Question 1

Acid dissociation constants for phosphoric acid are given below.  A buffer with a pH = 7.4 can best be made by using

Accepted Answer

To make a buffer with a pH of 7.4, we need a weak acid and its conjugate base with a pKa closest to 7.4. From the given pKa values of phosphoric acid, we can see that the closest pKa value is 7.21, which corresponds to the second dissociation step (H2PO4- ⇌ H+ + HPO42-). Therefore, we can use a mixture of NaH2PO4 and Na2HPO4 to make the buffer. Choice B is the only one that includes these two compounds. The addition of NaH2PO4 and Na2HPO4 (as well as NaHPO4) will increase the ionic strength of the buffer, but it will not significantly affect the pH. Choice A includes HS- and H2PO4-, but not the corresponding conjugate bases. Choice C includes NaHPO4 and Na3PO4, but not the corresponding weak acids. Choice D includes only NaH2PO4, which will not provide a sufficient buffer capacity. Choice E includes only Na2HPO4, which will not provide a sufficient buffer capacity either.

Question 2

What mass of sodium fluoride must be added to 250.mL of a 0.100 M HF solution to give a buffer solution having a pH of 3.50? [K_a(HF)= 7.1 × 10^-4]

Accepted Answer

The answer of What mass of sodium fluoride must be...

Question 3

Calculate the pH of the solution resulting from the addition of 25.0 mL of 0.20 M HCl to 50.0 mL of 0.10 M aniline (C₆H₅NH₂).K_b(C₆H₅NH₂)= 3.8 x 10^-10

Accepted Answer

The answer of Calculate the pH of the solution resulting...

Question 4

Calculate the pH at the equivalence point for the titration of 0.20 M HCl with 0.20 M NH₃ (K_b = 1.8 × 10^-5).

Accepted Answer

At the equivalence point, the solution contains NH4Cl, which is the conjugate acid of NH3. The pH is determined by the hydrolysis of NH4+, and using the Kb of NH3, we can find the Ka of NH4+ and calculate the pH, which is approximately 5.12.

Question 5

In which one of the following solutions will acetic acid have the greatest percent ionization?

Accepted Answer

Acetic acid will have the greatest percent ionization in a solution where it is not affected by the presence of a strong acid (like HCl) or its own salt (like sodium acetate, CH3COONa). In options B and C, the presence of HCl, a strong acid, will suppress the ionization of acetic acid due to the common ion effect. In options D and E, the presence of sodium acetate will also suppress the ionization of acetic acid due to the common ion effect. Therefore, acetic acid will have the greatest percent ionization in the solution where it is alone, which is option A.

Question 6

What mass of ammonium nitrate must be added to 350.mL of a 0.150 M solution of ammonia to give a buffer having a pH of 9.00? (K_b(NH₃)= 1.8 × 10^-5)

Accepted Answer

The correct answer is A because the mass of ammonium nitrate that must be added to 350.mL of a 0.150 M solution of ammonia to give a buffer having a pH of 9.00 is 7.6 g.

Question 7

You have 500.0 mL of a buffer solution containing 0.30 M acetic acid (CH₃COOH)and 0.20 M sodium acetate (CH₃COONa).What will the pH of this solution be after the addition of 20.0 mL of 1.00 M NaOH solution? [K_a = 1.8 × 10^-5]

Accepted Answer

First, we need to calculate the initial pH of the buffer solution:
pKa = 4.76 (from the dissociation constant of acetic acid)
pH = pKa + log([base]/[acid])
pH = 4.76 + log(0.20/0.30) = 4.66
Therefore, the initial pH of the buffer solution is 4.66.

Next, we need to calculate how much the pH will change after the addition of NaOH. To do this, we need to determine how much of the NaOH will react with the buffer. Since the buffer contains acetate ions, which are a weak base, they will react with the NaOH:

CH3COO- + NaOH → CH3COONa + H2O

The initial moles of acetate ions in the buffer is:
0.20 mol/L x 0.500 L = 0.100 mol

Since the NaOH is added in excess, all of the acetate ions will be converted to CH3COONa. The moles of NaOH added is:
1.00 mol/L x 0.020 L = 0.020 mol

Therefore, the new concentration of CH3COONa is:
0.100 mol + 0.020 mol = 0.120 mol
And the new concentration of CH3COOH is:
0.30 mol/L x 0.500 L = 0.150 mol

Now we can calculate the new pH of the buffer solution using the Henderson-Hasselbalch equation:
pH = pKa + log([base]/[acid])
pH = 4.76 + log(0.120/0.150) = 4.71

Therefore, the pH of the solution after the addition of NaOH is 4.71, which is closest to choice B.

Question 8

The pH at the equivalence point of a titration may differ from 7.0 due to

Accepted Answer

At the equivalence point of a titration, a salt is formed between the acid and base, and this salt can undergo hydrolysis, which can affect the pH. This can cause the pH to be different from 7.0. Therefore, choice E is the best answer. The initial concentration of the standard solution (A), the indicator used (B), the self-ionization of water (C), and the initial pH of the unknown (D) do not directly affect the pH at the equivalence point.

Question 9

Calculate the pH of the solution resulting from the addition of 10.0 mL of 0.10 M NaOH to 50.0 mL of 0.10 M HCN (K_a = 4.9 × 10^-10)solution.

Accepted Answer

The answer of Calculate the pH of the solution resulting...

Question 10

For which type of titration will the pH be neutral at the equivalence point?

Accepted Answer

In a strong acid vs. strong base titration, the pH will be neutral at the equivalence point because the strong acid and strong base will completely neutralize each other, resulting in a solution with a pH of 7. In a weak acid vs. strong base titration, the pH will be basic at the equivalence point because the strong base will fully neutralize the weak acid and the excess OH- ions will result in a basic solution. In a strong acid vs. weak base titration, the pH will be acidic at the equivalence point because the weak base will not be able to fully neutralize the strong acid and the excess H+ ions will result in an acidic solution. Therefore, the correct answer is A.

Question 11

Calculate the pH of the solution resulting from the addition of 75.0 mL of 0.15 M KOH to 35.0 mL of 0.20 M HCN (K_a (HCN)= 4.9 × 10^-10).

Accepted Answer

The answer of Calculate the pH of the solution resulting...

Question 12

For which type of titration will the pH be basic at the equivalence point?

Accepted Answer

In a weak acid-strong base titration, the equivalence point occurs when all the weak acid has reacted with the strong base to form its conjugate base, which is a strong base. Therefore, the pH will be basic at the equivalence point.

Question 13

Methyl red is a common acid-base indicator.It has a K_a equal to 6.3 × 10^-6.Its un-ionized form is red and its anionic form is yellow.What color would a methyl red solution have at pH = 7.8?

Accepted Answer

Methyl red has an acidic range of 4.4-6.2, where it appears red. At a pH of 7.8, which is basic, it would be above its acidic range and its anionic form would predominate, giving it a yellow color. Therefore, the correct answer is D) yellow.

Question 14

Starting with 0.750L of a buffer solution containing 0.30 M benzoic acid (C₆H₅COOH)and 0.35 M sodium benzoate (C₆H₅COONa), what will the pH of the solution be after the addition of 340.0 mL of 0.350M HCl? (K_a (C₆H₅COOH)= 6.5 x 10^-5)

Accepted Answer

First, we need to calculate the initial pH of the buffer solution. Since benzoic acid is a weak acid, we can use the Henderson-Hasselbalch equation:

pH = pKa + log([A-]/[HA])

where pKa is the dissociation constant of benzoic acid (6.5 x 10^-5), [A-] is the concentration of sodium benzoate, and [HA] is the concentration of benzoic acid.

pH = 6.5 x 10^-5 + log(0.35/0.30)
pH = 4.19

Next, we need to determine how the buffer solution will react to the addition of HCl. Since HCl is a strong acid, it will react completely with the sodium benzoate to form benzoic acid and sodium chloride:

HCl + NaC6H5COO -> HC6H5COOH + NaCl

This reaction will consume some of the benzoate ion, which will shift the equilibrium in the buffer solution to the left to produce more benzoate ion and benzoic acid. However, since we started with a buffer solution that has more benzoic acid than benzoate ion, the pH will decrease because we are producing more acidic benzoic acid.

To calculate the new pH, we need to determine the concentration of benzoic acid and benzoate ion after the reaction. We can use the stoichiometry of the reaction to determine that the amount of HCl added will be equal to the amount of sodium benzoate used up:

moles HCl = (0.340 L)(0.350 mol/L) = 0.119 mol

moles NaC6H5COO = 0.119 mol

Since we started with 0.2625 moles of NaC6H5COO, we can calculate the new concentration of sodium benzoate:

[NaC6H5COO] = (0.2625 mol - 0.119 mol)/(0.750 L + 0.340 L) = 0.168 M

Since we consumed 0.119 mol of sodium benzoate, we know that we produced 0.119 mol of benzoic acid:

[C6H5COOH] = 0.30 M + 0.119 mol/(0.750 L + 0.340 L) = 0.413 M

Now we can use the Henderson-Hasselbalch equation again to calculate the new pH:

pH = 6.5 x 10^-5 + log(0.168/0.413)
pH = 3.81

Therefore, the correct answer is (C).

Question 15

A titration of an acid and base to the equivalence point results in a noticeably acidic solution.It is likely this titration involves

Accepted Answer

When a strong acid is titrated with a weak base, the resulting solution at the equivalence point is acidic due to the weak base not being able to fully neutralize the strong acid.

Question 16

What is the pH at the equivalence point in the titration of 100 mL of 0.10 M HCN (K_a = 4.9 × 10^-10)with 0.10 M NaOH?

Accepted Answer

At the equivalence point, the weak acid HCN is completely converted to its conjugate base CN-. The pH is determined by the hydrolysis of CN-, which makes the solution basic. The pH is greater than 7, and for weak acids like HCN, it is typically around 11.

Question 17

What is the pH at the equivalence point in the titration of 100 mL of 0.10 M HCl with 0.10 M NaOH?

Accepted Answer

The equivalence point is reached when the moles of acid (HCl) are equal to the moles of base (NaOH). Since they have the same concentration, it will take 100 mL of 0.10 M NaOH to neutralize 100 mL of 0.10 M HCl. At this point, all of the HCl has been neutralized and converted to NaCl, which is a neutral salt. Therefore, the pH at the equivalence point will be 7.0.

Question 18

Starting with 0.250L of a buffer solution containing 0.250 M benzoic acid (C₆H₅COOH)and 0.20 M sodium benzoate (C₆H₅COONa), what will the pH of the solution be after the addition of 25.0 mL of 0.100M HCl? (K_a (C₆H₅COOH)= 6.5 x 10^-5)

Accepted Answer

The answer of Starting with 0.250L of a buffer solution...

Question 19

Calculate the percent ionization of cyanic acid, K_a = 2.0 × 10^-4, in a buffer solution that is 0.50 M HCNO and 0.10 M NaCNO.

Accepted Answer

HCNO is a weak acid, so it will produce H+ and CNO- ions in solution. NaCNO is a salt of the conjugate base of HCNO, so it will provide CNO- ions. The buffer solution will have a pH close to the pKa of HCNO, which is 3.48.

To calculate the percent ionization of HCNO, we need to determine the concentrations of the acid and the conjugate base ([HCNO] and [CNO-], respectively) and the concentration of undissociated HCNO ([H+] is negligible, since the pH is close to the pKa).

From the given concentrations of 0.50 M HCNO and 0.10 M NaCNO, we can assume that the initial concentration of CNO- is negligible compared to the amount produced by the dissociation of HCNO. Therefore, we can treat [CNO-] as equal to the concentration of HCNO that ionizes.

Let x be the extent of ionization of HCNO. Then:

HCNO ⇌ H+ + CNO-
Initial:      0.50 M          0             0
Change:     -x               +x           +x
Equilibrium: 0.50 - x        x            x

The equilibrium constant for this reaction is Ka = [H+][CNO-] / [HCNO] = x^2 / (0.50 - x).

Since x is small compared to 0.50, we can approximate (0.50 - x) as 0.50. Then:

Ka = x^2 / 0.50
x = sqrt(Ka * 0.50)

For HCNO, Ka = 1.2 x 10^-4, so:

x = sqrt(1.2 x 10^-4 * 0.50) = 0.012

The percent ionization is:

% ionization = (moles of ionized HCNO / initial moles of HCNO) x 100%
= (0.012 mol / 0.50 mol) x 100% = 2.4%.

However, we need to account for the effect of the added salt (NaCNO) on the ionization of HCNO. The presence of a common ion (CNO-) will shift the equilibrium to the left, reducing the ionization of HCNO.

We can use the common ion effect equation:

Ka' = Ka / (1 + [CNO-] / Ka)

To calculate the new equilibrium constant and ionization extent.

Since [CNO-] = 0.10 M and Ka = 1.2 x 10^-4:

Ka' = 1.2 x 10^-4 / (1 + 0.10 M / 1.2 x 10^-4) = 8.0 x 10^-6

x' = sqrt(Ka' * 0.50) = 0.006

The percent ionization with the added salt is:

% ionization = (moles of ionized HCNO / initial moles of HCNO) x 100%
= (0.006 mol / 0.50 mol) x 100% = 1.2%.

Therefore, the answer is (C) 0.20%.

Question 20

Calculate the pH of the solution resulting from the addition of 85.0 mL of 0.35 M HCl to 30.0 mL of 0.40 M aniline (C₆H₅NH₂).K_b(C₆H₅NH₂)= 3.8 x 10^-10

Accepted Answer

The answer of Calculate the pH of the solution resulting...

Quiz 16: Acid-Base Equilibria and Solubility Equilibria

Acid dissociation constants for phosphoric acid are given below. A buffer with a pH = 7.4 can best be made by using

What mass of sodium fluoride must be added to 250.mL of a 0.100 M HF solution to give a buffer solution having a pH of 3.50? [K_a(HF) = 7.1 × 10^-4]

Calculate the pH of the solution resulting from the addition of 25.0 mL of 0.20 M HCl to 50.0 mL of 0.10 M aniline (C₆H₅NH₂) .K_b(C₆H₅NH₂) = 3.8 x 10^-10

Calculate the pH at the equivalence point for the titration of 0.20 M HCl with 0.20 M NH₃ (K_b = 1.8 × 10^-5) .

In which one of the following solutions will acetic acid have the greatest percent ionization?

What mass of ammonium nitrate must be added to 350.mL of a 0.150 M solution of ammonia to give a buffer having a pH of 9.00? (K_b(NH₃) = 1.8 × 10^-5)

You have 500.0 mL of a buffer solution containing 0.30 M acetic acid (CH₃COOH) and 0.20 M sodium acetate (CH₃COONa) .What will the pH of this solution be after the addition of 20.0 mL of 1.00 M NaOH solution? [K_a = 1.8 × 10^-5]

The pH at the equivalence point of a titration may differ from 7.0 due to

Calculate the pH of the solution resulting from the addition of 10.0 mL of 0.10 M NaOH to 50.0 mL of 0.10 M HCN (K_a = 4.9 × 10^-10) solution.

For which type of titration will the pH be neutral at the equivalence point?

Calculate the pH of the solution resulting from the addition of 75.0 mL of 0.15 M KOH to 35.0 mL of 0.20 M HCN (K_a (HCN) = 4.9 × 10^-10) .

For which type of titration will the pH be basic at the equivalence point?

Methyl red is a common acid-base indicator.It has a K_a equal to 6.3 × 10^-6.Its un-ionized form is red and its anionic form is yellow.What color would a methyl red solution have at pH = 7.8?

Starting with 0.750L of a buffer solution containing 0.30 M benzoic acid (C₆H₅COOH) and 0.35 M sodium benzoate (C₆H₅COONa) , what will the pH of the solution be after the addition of 340.0 mL of 0.350M HCl? (K_a (C₆H₅COOH) = 6.5 x 10^-5)

A titration of an acid and base to the equivalence point results in a noticeably acidic solution.It is likely this titration involves

What is the pH at the equivalence point in the titration of 100 mL of 0.10 M HCN (K_a = 4.9 × 10^-10) with 0.10 M NaOH?

What is the pH at the equivalence point in the titration of 100 mL of 0.10 M HCl with 0.10 M NaOH?

Starting with 0.250L of a buffer solution containing 0.250 M benzoic acid (C₆H₅COOH) and 0.20 M sodium benzoate (C₆H₅COONa) , what will the pH of the solution be after the addition of 25.0 mL of 0.100M HCl? (K_a (C₆H₅COOH) = 6.5 x 10^-5)

Calculate the percent ionization of cyanic acid, K_a = 2.0 × 10^-4, in a buffer solution that is 0.50 M HCNO and 0.10 M NaCNO.

Calculate the pH of the solution resulting from the addition of 85.0 mL of 0.35 M HCl to 30.0 mL of 0.40 M aniline (C₆H₅NH₂) .K_b(C₆H₅NH₂) = 3.8 x 10^-10

Quiz 16: Acid-Base Equilibria and Solubility Equilibria

Acid dissociation constants for phosphoric acid are given below. A buffer with a pH = 7.4 can best be made by using

What mass of sodium fluoride must be added to 250.mL of a 0.100 M HF solution to give a buffer solution having a pH of 3.50? [Ka(HF) = 7.1 × 10-4]

Calculate the pH of the solution resulting from the addition of 25.0 mL of 0.20 M HCl to 50.0 mL of 0.10 M aniline (C6H5NH2) .Kb (C6H5NH2) = 3.8 x 10-10

Calculate the pH at the equivalence point for the titration of 0.20 M HCl with 0.20 M NH3 (Kb = 1.8 × 10-5) .

In which one of the following solutions will acetic acid have the greatest percent ionization?

What mass of ammonium nitrate must be added to 350.mL of a 0.150 M solution of ammonia to give a buffer having a pH of 9.00? (Kb(NH3) = 1.8 × 10-5)

You have 500.0 mL of a buffer solution containing 0.30 M acetic acid (CH3COOH) and 0.20 M sodium acetate (CH3COONa) .What will the pH of this solution be after the addition of 20.0 mL of 1.00 M NaOH solution? [Ka = 1.8 × 10-5]

The pH at the equivalence point of a titration may differ from 7.0 due to

Calculate the pH of the solution resulting from the addition of 10.0 mL of 0.10 M NaOH to 50.0 mL of 0.10 M HCN (Ka = 4.9 × 10-10) solution.

For which type of titration will the pH be neutral at the equivalence point?

Calculate the pH of the solution resulting from the addition of 75.0 mL of 0.15 M KOH to 35.0 mL of 0.20 M HCN (Ka (HCN) = 4.9 × 10-10) .

For which type of titration will the pH be basic at the equivalence point?

Methyl red is a common acid-base indicator.It has a Ka equal to 6.3 × 10-6.Its un-ionized form is red and its anionic form is yellow.What color would a methyl red solution have at pH = 7.8?

Starting with 0.750L of a buffer solution containing 0.30 M benzoic acid (C6H5COOH) and 0.35 M sodium benzoate (C6H5COONa) , what will the pH of the solution be after the addition of 340.0 mL of 0.350M HCl? (Ka (C6H5COOH) = 6.5 x 10-5)

A titration of an acid and base to the equivalence point results in a noticeably acidic solution.It is likely this titration involves

What is the pH at the equivalence point in the titration of 100 mL of 0.10 M HCN (Ka = 4.9 × 10-10) with 0.10 M NaOH?

What is the pH at the equivalence point in the titration of 100 mL of 0.10 M HCl with 0.10 M NaOH?

Starting with 0.250L of a buffer solution containing 0.250 M benzoic acid (C6H5COOH) and 0.20 M sodium benzoate (C6H5COONa) , what will the pH of the solution be after the addition of 25.0 mL of 0.100M HCl? (Ka (C6H5COOH) = 6.5 x 10-5)

Calculate the percent ionization of cyanic acid, Ka = 2.0 × 10-4, in a buffer solution that is 0.50 M HCNO and 0.10 M NaCNO.

Calculate the pH of the solution resulting from the addition of 85.0 mL of 0.35 M HCl to 30.0 mL of 0.40 M aniline (C6H5NH2) .Kb (C6H5NH2) = 3.8 x 10-10

What mass of sodium fluoride must be added to 250.mL of a 0.100 M HF solution to give a buffer solution having a pH of 3.50? [K_a(HF) = 7.1 × 10^-4]

Calculate the pH of the solution resulting from the addition of 25.0 mL of 0.20 M HCl to 50.0 mL of 0.10 M aniline (C₆H₅NH₂) .K_b(C₆H₅NH₂) = 3.8 x 10^-10

Calculate the pH at the equivalence point for the titration of 0.20 M HCl with 0.20 M NH₃ (K_b = 1.8 × 10^-5) .

What mass of ammonium nitrate must be added to 350.mL of a 0.150 M solution of ammonia to give a buffer having a pH of 9.00? (K_b(NH₃) = 1.8 × 10^-5)

You have 500.0 mL of a buffer solution containing 0.30 M acetic acid (CH₃COOH) and 0.20 M sodium acetate (CH₃COONa) .What will the pH of this solution be after the addition of 20.0 mL of 1.00 M NaOH solution? [K_a = 1.8 × 10^-5]

Calculate the pH of the solution resulting from the addition of 10.0 mL of 0.10 M NaOH to 50.0 mL of 0.10 M HCN (K_a = 4.9 × 10^-10) solution.

Calculate the pH of the solution resulting from the addition of 75.0 mL of 0.15 M KOH to 35.0 mL of 0.20 M HCN (K_a (HCN) = 4.9 × 10^-10) .

Methyl red is a common acid-base indicator.It has a K_a equal to 6.3 × 10^-6.Its un-ionized form is red and its anionic form is yellow.What color would a methyl red solution have at pH = 7.8?

Starting with 0.750L of a buffer solution containing 0.30 M benzoic acid (C₆H₅COOH) and 0.35 M sodium benzoate (C₆H₅COONa) , what will the pH of the solution be after the addition of 340.0 mL of 0.350M HCl? (K_a (C₆H₅COOH) = 6.5 x 10^-5)

What is the pH at the equivalence point in the titration of 100 mL of 0.10 M HCN (K_a = 4.9 × 10^-10) with 0.10 M NaOH?

Starting with 0.250L of a buffer solution containing 0.250 M benzoic acid (C₆H₅COOH) and 0.20 M sodium benzoate (C₆H₅COONa) , what will the pH of the solution be after the addition of 25.0 mL of 0.100M HCl? (K_a (C₆H₅COOH) = 6.5 x 10^-5)

Calculate the percent ionization of cyanic acid, K_a = 2.0 × 10^-4, in a buffer solution that is 0.50 M HCNO and 0.10 M NaCNO.

Calculate the pH of the solution resulting from the addition of 85.0 mL of 0.35 M HCl to 30.0 mL of 0.40 M aniline (C₆H₅NH₂) .K_b(C₆H₅NH₂) = 3.8 x 10^-10