Question 1

If the pKa of HCHO₂ is 3.74 and the pH of an HCHO₂/NaCHO₂solution is 3.11,which of the following is TRUE?

Accepted Answer

The pH of the solution is lower than the pKa of the acid, indicating that the solution is more acidic than the acid's conjugate base/acid ratio at equilibrium, which means that more of the acid exists in its protonated form, HCHO₂. Therefore, [HCHO₂] must be greater than [NaCHO₂]. Choice D is the only option that reflects this relationship. Furthermore, since the acid is present in its protonated form, it can be used to make a buffer with a pH lower than 3.74.

Question 2

A 1.00 L buffer solution is 0.250 M in HF and 0.250 M in LiF.Calculate the pH of the solution after the addition of 0.150 moles of solid LiOH.Assume no volume change upon the addition of base.The K_a for HF is 3.5 × 10^-4.

Accepted Answer

The addition of LiOH to the buffer solution will result in the formation of LiF and water:

LiOH + HF → LiF + H2O

Since LiOH is a strong base, it will react completely with the weak acid HF, leaving LiF as the only remaining salt in solution. We can use the Henderson-Hasselbalch equation to calculate the pH of the solution after this reaction:

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

where pKa is the dissociation constant of HF (3.5 × 10^-4), [A-] is the concentration of F- (0.250 - 0.150 = 0.100 M), and [HA] is the concentration of HF (0.250 M).

pH = -log(3.5 × 10^-4) + log(0.100/0.250) = 4.06

Therefore, the pH of the solution after the addition of 0.150 moles of LiOH is 4.06.

Question 3

Calculate the pH of a solution formed by mixing 250.0 mL of 0.15 M NH₄Cl with 200.0 mL of 0.12 M NH₃.The K_b for NH₃ is 1.8 × 10^-5.

Accepted Answer

The correct answer is A because the pH of the solution is 9.06. This can be calculated using the Henderson-Hasselbalch equation: pH = pKa + log([A-]/[HA]), where pKa is the dissociation constant of the weak acid (NH4+), [A-] is the concentration of the conjugate base (NH3), and [HA] is the concentration of the weak acid (NH4Cl). In this case, the pKa of NH4+ is 9.25, [A-] is 0.12 M, and [HA] is 0.15 M. Plugging these values into the equation gives a pH of 9.06.

Question 4

If the pKa of HCHO₂ is 3.74 and the pH of an HCHO₂/NaCHO₂solution is 3.74,which of the following is TRUE?

Accepted Answer

At a pH equal to the pKa of the acid, the acid and its conjugate base are present in equal amounts. Therefore, in this case, [HCHO₂] = [NaCHO₂]. Choice B is the only option that reflects this relationship between the acid and its conjugate base at a pH equal to the pKa.

Question 5

If the pKa of HCHO₂ is 3.74 and the pH of an HCHO₂/NaCHO₂solution is 3.89,which of the following is TRUE?

Accepted Answer

Since the pH (3.89) is higher than the pKa (3.74), it means the solution is more basic relative to the acid's dissociation point. Therefore, there are more conjugate base ions (NaCHO₂) than there are undissociated acid molecules (HCHO₂).

Question 6

Calculate the pH of a solution formed by mixing 150.0 mL of 0.10 M HC₇H₅O₂ with 100.0 mL of 0.30 M NaC₇H₅O₂.The K_a for HC₇H₅O₂ is 6.5 × 10^-5.

Accepted Answer

First, we need to calculate the moles of each species present in the solution.

Moles of HCS1U1B17S1U1B0HS1U1B15S1U1B0OS1U1B12S1U1B0:

0.10 mol/L * 0.150 L = 0.015 mol

Moles of NaCS1U1B17S1U1B0HS1U1B15S1U1B0OS1U1B12S1U1B0:

0.30 mol/L * 0.100 L = 0.030 mol

Now, we need to determine which species will act as the acid and which as the base in this solution. Since HCS1U1B17S1U1B0HS1U1B15S1U1B0OS1U1B12S1U1B0 is a weak acid, it will donate a proton to water and act as the acid. NaCS1U1B17S1U1B0HS1U1B15S1U1B0OS1U1B12S1U1B0 will accept the proton and act as the base.

HCS1U1B17S1U1B0HS1U1B15S1U1B0OS1U1B12S1U1B0 + H2O --> H3O+ + CS1U1B17S1U1B0HS1U1B15S1U1B0OS1U1B12S1U1B0

Using the Ka expression for HCS1U1B17S1U1B0HS1U1B15S1U1B0OS1U1B12S1U1B0, we can solve for the concentration of H3O+ and thus the pH:

Ka = [H3O+][CS1U1B17S1U1B0HS1U1B15S1U1B0OS1U1B12S1U1B0] / [HCS1U1B17S1U1B0HS1U1B15S1U1B0OS1U1B12S1U1B0]

6.5 × 10S1U1P1-5 = x^2 / (0.015 - x)

Assuming x << 0.015, we can simplify this to:

6.5 × 10S1U1P1-5 = x^2 / 0.015

x = 2.02 × 10S1U1P1-4

pH = -log[H3O+] = -log(2.02 × 10S1U1P1-4) = 4.49

Therefore, the answer is (C).

Question 7

In the blood,the concentration of bicarbonate is ________ times higher than the concentration of carbonic acid.

Accepted Answer

The concentration of bicarbonate in the blood is approximately 20 times higher than that of carbonic acid, maintaining the acid-base balance.

Question 8

A 1.00 L buffer solution is 0.150 M in HC₇H₅O₂ and 0.250 M in LiC₇H₅O₂.Calculate the pH of the solution after the addition of 100.0 mL of 1.00 M HCl.The K_a for HC₇H₅O₂ is 6.5 × 10^-5.

Accepted Answer

The answer of A 1.00 L buffer solution is 0.150...

Question 9

A 1.50 L buffer solution is 0.250 M in HF and 0.250 M in NaF.Calculate the pH of the solution after the addition of 0.100 moles of solid NaOH.Assume no volume change upon the addition of base.The K_a for HF is 3.5 × 10^-4.

Accepted Answer

Before the addition of NaOH, the buffer solution consists of HF and NaF in equal concentration, so the pH can be calculated using the Henderson-Hasselbalch equation:

pH = pKa + log([A-]/[HA])
where pKa = -log(Ka) = 3.5, [HA] = 0.250 M (concentration of HF), and [A-] = 0.250 M (concentration of NaF).

pH = 3.5 + log(0.250/0.250)
pH = 3.5 + log(1)
pH = 3.5

After the addition of 0.100 moles of NaOH, the NaOH will react with the HF in the buffer solution. The balanced chemical equation for the reaction is:

HF + NaOH -> NaF + H2O

NaOH is a strong base and will completely ionize in water, so we can assume that 0.100 moles of OH- are added to the buffer solution. The HF will react with the OH- to produce water and F-, according to the equation:

HF + OH- -> F- + H2O

Since the HF concentration is much larger than the initial OH- concentration (0.100 moles in 1.50 L is equivalent to 0.067 M), the HF will be completely consumed before the NaF is significantly affected. This means that the buffer solution will no longer be in equilibrium, and we will need to calculate the pH using the concentrations of the remaining species.

First, we need to calculate the amount of HF that reacts with the OH-:

0.067 M HF * V = 0.100 mol OH-
V = 1.493 L

This means that 1.493 L of the buffer solution reacts with the OH-, leaving 0.007 L (or 7 mL) of the original buffer solution unreacted. The concentration of F- remains at 0.250 M, since NaF is a salt and does not react with the OH-. The concentration of HF is now:

moles HF = (0.067 M HF) * (1.493 L HF - 0.100 L OH-) = 0.0928 mol HF
concentration HF = 0.0928 mol HF / 0.007 L solution = 13.3 M

The concentration of H+ can be calculated from the concentration of HF using the Ka value:

Ka = [H+][F-] / [HF]

[H+] = Ka * [HF] / [F-] = (3.5 x 10^-4) * (13.3 M) / (0.250 M) = 0.0187 M

pH = -log[H+] = -log(0.0187) = 1.73

However, this pH value is not correct because we assumed that the volume was constant upon the addition of NaOH. In reality, the addition of NaOH will increase the volume of the solution slightly, which will dilute the concentrations of HF and F-. In order to take this into account, we need to recalculate the concentrations of HF, F-, and H+ using the new total volume:

Vfinal = Vinitial + 0.100 L NaOH = 1.50 L + 0.100 L = 1.60 L

The new concentration of HF is:

mol HF = 0.0928 mol / 1.50 L = 0.0619 M
new [HF] = 0.0619 M * (1.50 L / 1.60 L) = 0.0582 M

The new concentration of F- is:

new [F-] = 0.250 M * (1.50 L / 1.60 L) = 0.234 M

The new concentration of H+ can be calculated using the new [HF] and [F-]:

Ka = [H+][F-] / [HF]
[H+] = Ka * [HF] / [F-] = (3.5 x 10^-4) * (0.0582 M) / (0.234 M) = 8.76 x 10^-5 M

pH = -log[H+] = -log(8.76 x 10^-5) = 3.69

Therefore, the correct answer is C.

Question 10

Identify the structure for antifreeze.

Accepted Answer

Ethylene glycol (HOCH2CH2OH) is commonly used as an antifreeze. Option C represents the structure of ethylene glycol, with two hydroxyl groups (-OH) on a two-carbon chain, making it an effective antifreeze.

Question 11

Calculate the pH of a buffer that is 0.225 M HC₂H₃O₂ and 0.162 M KC₂H₃O₂.The K_a for HC₂H₃O₂ is 1.8 × 10^-5.

Accepted Answer

The answer of Calculate the pH of a buffer that...

Question 12

When titrating a weak monoprotic acid with NaOH at 25°C,the

Accepted Answer

When titrating a weak monoprotic acid with NaOH, the equivalence point occurs when all of the acid molecules have reacted with the base molecules to form a salt and water. At this point, the pH will be greater than 7 because the salt (formed from the acid and base) will have a basic anion (from the base) and an acidic cation (from the acid). This will result in a slightly basic solution. Therefore, choice C is the correct answer.

Question 13

A 1.50 L buffer solution is 0.250 M in HF and 0.250 M in NaF.Calculate the pH of the solution after the addition of 0.0500 moles of solid NaOH.Assume no volume change upon the addition of base.The K_a for HF is 3.5 × 10^-4.

Accepted Answer

The addition of NaOH to the buffer solution will react with HF to form NaF and water. Since 0.0500 moles of NaOH are added, the concentration of HF will decrease by 0.0500 moles/L, and the concentration of NaF will increase by the same amount. The new concentrations in the 1.50 L solution are:- HF: $0.250 - \frac{0.0500}{1.50} = 0.2167$ M- NaF: $0.250 + \frac{0.0500}{1.50} = 0.2833$ MUsing the Henderson-Hasselbalch equation: $pH = pKa + \log\left(\frac{[A^-]}{[HA]}\right)$Given $pKa = -\log(3.5 \times 10^{-4}) = 3.46$,$pH = 3.46 + \log\left(\frac{0.2833}{0.2167}\right) = 3.46 + \log(1.307) = 3.46 + 0.116 = 3.576$Rounded to two decimal places, the pH is approximately 3.57.

Question 14

When titrating a strong monoprotic acid and KOH at 25°C,the

Accepted Answer

In the titration of a strong monoprotic acid with a strong base like KOH, the pH at the equivalence point is 7 because the acid and base completely neutralize each other, forming water and a neutral salt.

Question 15

Which of the following is TRUE?

Accepted Answer

The equivalence point of an acid-base titration is defined as the point where the moles of acid and moles of base are equal. This means that the amount of acid equals the amount of base at this point.

Option B is not true because the pH at the equivalence point depends on the particular acid-base system being titrated. It can be higher, lower or equal to 7.

Option C is not true because an indicator is a substance that undergoes a color change as the pH of the solution changes. So, it is pH sensitive.

Option D is not true because a titration curve is a plot of pH vs. the volume of titrant added.

Question 16

Calculate the pH of a solution formed by mixing 250.0 mL of 0.15 M NH₄Cl with 100.0 mL of 0.20 M NH₃.The K_b for NH₃ is 1.8 × 10^-5.

Accepted Answer

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

Question 17

Which of the following is TRUE?

Accepted Answer

A buffer is most resistant to pH change when [acid] = [conjugate base]. This is because when the concentrations of acid and conjugate base are equal, they can react and neutralize any added acid or base, thus maintaining the pH of the solution. The other options are not true: A) An effective buffer has a [base]/[acid] ratio in the range of 10 - 100 - this is just a guideline, not a strict rule. C) An effective buffer has very small absolute concentrations of acid and conjugate base - this is not necessarily true, as long as the concentrations are not too low or too high. D) A buffer cannot be destroyed by adding too much strong base. It can only be destroyed by adding too much strong acid - this is also not true, as adding too much strong base can shift the equilibrium towards the basic form, destroying the buffer.

Question 18

Define buffer capacity.

Accepted Answer

Buffer capacity is the measure of the ability of a buffer solution to resist changes in pH upon addition of acid or base. It is determined by the concentration of the buffer components and their relative acid-base strengths. Buffer capacity is the amount of acid or base that can be added to a buffer solution without significantly changing its pH or destroying its effectiveness. Therefore, option A is the best choice.

Question 19

A 1.00 L buffer solution is 0.250 M in HF and 0.250 M in NaF.Calculate the pH of the solution after the addition of 100.0 mL of 1.00 M HCl.The K_a for HF is 3.5 × 10^-4.

Accepted Answer

The correct answer is A because the pH of the solution after the addition of 100.0 mL of 1.00 M HCl can be calculated using the Henderson-Hasselbalch equation:pH = pKa + log([A-]/[HA])where pKa is the dissociation constant of the acid, [A-] is the concentration of the conjugate base, and [HA] is the concentration of the acid.In this case, the pKa of HF is 3.5 × 10^-4, the concentration of the conjugate base (NaF) is 0.250 M, and the concentration of the acid (HF) is 0.250 M.Substituting these values into the Henderson-Hasselbalch equation, we get:pH = 3.5 × 10^-4 + log(0.250/0.250) = 3.5 × 10^-4Therefore, the pH of the solution after the addition of 100.0 mL of 1.00 M HCl is 3.5 × 10^-4.

Question 20

Calculate the pH of a solution formed by mixing 250.0 mL of 0.15 M HCHO₂ with 100.0 mL of 0.20 M LiCHO₂.The K_a for HCHO₂ is 1.8 × 10^-4.

Accepted Answer

First, we need to write the balanced chemical equation of the reaction between HCHO₂ and LiCHO₂ to determine which species will act as the acid and which will act as the base: HCHO₂ + LiCHO₂ → CHO₂Li + H2O In this reaction, HCHO₂ acts as the acid (donates a proton) and LiCHO₂ acts as the base (accepts a proton) to form CHO₂Li, which is a salt. Next, we need to calculate the moles of acid (HCHO₂) and base (LiCHO₂) in the mixture: moles of HCHO₂ = (0.15 mol/L) x (0.250 L) = 0.0375 mol moles of LiCHO₂ = (0.20 mol/L) x (0.100 L) = 0.0200 mol Since we have more acid than base, the remaining HCHO₂ will be present in the solution. Using the Henderson-Hasselbalch equation, we can calculate the pH of the solution: pH = pKa + log([base]/[acid]) pKa = -log(Ka) = -log(1.8 × 10^-4) = 3.74 [base] = moles of CHO₂Li / total volume of solution = (0.0375 mol) / (0.250 L + 0.100 L) = 0.107 M [acid] = moles of HCHO₂ remaining / total volume of solution = (0.0375 - 0.0200) mol / (0.250 L + 0.100 L) = 0.0675 M pH = 3.74 + log(0.107/0.0675) = 3.47 Therefore, the pH of the solution is 3.47, which is best represented by choice D.

Quiz 17: Aqueous Ionic Equilibrium

If the pKa of HCHO2 is 3.74 and the pH of an HCHO2/NaCHO2 solution is 3.11,which of the following is TRUE?

A 1.00 L buffer solution is 0.250 M in HF and 0.250 M in LiF.Calculate the pH of the solution after the addition of 0.150 moles of solid LiOH.Assume no volume change upon the addition of base.The Ka for HF is 3.5 × 10-4.

Calculate the pH of a solution formed by mixing 250.0 mL of 0.15 M NH4Cl with 200.0 mL of 0.12 M NH3.The Kb for NH3 is 1.8 × 10-5.

If the pKa of HCHO2 is 3.74 and the pH of an HCHO2/NaCHO2 solution is 3.74,which of the following is TRUE?

If the pKa of HCHO2 is 3.74 and the pH of an HCHO2/NaCHO2 solution is 3.89,which of the following is TRUE?

Calculate the pH of a solution formed by mixing 150.0 mL of 0.10 M HC7H5O2 with 100.0 mL of 0.30 M NaC7H5O2.The Ka for HC7H5O2 is 6.5 × 10-5.

In the blood,the concentration of bicarbonate is ________ times higher than the concentration of carbonic acid.

A 1.00 L buffer solution is 0.150 M in HC7H5O2 and 0.250 M in LiC7H5O2.Calculate the pH of the solution after the addition of 100.0 mL of 1.00 M HCl.The Ka for HC7H5O2 is 6.5 × 10-5.

A 1.50 L buffer solution is 0.250 M in HF and 0.250 M in NaF.Calculate the pH of the solution after the addition of 0.100 moles of solid NaOH.Assume no volume change upon the addition of base.The Ka for HF is 3.5 × 10-4.

Identify the structure for antifreeze.

Calculate the pH of a buffer that is 0.225 M HC2H3O2 and 0.162 M KC2H3O2.The Ka for HC2H3O2 is 1.8 × 10-5.

When titrating a weak monoprotic acid with NaOH at 25°C,the

A 1.50 L buffer solution is 0.250 M in HF and 0.250 M in NaF.Calculate the pH of the solution after the addition of 0.0500 moles of solid NaOH.Assume no volume change upon the addition of base.The Ka for HF is 3.5 × 10-4.

When titrating a strong monoprotic acid and KOH at 25°C,the

Which of the following is TRUE?

Calculate the pH of a solution formed by mixing 250.0 mL of 0.15 M NH4Cl with 100.0 mL of 0.20 M NH3.The Kb for NH3 is 1.8 × 10-5.

Which of the following is TRUE?

Define buffer capacity.

A 1.00 L buffer solution is 0.250 M in HF and 0.250 M in NaF.Calculate the pH of the solution after the addition of 100.0 mL of 1.00 M HCl.The Ka for HF is 3.5 × 10-4.

Calculate the pH of a solution formed by mixing 250.0 mL of 0.15 M HCHO2 with 100.0 mL of 0.20 M LiCHO2.The Ka for HCHO2 is 1.8 × 10-4.

If the pKa of HCHO₂ is 3.74 and the pH of an HCHO₂/NaCHO₂solution is 3.11,which of the following is TRUE?

A 1.00 L buffer solution is 0.250 M in HF and 0.250 M in LiF.Calculate the pH of the solution after the addition of 0.150 moles of solid LiOH.Assume no volume change upon the addition of base.The K_a for HF is 3.5 × 10^-4.

Calculate the pH of a solution formed by mixing 250.0 mL of 0.15 M NH₄Cl with 200.0 mL of 0.12 M NH₃.The K_b for NH₃ is 1.8 × 10^-5.

If the pKa of HCHO₂ is 3.74 and the pH of an HCHO₂/NaCHO₂solution is 3.74,which of the following is TRUE?

If the pKa of HCHO₂ is 3.74 and the pH of an HCHO₂/NaCHO₂solution is 3.89,which of the following is TRUE?

Calculate the pH of a solution formed by mixing 150.0 mL of 0.10 M HC₇H₅O₂ with 100.0 mL of 0.30 M NaC₇H₅O₂.The K_a for HC₇H₅O₂ is 6.5 × 10^-5.

A 1.00 L buffer solution is 0.150 M in HC₇H₅O₂ and 0.250 M in LiC₇H₅O₂.Calculate the pH of the solution after the addition of 100.0 mL of 1.00 M HCl.The K_a for HC₇H₅O₂ is 6.5 × 10^-5.

A 1.50 L buffer solution is 0.250 M in HF and 0.250 M in NaF.Calculate the pH of the solution after the addition of 0.100 moles of solid NaOH.Assume no volume change upon the addition of base.The K_a for HF is 3.5 × 10^-4.

Calculate the pH of a buffer that is 0.225 M HC₂H₃O₂ and 0.162 M KC₂H₃O₂.The K_a for HC₂H₃O₂ is 1.8 × 10^-5.

A 1.50 L buffer solution is 0.250 M in HF and 0.250 M in NaF.Calculate the pH of the solution after the addition of 0.0500 moles of solid NaOH.Assume no volume change upon the addition of base.The K_a for HF is 3.5 × 10^-4.

Calculate the pH of a solution formed by mixing 250.0 mL of 0.15 M NH₄Cl with 100.0 mL of 0.20 M NH₃.The K_b for NH₃ is 1.8 × 10^-5.

A 1.00 L buffer solution is 0.250 M in HF and 0.250 M in NaF.Calculate the pH of the solution after the addition of 100.0 mL of 1.00 M HCl.The K_a for HF is 3.5 × 10^-4.

Calculate the pH of a solution formed by mixing 250.0 mL of 0.15 M HCHO₂ with 100.0 mL of 0.20 M LiCHO₂.The K_a for HCHO₂ is 1.8 × 10^-4.