10^-5) ≈ 4.66. Substituting the values, pH = 1/2(4.66 - log(0.030)) ≈ 3.09.

Question 16

Potassium hydrogen phthalate (molar mass = 204.2 g/mol)is one of the most commonly used acids for standardizing solutions containing bases.KHP is a monoprotic weak acid with K_a = 3.91 × 10^-6.Calculate the pH of the solution that results when 0.40 g of KHP is dissolved in enough water to produce 25.0 mL of solution.

Accepted Answer

First, calculate the molarity of KHP in the solution: $ \frac{0.40 \, \text{g}}{204.2 \, \text{g/mol}} = 0.00196 \, \text{mol} $, and when dissolved in 25.0 mL (or 0.025 L) of water, the molarity is $ \frac{0.00196 \, \text{mol}}{0.025 \, \text{L}} = 0.0784 \, \text{M} $.Given that KHP is a weak acid with $ K_a = 3.91 \times 10^{-6} $, we can use the acid dissociation constant to find the pH. For a weak acid $ HA $, the dissociation is $ HA \rightleftharpoons H^+ + A^- $, and at equilibrium, $ K_a = \frac{[H^+][A^-]}{[HA]} $. Assuming $ x $ is the concentration of $ H^+ $ (and $ A^- $ since they are produced in a 1:1 ratio), and $ 0.0784 - x $ is the concentration of $ HA $ at equilibrium, we can set up the equation $ 3.91 \times 10^{-6} = \frac{x^2}{0.0784 - x} $.Given the small $ K_a $ value, we can approximate $ 0.0784 - x \approx 0.0784 $ to simplify the calculation, leading to $ x^2 = 3.91 \times 10^{-6} \times 0.0784 $. Solving for $ x $, which is the $ [H^+] $, we find $ x \approx 7.0 \times 10^{-4} $. The pH is then $ -\log(7.0 \times 10^{-4}) $, which is approximately 3.26.

Question 17

Vinegar is a 5.0% solution by weight of acetic acid (CH₃CO₂H)in water.Given that the pH for acetic acid is 2.41,the K_a = 1.8 × 10^-5and assuming the density of vinegar to be 1.00 g/cm³,what is the percent dissociation of acetic acid in vinegar?

Accepted Answer

The percent dissociation of acetic acid in vinegar can be calculated using the given information and the formula for percent dissociation. The dissociation constant (Ka) for acetic acid is given as 1.8 × 10^-5. The pH of the solution is 2.41, which means the concentration of H+ ions ([H+]) is 10^-2.41. Since the density of vinegar is 1.00 g/cm^3 and it is a 5.0% solution by weight, the concentration of acetic acid in the solution is 0.05 g/mL or 50 g/L. Given the molar mass of acetic acid (CH3COOH) is approximately 60 g/mol, the molar concentration of acetic acid is 50/60 = 0.833 mol/L. The percent dissociation is calculated as ([H+]/initial concentration of acetic acid) × 100 = (10^-2.41/0.833) × 100 ≈ 0.47%.

Question 18

What is the hydronium ion concentration of a 0.100 M hypochlorous acid solution with K_a = 3.5 × 10^-8? The equation for the dissociation of hypochlorous acid is: HOCl(aq)+ H₂O(l)⇌ H₃O⁺(aq)+ OCl^-(aq).

Accepted Answer

The answer of What is the hydronium ion concentration of...

Question 19

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 20

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 $pH = 14 - pOH = 14 - 2.87 = 11.13$.

Quiz 15: Aqueous Equilibria: Acids and Bases

What is the percent dissociation of a benzoic acid solution with pH = 2.59? The acid dissociation constant for this monoprotic acid is 6.5 × 10^-5.

A tablet containing 500.0 mg of aspirin (acetylsalicylic acid or HC₉H₇O₄) was dissolved in enough water to make 100 mL of solution.Given that K_a = 3.0 × 10^-4 for aspirin,what is the pH of the solution?

The pH of 0.150 M CH₃CO₂H,acetic acid,is 2.78.What is the value of K_a for acetic acid?

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.

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.

What is the strongest monoprotic acid of the following set if all the acids are at 0.100 M concentration?

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?

Vinegar is a 5.0% solution by weight of acetic acid (CH₃CO₂H) in water.Given that K_a = 1.8 × 10^-5 for acetic acid and assuming the density of vinegar to be 1.00 g/cm³,what is the pH of this vinegar solution?

What is the hydronium ion concentration of a 0.100 M acetic acid solution with a K_a = 1.8 × 10^-5? The equation for the dissociation of acetic acid is: CH₃CO₂H(aq) + H₂O(l) ⇌ H₃O⁺(aq) + CH₃CO₂^-(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?

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.

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

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

Para-Aminobenzoic acid (PABA) ,p-H₂NC₆H₄(COOH) ,is used in some sunscreens and hair conditioning products.Calculate the pH of an aqueous solution with [PABA] = 0.030 M and K_a = 2.2 × 10^-5.

Vinegar is a 5.0% solution by weight of acetic acid (CH₃CO₂H) in water.Given that the pH for acetic acid is 2.41,the K_a = 1.8 × 10^-5and assuming the density of vinegar to be 1.00 g/cm³,what is the percent dissociation of acetic acid in vinegar?

What is the hydronium ion concentration of a 0.100 M hypochlorous acid solution with K_a = 3.5 × 10^-8? The equation for the dissociation of hypochlorous acid is: HOCl(aq) + H₂O(l) ⇌ H₃O⁺(aq) + OCl^-(aq) .

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

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) .

Quiz 15: Aqueous Equilibria: Acids and Bases

What is the percent dissociation of a benzoic acid solution with pH = 2.59? The acid dissociation constant for this monoprotic acid is 6.5 × 10-5.

A tablet containing 500.0 mg of aspirin (acetylsalicylic acid or HC9H7O4) was dissolved in enough water to make 100 mL of solution.Given that Ka = 3.0 × 10-4 for aspirin,what is the pH of the solution?

The pH of 0.150 M CH3CO2H,acetic acid,is 2.78.What is the value of Ka for acetic acid?

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.

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.

What is the strongest monoprotic acid of the following set if all the acids are at 0.100 M concentration?

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?

Vinegar is a 5.0% solution by weight of acetic acid (CH3CO2H) in water.Given that Ka = 1.8 × 10-5 for acetic acid and assuming the density of vinegar to be 1.00 g/cm3,what is the pH of this vinegar solution?

What is the hydronium ion concentration of a 0.100 M acetic acid solution with a Ka = 1.8 × 10-5? The equation for the dissociation of acetic acid is: CH3CO2H(aq) + H2O(l) ⇌ H3O+(aq) + CH3CO2-(aq) .

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?

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.

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.

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

Para-Aminobenzoic acid (PABA) ,p-H2NC6H4(COOH) ,is used in some sunscreens and hair conditioning products.Calculate the pH of an aqueous solution with [PABA] = 0.030 M and Ka = 2.2 × 10-5.

Vinegar is a 5.0% solution by weight of acetic acid (CH3CO2H) in water.Given that the pH for acetic acid is 2.41,the Ka = 1.8 × 10-5 and assuming the density of vinegar to be 1.00 g/cm3,what is the percent dissociation of acetic acid in vinegar?

What is the hydronium ion concentration of a 0.100 M hypochlorous acid solution with Ka = 3.5 × 10-8? The equation for the dissociation of hypochlorous acid is: HOCl(aq) + H2O(l) ⇌ H3O+(aq) + OCl-(aq) .

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

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 percent dissociation of a benzoic acid solution with pH = 2.59? The acid dissociation constant for this monoprotic acid is 6.5 × 10^-5.

A tablet containing 500.0 mg of aspirin (acetylsalicylic acid or HC₉H₇O₄) was dissolved in enough water to make 100 mL of solution.Given that K_a = 3.0 × 10^-4 for aspirin,what is the pH of the solution?

The pH of 0.150 M CH₃CO₂H,acetic acid,is 2.78.What is the value of K_a for acetic acid?

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.

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.

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?

Vinegar is a 5.0% solution by weight of acetic acid (CH₃CO₂H) in water.Given that K_a = 1.8 × 10^-5 for acetic acid and assuming the density of vinegar to be 1.00 g/cm³,what is the pH of this vinegar solution?

What is the hydronium ion concentration of a 0.100 M acetic acid solution with a K_a = 1.8 × 10^-5? The equation for the dissociation of acetic acid is: CH₃CO₂H(aq) + H₂O(l) ⇌ H₃O⁺(aq) + CH₃CO₂^-(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?

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.

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

Para-Aminobenzoic acid (PABA) ,p-H₂NC₆H₄(COOH) ,is used in some sunscreens and hair conditioning products.Calculate the pH of an aqueous solution with [PABA] = 0.030 M and K_a = 2.2 × 10^-5.

Vinegar is a 5.0% solution by weight of acetic acid (CH₃CO₂H) in water.Given that the pH for acetic acid is 2.41,the K_a = 1.8 × 10^-5and assuming the density of vinegar to be 1.00 g/cm³,what is the percent dissociation of acetic acid in vinegar?

What is the hydronium ion concentration of a 0.100 M hypochlorous acid solution with K_a = 3.5 × 10^-8? The equation for the dissociation of hypochlorous acid is: HOCl(aq) + H₂O(l) ⇌ H₃O⁺(aq) + OCl^-(aq) .

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

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) .