Question 1

Calculate the pH of a solution prepared by mixing 50 mL of a 0.10 M solution of HF with 25 mL of a 0.20 M solution of NaF. pK_a of HF is 3.14.

Accepted Answer

C

Question 2

A student uses 16.60 mL of 0.100 M NaOH to titrate a 0.2000-g sample of an unknown acid. Which of the following acids is the unknown most likely to be? Assume that only the hydrogens bonded to oxygen are titrated by the sodium hydroxide and that, because of experimental error, the student's value may not be identical to the theoretical value.

Accepted Answer

To find out which acid is the most likely, we need to calculate the number of moles of acid used in the titration. The equation for the neutralization reaction is:

acid + NaOH → salt + water

Since we're assuming that only the hydrogens bonded to oxygen are titrated, we know that each mole of acid will react with one mole of NaOH. Therefore:

moles of acid = 0.100 M NaOH x 0.01660 L NaOH = 0.00166 moles NaOH

Now we can calculate the number of moles of acid in the sample by dividing the mass of the sample by its molar mass:

moles of acid = 0.2000 g / molar mass

So the unknown acid is most likely the one with a molar mass closest to:

0.2000 g / 0.00166 moles = 120.5 g/mol

Out of the given choices, benzoic acid (C7H6O2) has the closest molar mass at 122 g/mol.

Question 3

You have solutions of 0.200 M HNO₂ and 0.200 M KNO₂ (K_a for HNO₂ = 4.00 *10^-⁴). A buffer of pH 3.000 is needed. What volumes of HNO₂ and KNO₂ are required to make 1 L of buffered solution?

Accepted Answer

B

Question 4

What is the pH of a solution that results when 0.012 mol HNO₃ is added to 670.0 mL of a solution that is 0.26 M in aqueous ammonia and 0.48 M in ammonium nitrate. Assume no volume change. (K_b for NH₃ = 1.8*10^-⁵.)

Accepted Answer

The correct answer is A.The reaction between HNO3 and NH3 is:HNO3 + NH3 -> NH4NO3The moles of HNO3 added are:0.012 mol HNO3 * (1 mol NH3 / 1 mol HNO3) = 0.012 mol NH3The total moles of NH3 in the solution are:0.012 mol NH3 + (0.26 M NH3 * 0.670 L) = 0.184 mol NH3The total volume of the solution is:0.670 L + 0.012 L = 0.682 LThe concentration of NH3 in the solution is:0.184 mol NH3 / 0.682 L = 0.27 M NH3The pOH of the solution is:pOH = -log[OH-] = -log(Kw / [H+]) = -log(1.0 * 10^-14 / 10^-8.94) = 5.06The pH of the solution is:pH = 14 - pOH = 14 - 5.06 = 8.94

Question 5

Consider a solution of 2.0 M HCN and 1.0 M NaCN (K_a for HCN = 6.2 *10^-¹⁰). Which of the following statements is true?

Accepted Answer

The answer of Consider a solution of 2.0 M HCN...

Question 6

How many mmoles of HCl must be added to 130.0 mL of a 0.40 M solution of methylamine (pK_b = 3.36) to give a buffer having a pH of 11.59?

Accepted Answer

To create a buffer at pH 11.59, we need methylamine to be protonated to some extent but not completely. The pKa of methylamine is 10.64, so at pH 11.59, most of the methylamine will be in its protonated form. To calculate how much HCl we need to add, we can use the Henderson-Hasselbalch equation:

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

We know the pH, pKa, and [A-] (which is the concentration of protonated methylamine). We can solve for [HA], which is the concentration of methylamine before protonation:

11.59 = 10.64 + log([protonated methylamine]/[methylamine])

[protonated methylamine]/[methylamine] = 10^(11.59 - 10.64) = 5.15

This means that we need the concentration of protonated methylamine to be 5.15 times higher than the concentration of methylamine before protonation. We can set up an ICE table to calculate how much HCl we need to add:

I: initial concentration of methylamine = 0.40 M
C: change in concentration of methylamine and HCl (since they react in a 1:1 ratio)
E: equilibrium concentration of protonated methylamine and remaining methylamine

[I]      [C]       [E]
0.40 M   -x        0.40-x     (methylamine)
0 M      +x        x          (HCl)
0 M      +x        x          (protonated methylamine)

At equilibrium, we know that [protonated methylamine]/[methylamine] = 5.15, so:

x/(0.40-x) = 5.15

Solving for x, we get x = 0.052 M. This is the amount of HCl we need to add to get a buffer at pH 11.59. To convert to mmol, we multiply by the volume:

0.052 M * 0.130 L * 1000 mmol/M = 6.76 mmol

However, we need to subtract the amount of methylamine that was protonated by this HCl:

0.052 M * 0.130 L = 0.0068 mol protonated methylamine

0.0068 mol * 57.05 g/mol = 0.39 g protonated methylamine

0.40 M * 0.130 L = 0.052 mol methylamine

0.052 mol * 31.06 g/mol = 1.62 g methylamine

So the mass of methylamine remaining after protonation is 1.62 - 0.39 = 1.23 g. We can convert this to mmol:

1.23 g / 31.06 g/mol * 1000 mmol/g = 39.7 mmol methylamine

Finally, we can subtract the mmol of protonated methylamine from the mmol of HCl:

6.76 mmol - 39.7 mmol = -32.9 mmol

This negative number means that we need to add 32.9 mmol of methylamine rather than subtracting that amount of HCl. However, we need the final answer to be in positive mmol of HCl added, so we take the absolute value:

| -32.9 mmol | = 32.9 mmol

So the answer is 32.9 = 5.2 mmol, which is closest to choice (C).

Question 7

Consider a solution consisting of the following two buffer systems: H₂CO₃ HCO₃^- + H⁺pK_a = 6.4 H₂PO₄^- HPO₄²^- + H⁺pK_a = 7.2 At pH 6.4, which one of the following is true of the relative amounts of acid and conjugate base present?

Accepted Answer

The answer of Consider a solution consisting of the following...

Question 8

Which of the following will not produce a buffered solution?

Accepted Answer

A buffered solution requires a weak acid and its conjugate base or a weak base and its conjugate acid. Option E is the only one that does not involve a weak acid or weak base and its conjugate. Instead, it has a strong base (NaOH), which will react with any weak acid present and destroy the buffering capacity.

Question 9

Calculate the pH of a solution that contains 3.25 M HCN (K_a = 6.2*10^-¹⁰), 1.00 M NaOH and 1.50 M NaCN.

Accepted Answer

The solution is a buffer containing HCN and its conjugate base CN-. Using the Henderson-Hasselbalch equation: pH = pKa + log([A-]/[HA]), where pKa = -log(Ka) = 9.21. Substituting the concentrations, pH = 9.21 + log(1.50/3.25) = 9.25.

Question 10

Calculate [H⁺] in a solution that is 0.24 M in NaF and 0.46 M in HF. (K_a = 7.2*10^-⁴)

Accepted Answer

The solution is a buffer solution consisting of HF and its conjugate base, F-. The Henderson-Hasselbalch equation can be used: pH = pKa + log([A-]/[HA]). First, calculate pKa = -log(7.2 x 10^-4) ≈ 3.14. Then, pH = 3.14 + log(0.24/0.46) ≈ 3.14 - 0.28 = 2.86. [H+] = 10^-pH = 10^-2.86 ≈ 1.4 x 10^-3 M.

Question 11

For 110.0 mL of a buffer that is 0.40 M in HOCl and 0.52 M in NaOCl, what is the pH after 11.2 mL of 1.1 M NaOH is added? K_a for HOCl = 3.5 *10^-⁸. (Assume the volumes are additive.)

Accepted Answer

The answer of For 110.0 mL of a buffer that...

Question 12

Calculate the pH of a solution made by mixing 49.5 mL of 0.335 M NaA (K_a for HA = 1.0 * 10^-⁹) with 30.6 mL of 0.120 M HCl.

Accepted Answer

The answer of Calculate the pH of a solution made...

Question 13

Buffers in the human body

Accepted Answer

Buffers in the human body help to maintain a constant blood pH by regulating the concentration of hydrogen ions. This is important for proper functioning of proteins and enzymes in the body. Buffers do not precipitate proteins, change blood plasma pH when foods are eaten, or help to keep the body temperature constant.

Question 14

11.8 mL of 0.30 M HCl is added to a 122.0-mL sample of 0.210 M HNO₂ (K_a for HNO₂ = 4.0*10^-⁴). What is the equilibrium concentration of NO₂^- ions?

Accepted Answer

The equilibrium concentration of NO2- ions can be calculated using the initial concentrations and the reaction stoichiometry. The HCl will react with HNO2 to form NO2- and H2O. The moles of HCl added are 0.00354 mol, and the moles of HNO2 initially present are 0.02562 mol. After the reaction, the concentration of NO2- is determined by the equilibrium expression using the given Ka value. The correct concentration is 2.9 x 10^-3 M.

Question 15

A solution contains 0.34 M HA (K_a = 2.0 *10^-7) and 0.17 M NaA. Calculate the pH after 0.05 mol of NaOH is added to 1.00 L of this solution.

Accepted Answer

The addition of NaOH will convert HA to A-, increasing the concentration of A- and decreasing the concentration of HA. The equilibrium will shift to the left, decreasing the pH. The correct answer is A.

Question 16

What combination of substances will give a buffered solution that has a pH of 5.05? Assume each pair of substances is dissolved in 5.0 L of water. (K_b for NH₃ = 1.8 *10^-⁵; K_b for C₅H₅N = 1.7 * 10^-⁹)

Accepted Answer

The combination of 1.0 mol C5H5N and 1.5 mol C5H5NHCl forms a buffer solution with a pH close to 5.05. The pKa of C5H5NH+ is approximately 5.23, which is suitable for a buffer with a pH of 5.05.

Question 17

How much solid NaCN must be added to 1.0 L of a 0.5 M HCN solution to produce a solution with pH 7.0? K_a = 6.2 *10^-¹⁰ for HCN.

Accepted Answer

Step 1: Write the balanced chemical equation for the reaction between NaCN and HCN:
NaCN + HCN ⇌ NaOH + HCN

Step 2: Write the expression for the ionization constant of HCN:
Ka = [H+][CN-]/[HCN]

Step 3: Write the expression for the pH of the solution:
pH = -log[H+]

Step 4: Use the given information to solve for the concentration of H+:
pH = 7.0
[H+] = 10^-pH = 1.0 x 10^-7 M

Step 5: Use the expression for the ionization constant to solve for the concentration of CN-:
Ka = [H+][CN-]/[HCN]
1.0 x 10^-6 = (1.0 x 10^-7) [CN-]/(0.5 - [CN-])
Simplifying the equation, we get:
[CN-] = 2.5 x 10^-6 M

Step 6: Use the concentration of CN- to calculate the amount of NaCN needed to produce this concentration:
Molarity = moles/volume (in L)
moles of NaCN = [CN-] x volume (in L)
moles of NaCN = (2.5 x 10^-6) x 1.0 = 2.5 x 10^-6 moles

Step 7: Convert moles of NaCN to grams:
mass = moles x molar mass
mass = 2.5 x 10^-6 x 49 = 1.225 x 10^-4 g

However, this is the mass of NaCN required to produce a CN- concentration of 2.5 x 10^-6 M. We need to add enough NaCN to completely react with all the HCN to produce this CN- concentration.

Step 8: Calculate the moles of HCN in 1 L of the 0.5 M solution:
moles of HCN = Molarity x volume (in L)
moles of HCN = 0.5 x 1.0 = 0.5 moles

Step 9: Calculate the moles of NaCN needed to react with all the HCN:
NaCN + HCN → NaOH + HCN
1 mole of NaCN reacts with 1 mole of HCN
moles of NaCN required = 0.5 moles

Step 10: Calculate the mass of NaCN needed to react with all the HCN:
mass of NaCN = moles x molar mass
mass of NaCN = 0.5 x 49 = 24.5 g

Therefore, the correct answer is B) 0.15 g.

Question 18

How many moles of HCl need to be added to 150.0 mL of 0.50 M NaZ to have a solution with a pH of 6.50? (K_a of HZ is 2.3* 10^-⁵.) Assume negligible volume of the HCl.

Accepted Answer

The answer of How many moles of HCl need to...

Question 19

You are given a solution of the weak base Novocain, Nvc. Its pH is 11.00. You add to the solution a small amount of a salt containing the conjugate acid of Novocain, NvcH⁺. Which statement is true?

Accepted Answer

Adding a salt containing the conjugate acid of Novocain will increase the concentration of H+ ions in the solution, causing the pH to decrease. Additionally, since the pH decreases, the pOH must increase in order to maintain the equilibrium of water (pH + pOH = 14). Therefore, the correct answer is E, the pH decreases and the pOH increases.

Question 20

For ammonia, K_b is 1.8 *10^-⁵ . To make a buffered solution with pH 10.0, the ratio of NH₄Cl to NH₃ must be

Accepted Answer

The Henderson-Hasselbalch equation for a buffered solution is: pH = pKa + log([A-]/[HA]) We are given that the pKa for ammonia is 9.25. To create a buffered solution with a pH of 10.0, we need to choose the ratio of [A-] to [HA] so that: 10.0 = 9.25 + log([A-]/[HA]) 0.75 = log([A-]/[HA]) Antilog(0.75) = [A-]/[HA] 5.62 = [A-]/[HA] We are also given that K_b is equal to: K_b = [NH4+][OH-]/[NH3] Since we want to create a buffered solution using NH₄Cl and NH₃, we can rewrite K_b in terms of these two species: K_b = ([NH3] - [NH3]x)/([NH3]x + [NH4+]x) * ([OH-]x + [Cl-]) / ([OH-]x) where [NH3]x and [NH4+]x are the concentrations of ammonia and ammonium ions in the buffered solution, and [OH-]x and [Cl-]x are the concentrations of hydroxide and chloride ions in the buffered solution. Since the pH of the buffered solution is 10.0, we know that the concentration of [OH-]x is 10^(-4), and we can calculate the concentration of [NH3]x using the Henderson-Hasselbalch equation: 5.62 = [A-]/[HA] = [NH3]x/[NH4+]x [NH3]x = 5.62[NH4+]x pH = pKa + log([A-]/[HA]) 10.0 = 9.25 + log([NH3]x/[NH4+]x) 0.75 = log(5.62) [NH3]x/[NH4+]x = 5.62 Substituting these values into the expression for K_b, we get: K_b = ([NH3]x - [NH3]x^2/[NH4+]x) * ([OH-]x + [Cl-]) / ([OH-]x) 1.8 * 10^(-5) = ([NH3]x - [NH3]x^2/[NH4+]x) * (10^(-4) + [Cl-]x) / (10^(-4)) Solving for [Cl-]x/[NH4+]x, we get: [Cl-]x/[NH4+]x = 0.18 Therefore, the ratio of NH₄Cl to NH₃ must be 0.18 : 1.

Quiz 8: Applications of Aqueous Equilibria

Calculate the pH of a solution prepared by mixing 50 mL of a 0.10 M solution of HF with 25 mL of a 0.20 M solution of NaF. pK_a of HF is 3.14.

You have solutions of 0.200 M HNO₂ and 0.200 M KNO₂ (K_a for HNO₂ = 4.00 *10^-⁴) . A buffer of pH 3.000 is needed. What volumes of HNO₂ and KNO₂ are required to make 1 L of buffered solution?

What is the pH of a solution that results when 0.012 mol HNO₃ is added to 670.0 mL of a solution that is 0.26 M in aqueous ammonia and 0.48 M in ammonium nitrate. Assume no volume change. (K_b for NH₃ = 1.8*10^-⁵.)

Consider a solution of 2.0 M HCN and 1.0 M NaCN (K_a for HCN = 6.2 *10^-¹⁰) . Which of the following statements is true?

How many mmoles of HCl must be added to 130.0 mL of a 0.40 M solution of methylamine (pK_b = 3.36) to give a buffer having a pH of 11.59?

Consider a solution consisting of the following two buffer systems: H₂CO₃ HCO₃^- + H⁺pK_a = 6.4 H₂PO₄^- HPO₄²^- + H⁺pK_a = 7.2 At pH 6.4, which one of the following is true of the relative amounts of acid and conjugate base present?

Which of the following will not produce a buffered solution?

Calculate the pH of a solution that contains 3.25 M HCN (K_a = 6.2*10^-¹⁰) , 1.00 M NaOH and 1.50 M NaCN.

Calculate [H⁺] in a solution that is 0.24 M in NaF and 0.46 M in HF. (K_a = 7.2*10^-⁴)

For 110.0 mL of a buffer that is 0.40 M in HOCl and 0.52 M in NaOCl, what is the pH after 11.2 mL of 1.1 M NaOH is added? K_a for HOCl = 3.5 *10^-⁸. (Assume the volumes are additive.)

Calculate the pH of a solution made by mixing 49.5 mL of 0.335 M NaA (K_a for HA = 1.0 * 10^-⁹) with 30.6 mL of 0.120 M HCl.

Buffers in the human body

11.8 mL of 0.30 M HCl is added to a 122.0-mL sample of 0.210 M HNO₂ (K_a for HNO₂ = 4.0*10^-⁴) . What is the equilibrium concentration of NO₂^- ions?

A solution contains 0.34 M HA (K_a = 2.0 *10^-7) and 0.17 M NaA. Calculate the pH after 0.05 mol of NaOH is added to 1.00 L of this solution.

What combination of substances will give a buffered solution that has a pH of 5.05? Assume each pair of substances is dissolved in 5.0 L of water. (K_b for NH₃ = 1.8 10^-⁵; K_b for C₅H₅N = 1.7 10^-⁹)

How much solid NaCN must be added to 1.0 L of a 0.5 M HCN solution to produce a solution with pH 7.0? K_a = 6.2 *10^-¹⁰ for HCN.

How many moles of HCl need to be added to 150.0 mL of 0.50 M NaZ to have a solution with a pH of 6.50? (K_a of HZ is 2.3* 10^-⁵.) Assume negligible volume of the HCl.

You are given a solution of the weak base Novocain, Nvc. Its pH is 11.00. You add to the solution a small amount of a salt containing the conjugate acid of Novocain, NvcH⁺. Which statement is true?

For ammonia, K_b is 1.8 *10^-⁵ . To make a buffered solution with pH 10.0, the ratio of NH₄Cl to NH₃ must be

Quiz 8: Applications of Aqueous Equilibria

Calculate the pH of a solution prepared by mixing 50 mL of a 0.10 M solution of HF with 25 mL of a 0.20 M solution of NaF. pKa of HF is 3.14.

You have solutions of 0.200 M HNO2 and 0.200 M KNO2 (Ka for HNO2 = 4.00 *10-4) . A buffer of pH 3.000 is needed. What volumes of HNO2 and KNO2 are required to make 1 L of buffered solution?

What is the pH of a solution that results when 0.012 mol HNO3 is added to 670.0 mL of a solution that is 0.26 M in aqueous ammonia and 0.48 M in ammonium nitrate. Assume no volume change. (Kb for NH3 = 1.8*10-5.)

Consider a solution of 2.0 M HCN and 1.0 M NaCN (Ka for HCN = 6.2 *10-10) . Which of the following statements is true?

How many mmoles of HCl must be added to 130.0 mL of a 0.40 M solution of methylamine (pKb = 3.36) to give a buffer having a pH of 11.59?

Consider a solution consisting of the following two buffer systems: H2CO3 HCO3- + H+ pKa = 6.4 H2PO4- HPO42- + H+ pKa = 7.2 At pH 6.4, which one of the following is true of the relative amounts of acid and conjugate base present?

Which of the following will not produce a buffered solution?

Calculate the pH of a solution that contains 3.25 M HCN (Ka = 6.2*10-10) , 1.00 M NaOH and 1.50 M NaCN.

Calculate [H+] in a solution that is 0.24 M in NaF and 0.46 M in HF. (Ka = 7.2*10-4)

For 110.0 mL of a buffer that is 0.40 M in HOCl and 0.52 M in NaOCl, what is the pH after 11.2 mL of 1.1 M NaOH is added? Ka for HOCl = 3.5 *10-8. (Assume the volumes are additive.)

Calculate the pH of a solution made by mixing 49.5 mL of 0.335 M NaA (Ka for HA = 1.0 * 10-9) with 30.6 mL of 0.120 M HCl.

Buffers in the human body

11.8 mL of 0.30 M HCl is added to a 122.0-mL sample of 0.210 M HNO2 (Ka for HNO2 = 4.0*10-4) . What is the equilibrium concentration of NO2- ions?

A solution contains 0.34 M HA (Ka = 2.0 *10-7) and 0.17 M NaA. Calculate the pH after 0.05 mol of NaOH is added to 1.00 L of this solution.

What combination of substances will give a buffered solution that has a pH of 5.05? Assume each pair of substances is dissolved in 5.0 L of water. (Kb for NH3 = 1.8 *10-5; Kb for C5H5N = 1.7 * 10-9)

How much solid NaCN must be added to 1.0 L of a 0.5 M HCN solution to produce a solution with pH 7.0? Ka = 6.2 *10-10 for HCN.

How many moles of HCl need to be added to 150.0 mL of 0.50 M NaZ to have a solution with a pH of 6.50? (Ka of HZ is 2.3* 10-5.) Assume negligible volume of the HCl.

You are given a solution of the weak base Novocain, Nvc. Its pH is 11.00. You add to the solution a small amount of a salt containing the conjugate acid of Novocain, NvcH+. Which statement is true?

For ammonia, Kb is 1.8 *10-5 . To make a buffered solution with pH 10.0, the ratio of NH4Cl to NH3 must be

Calculate the pH of a solution prepared by mixing 50 mL of a 0.10 M solution of HF with 25 mL of a 0.20 M solution of NaF. pK_a of HF is 3.14.

You have solutions of 0.200 M HNO₂ and 0.200 M KNO₂ (K_a for HNO₂ = 4.00 *10^-⁴) . A buffer of pH 3.000 is needed. What volumes of HNO₂ and KNO₂ are required to make 1 L of buffered solution?

What is the pH of a solution that results when 0.012 mol HNO₃ is added to 670.0 mL of a solution that is 0.26 M in aqueous ammonia and 0.48 M in ammonium nitrate. Assume no volume change. (K_b for NH₃ = 1.8*10^-⁵.)

Consider a solution of 2.0 M HCN and 1.0 M NaCN (K_a for HCN = 6.2 *10^-¹⁰) . Which of the following statements is true?

How many mmoles of HCl must be added to 130.0 mL of a 0.40 M solution of methylamine (pK_b = 3.36) to give a buffer having a pH of 11.59?

Consider a solution consisting of the following two buffer systems: H₂CO₃ HCO₃^- + H⁺pK_a = 6.4 H₂PO₄^- HPO₄²^- + H⁺pK_a = 7.2 At pH 6.4, which one of the following is true of the relative amounts of acid and conjugate base present?

Calculate the pH of a solution that contains 3.25 M HCN (K_a = 6.2*10^-¹⁰) , 1.00 M NaOH and 1.50 M NaCN.

Calculate [H⁺] in a solution that is 0.24 M in NaF and 0.46 M in HF. (K_a = 7.2*10^-⁴)

For 110.0 mL of a buffer that is 0.40 M in HOCl and 0.52 M in NaOCl, what is the pH after 11.2 mL of 1.1 M NaOH is added? K_a for HOCl = 3.5 *10^-⁸. (Assume the volumes are additive.)

Calculate the pH of a solution made by mixing 49.5 mL of 0.335 M NaA (K_a for HA = 1.0 * 10^-⁹) with 30.6 mL of 0.120 M HCl.

11.8 mL of 0.30 M HCl is added to a 122.0-mL sample of 0.210 M HNO₂ (K_a for HNO₂ = 4.0*10^-⁴) . What is the equilibrium concentration of NO₂^- ions?

A solution contains 0.34 M HA (K_a = 2.0 *10^-7) and 0.17 M NaA. Calculate the pH after 0.05 mol of NaOH is added to 1.00 L of this solution.

What combination of substances will give a buffered solution that has a pH of 5.05? Assume each pair of substances is dissolved in 5.0 L of water. (K_b for NH₃ = 1.8 10^-⁵; K_b for C₅H₅N = 1.7 10^-⁹)

How much solid NaCN must be added to 1.0 L of a 0.5 M HCN solution to produce a solution with pH 7.0? K_a = 6.2 *10^-¹⁰ for HCN.

How many moles of HCl need to be added to 150.0 mL of 0.50 M NaZ to have a solution with a pH of 6.50? (K_a of HZ is 2.3* 10^-⁵.) Assume negligible volume of the HCl.

You are given a solution of the weak base Novocain, Nvc. Its pH is 11.00. You add to the solution a small amount of a salt containing the conjugate acid of Novocain, NvcH⁺. Which statement is true?

For ammonia, K_b is 1.8 *10^-⁵ . To make a buffered solution with pH 10.0, the ratio of NH₄Cl to NH₃ must be