Question 1

A 25.0-mL sample of 1.00 M NH₃is titrated with 0.15 M HCl. What is the pH of the solution after 15.00 mL of acid have been added to the ammonia solution? K_b = 1.8 × 10^-5

Accepted Answer

The answer of A 25.0-mL sample of 1.00 M NH<sub>3...

Question 2

At the equivalence point in an acid-base titration

Accepted Answer

At the equivalence point in an acid-base titration, the amounts of acid and base which have been combined are in their stoichiometric ratio. This means that all of the acid has been neutralized by an equal amount of base, and the solution is no longer acidic or basic. The pH at the equivalence point depends on the strength of the acid and base used in the titration, but it is not necessarily 7.0. The other answer choices do not describe what happens at the equivalence point accurately.

Question 3

A 20.0-mL sample of 0.30 M HBr is titrated with 0.15 M NaOH. What is the pH of the solution after 40.3 mL of NaOH have been added to the acid?

Accepted Answer

The reaction between HBr and NaOH is a neutralization reaction, where HBr (a strong acid) reacts with NaOH (a strong base) to form water and NaBr (a salt). The stoichiometry of the reaction is 1:1. Initially, there are $20.0 \, \text{mL} \times 0.30 \, \text{M} = 6.0 \, \text{mmol}$ of HBr. $40.3 \, \text{mL} \times 0.15 \, \text{M} = 6.045 \, \text{mmol}$ of NaOH are added, which is slightly more than the amount of HBr. This means all the HBr will react, and there will be a small excess of NaOH. The excess NaOH is $6.045 \, \text{mmol} - 6.0 \, \text{mmol} = 0.045 \, \text{mmol}$. The total volume after mixing is $20.0 \, \text{mL} + 40.3 \, \text{mL} = 60.3 \, \text{mL}$. The concentration of the excess NaOH is $0.045 \, \text{mmol} / 60.3 \, \text{mL} = 0.000746 \, \text{M}$. The pH of a solution of NaOH at this concentration can be found by calculating the pOH first: $pOH = -\log[OH^-] = -\log(0.000746)$, and then converting to pH: $pH = 14 - pOH$. This calculation gives a pH close to 10.87, indicating the solution is basic due to the excess NaOH.

Question 4

When a strong acid is titrated with a strong base, the pH at the equivalence point

Accepted Answer

At the equivalence point, the moles of acid are equal to the moles of base, resulting in a neutral solution with a pH of 7.0.

Question 5

When a strong acid is titrated with a weak base, the pH at the equivalence point

Accepted Answer

When a strong acid is titrated with a weak base, the resulting salt will be acidic. Therefore, the pH at the equivalence point will be less than 7.0.

Question 6

A 20.0-mL sample of 0.25 M HNO₃is titrated with 0.15 M NaOH. What is the pH of the solution after 30.0 mL of NaOH have been added to the acid?

Accepted Answer

The answer of A 20.0-mL sample of 0.25 M HNO<sub>3...

Question 7

Which of the following indicators would be the best to use when 0.050 M benzoic acid (K_a = 6.6 × 10^-5) is titrated with 0.05 M NaOH?

Accepted Answer

The answer of Which of the following indicators would be...

Question 8

When a weak acid is titrated with a weak base, the pH at the equivalence point

Accepted Answer

When a weak acid is titrated with a weak base, the pH at the equivalence point is determined by the relative strengths of the two conjugate acid-base pairs (the acid and its conjugate base, and the base and its conjugate acid). The equilibrium constants for these pairs, represented by K_a and K_b, respectively, will determine whether the resulting solution is acidic, basic, or neutral at the equivalence point. Therefore, the pH cannot be accurately predicted without knowledge of these equilibrium constants.

Question 9

A 25.0-mL sample of 0.35 M HCOOH is titrated with 0.20 M KOH. What is the pH of the solution after 25.0 mL of KOH has been added to the acid? K_a = 1.77 × 10^-4

Accepted Answer

Before adding any KOH, the solution is just 0.35 M HCOOH. This is a weak acid, so we need to use the Ka expression to find the initial concentration of H+ ions:

Ka = [H+][HCOO-]/[HCOOH]

At equilibrium, the concentration of HCOO- and H+ will be equal, so we can write:

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

Solving for [H+], we get:

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

[H+] = sqrt(1.77x10^-4 * 0.35) = 0.012 M

Now we can use the titration information to find out how much of the HCOOH has been neutralized. When we add 25.0 mL of 0.20 M KOH, we are adding:

0.020 L * 0.20 mol/L = 0.004 mol KOH

Since the stoichiometry of the reaction is 1:1 between KOH and HCOOH, we know that 0.004 mol of HCOOH has been neutralized. To find out how much HCOOH is left, we can subtract:

moles HCOOH at start - moles HCOOH neutralized = moles HCOOH remaining

0.35 mol/L * 0.0250 L - 0.004 mol = 0.00625 mol remaining

Now we need to find the concentration of HCOOH and HCOO- in the solution at this point. We know that the total volume is 25.0 + 25.0 = 50.0 mL = 0.0500 L. So the new concentration of HCOOH is:

[HCOOH] = 0.00625 mol / 0.0500 L = 0.125 M

And the new concentration of HCOO- is also 0.125 M (since KOH is a strong base and completely dissociates). Now we can use the Kb expression for HCOO- to find the concentration of OH- ions:

Kb = [OH-][HCOOH] / [HCOO-]

Kb = [OH-]^2 / [HCOO-]

Solving for [OH-], we get:

[OH-] = sqrt(Kb*[HCOO-])

Kb is the same as Kw/Ka, so we can use:

Kw/Ka = 1.00x10^-14 / 1.77x10^-4 = 5.65x10^-11

Kb = 5.65x10^-11 / 1.77x10^-4 = 3.19x10^-7

[OH-] = sqrt(3.19x10^-7 * 0.125) = 3.00x10^-4 M

Finally, we can use the fact that Kw = [H+][OH-] to find the pH:

Kw = 1.00x10^-14 = [H+][OH-]

[H+] = 1.00x10^-14 / [OH-] = 1.00x10^-14 / 3.00x10^-4 = 3.33x10^-11 M

pH = -log[H+] = -log(3.33x10^-11) = 3.88

Question 10

A 35.0-mL sample of 0.20 M LiOH is titrated with 0.25 M HCl. What is the pH of the solution after 23.0 mL of HCl have been added to the base?

Accepted Answer

The reaction between LiOH and HCl is a neutralization reaction, where LiOH (a strong base) reacts with HCl (a strong acid) to form LiCl (a salt) and water. The stoichiometry of the reaction is 1:1. First, calculate the moles of LiOH and HCl:Moles of LiOH = 0.035 L * 0.20 mol/L = 0.007 molMoles of HCl added = 0.023 L * 0.25 mol/L = 0.00575 molSince less moles of HCl are added than there are moles of LiOH, not all the LiOH will react. The remaining moles of LiOH:Remaining moles of LiOH = 0.007 mol - 0.00575 mol = 0.00125 molThe volume of the solution after adding HCl:Total volume = 35.0 mL + 23.0 mL = 58.0 mL = 0.058 LThe concentration of LiOH after the reaction:Concentration of LiOH = 0.00125 mol / 0.058 L = 0.02155 MSince LiOH is a strong base, its pH can be calculated using the formula:$$ \text{pOH} = -\log[\text{OH}^-] $$$$ \text{pH} = 14 - \text{pOH} $$$$ \text{pH} = 14 - (-\log[0.02155]) $$$$ \text{pH} \approx 12.33 $$Therefore, the pH of the solution after adding 23.0 mL of HCl is approximately 12.33.

Question 11

A 10.0-mL sample of 0.75 M CH₃CH₂COOH is titrated with 0.30 M NaOH. What is the pH of the solution after 22.0 mL of NaOH have been added to the acid? K_a = 1.3 × 10^-5

Accepted Answer

The answer of A 10.0-mL sample of 0.75 M CH₃CH₂COOH...

Question 12

The indicator propyl red has K_a = 3.3 × 10^-6. What would be the approximate pH range over which it would change color?

Accepted Answer

The answer of The indicator propyl red has K_a =...

Question 13

When a weak acid is titrated with a strong base, the pH at the equivalence point

Accepted Answer

A weak acid and strong base titration results in the formation of a basic solution at the equivalence point. The pH will be greater than 7.0 due to the presence of excess hydroxide ions (OH-) in solution.

Question 14

A 20.0-mL sample of 0.30 M HClO was titrated with 0.30 M NaOH. The following data were collected during the titration. What is the K_a for HClO?

Accepted Answer

The balanced chemical equation for the reaction between HClO and NaOH is:

HClO + NaOH -> NaClO + H2O

Since the initial concentration of HClO is 0.30 M and the volume of the solution is 20.0 mL (0.0200 L), the initial moles of HClO can be calculated as:

moles of HClO = (0.30 M) x (0.0200 L) = 0.00600 moles

During the titration, NaOH is added until the equivalence point is reached, which occurs when all of the HClO has reacted with NaOH. At the equivalence point, the moles of NaOH added are equal to the moles of HClO present in the initial solution. Therefore, the moles of NaOH added can be calculated as:

moles of NaOH = (0.30 M) x (0.0230 L) = 0.00690 moles

The volume of NaOH required to reach the equivalence point is 23.0 mL (0.0230 L).

From the balanced chemical equation, it can be seen that one mole of NaOH reacts with one mole of HClO. Therefore, the moles of HClO at the equivalence point can be calculated as:

moles of HClO = 0.00690 moles

The volume of the solution at the equivalence point can be calculated as:

volume of solution at equivalence point = 20.0 mL + 23.0 mL = 43.0 mL = 0.0430 L

The concentration of HClO at the equivalence point can be calculated as:

[HClO] = moles of HClO / volume of solution at equivalence point
[HClO] = 0.00690 moles / 0.0430 L
[HClO] = 0.161 M

The Ka expression for the ionization of HClO is:

Ka = [H3O+][ClO-] / [HClO]

At the equivalence point, [HClO] = 0.161 M, and [OH-] = [H3O+] since the solution is neutral. The concentration of [OH-] can be calculated from the amount of NaOH added:

moles of NaOH = (0.30 M) x (0.0230 L) = 0.00690 moles
moles of H3O+ = moles of OH- = 0.00690 moles
[OH-] = moles of OH- / volume of solution at equivalence point
[OH-] = 0.00690 moles / 0.0430 L
[OH-] = 0.161 M

Substituting the values into the Ka expression gives:

Ka = (0.161 M)^2 / 0.161 M = 0.161 M

Therefore, the pKa is:

pKa = -log(Ka) = -log(0.161) = 0.791

Using the relationship between pKa and pKb for conjugate acid-base pairs:

pKa + pKb = 14.00

the pKb can be calculated as:

pKb = 14.00 - pKa = 14.00 - 0.791 = 13.209

Finally, the Kb can be calculated using:

Kb = 10^(-pKb) = 3.5 x 10^(-14)

The correct answer is B: 3.5 × 10^-8.

Question 15

When 20.0 mL of 0.15 M hydrochloric acid is mixed with 20.0 mL of 0.10 M sodium hydroxide, the pH of the resulting solution is

Accepted Answer

The reaction between hydrochloric acid (HCl) and sodium hydroxide (NaOH) is a neutralization reaction, producing water and sodium chloride (NaCl). The moles of HCl are $20.0 \, \text{mL} \times 0.15 \, \text{M} = 3.0 \times 10^{-3} \, \text{moles}$, and the moles of NaOH are $20.0 \, \text{mL} \times 0.10 \, \text{M} = 2.0 \times 10^{-3} \, \text{moles}$. After the reaction, there are $3.0 \times 10^{-3} - 2.0 \times 10^{-3} = 1.0 \times 10^{-3} \, \text{moles}$ of HCl left unreacted. The total volume of the solution is $20.0 \, \text{mL} + 20.0 \, \text{mL} = 40.0 \, \text{mL}$ or $0.040 \, \text{L}$. The concentration of HCl remaining is $\frac{1.0 \times 10^{-3} \, \text{moles}}{0.040 \, \text{L}} = 0.025 \, \text{M}$. The pH is calculated using the formula $\text{pH} = -\log[H^+]$, where $[H^+] = 0.025 \, \text{M}$. Thus, $\text{pH} = -\log(0.025) \approx 1.60$.

Question 16

Which one of the following is the best representation of the titration curve which will be obtained in the titration of a weak base (0.10 mol L^-1) with HCl of the same concentration?

Accepted Answer

The answer of Which one of the following is the...

Question 17

A 25.0-mL sample of 0.10 M C₂H₃NH₂(ethylamine) is titrated with 0.15 M HCl. What is the pH of the solution after 9.00 mL of acid have been added to the amine? K_b = 6.5 × 10^-4

Accepted Answer

The answer of A 25.0-mL sample of 0.10 M C₂H₃NH_2...

Question 18

A 50.0-mL sample of 0.50 M HCl is titrated with 0.50 M NaOH. What is the pH of the solution after 28.0 mL of NaOH have been added to the acid?

Accepted Answer

The answer of A 50.0-mL sample of 0.50 M HCl...

Question 19

Which one of the following is the best representation of the titration curve which will be obtained in the titration of a weak acid (0.10 mol L^-1) with a strong base of the same concentration?

Accepted Answer

The answer of Which one of the following is the...

Question 20

A diprotic acid H₂A has K_a1 = 1 × 10^-4 and K_a2 = 1 × 10^-8. The corresponding base A^2- is titrated with aqueous HCl, both solutions being 0.1 mol L^-1. Which one of the following diagrams best represents the titration curve which will be seen?

Accepted Answer

The answer of A diprotic acid H₂A has K_a1 =...

Quiz 19: Ionic Equilibria in Aqueous Systems

A 25.0-mL sample of 1.00 M NH3 is titrated with 0.15 M HCl. What is the pH of the solution after 15.00 mL of acid have been added to the ammonia solution? Kb = 1.8 × 10-5

At the equivalence point in an acid-base titration

A 20.0-mL sample of 0.30 M HBr is titrated with 0.15 M NaOH. What is the pH of the solution after 40.3 mL of NaOH have been added to the acid?

When a strong acid is titrated with a strong base, the pH at the equivalence point

When a strong acid is titrated with a weak base, the pH at the equivalence point

A 20.0-mL sample of 0.25 M HNO3 is titrated with 0.15 M NaOH. What is the pH of the solution after 30.0 mL of NaOH have been added to the acid?

Which of the following indicators would be the best to use when 0.050 M benzoic acid (Ka = 6.6 × 10-5) is titrated with 0.05 M NaOH?

When a weak acid is titrated with a weak base, the pH at the equivalence point

A 25.0-mL sample of 0.35 M HCOOH is titrated with 0.20 M KOH. What is the pH of the solution after 25.0 mL of KOH has been added to the acid? Ka = 1.77 × 10-4

A 35.0-mL sample of 0.20 M LiOH is titrated with 0.25 M HCl. What is the pH of the solution after 23.0 mL of HCl have been added to the base?

A 10.0-mL sample of 0.75 M CH3CH2COOH is titrated with 0.30 M NaOH. What is the pH of the solution after 22.0 mL of NaOH have been added to the acid? Ka = 1.3 × 10-5

The indicator propyl red has Ka = 3.3 × 10-6. What would be the approximate pH range over which it would change color?