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AP 단원별 문제

Ap chem 프린스턴 화학 summary part 8

IB, AP, A Level 화학 2024. 10. 31. 04:08

Use the following information to answer questions 1–5.

A student titrates 20.0 mL of 1.0 M NaOH with 2.0 M formic acid, HCO2H

(Ka = 1.8 × 10-4). Formic acid is a monoprotic acid.

Question 1:

A solution of 0.1 M sodium hydroxide (NaOH) is being titrated into 20.0 mL of formic acid (HCOOH) of unknown concentration. It takes 10.0 mL, 0.05 M of the NaOH solution to reach the equivalence point. What is the concentration of the formic acid solution?

Options:

  • (A) 0.050 M
  • (B) 0.100 M
  • (C) 0.200 M
  • (D) 0.500 M
View Answer 
Answer Explanation:
To find the concentration of the formic acid, use the titration formula:
\[
C_{\text{acid}} \times V_{\text{acid}} = C_{\text{base}} \times V_{\text{base}}
\]
\[
C_{\text{acid}} \times 20.0\,\text{mL} = 0.1\,\text{M} \times 10.0\,\text{mL}
\]
\[
C_{\text{acid}} = \frac{0.1\,\text{M} \times 10.0\,\text{mL}}{20.0\,\text{mL}} = 0.050\,\text{M}
\]
Correct Answer: (A)

Question 2:

At the equivalence point, is the solution acidic, basic, or neutral? Why?

Options:

  • (A) Acidic; the strong acid dissociates more than the weak base
  • (B) Basic; the only acidic or basic ion present at equilibrium is the conjugate base
  • (C) Basic; the higher concentration of the base is the determining factor
  • (D) Neutral; equal moles of both acid and base are present
View Answer 
Correct Answer: (B)

Question 3:

If the formic acid were replaced with a strong acid such as HCl at the same concentration (2.0 M), how would that change the volume needed to reach the equivalence point?

Options:

  • (A) The change would reduce the amount, as the acid now fully dissociates
  • (B) The change would reduce the amount, because the base will be more strongly attracted to the acid.
  • (C) The change would increase the amount, because the reaction will now go to completion instead of equilibrium
  • (D) Changing the strength of the acid will not change the volume needed to reach equivalence.
View Answer 
Correct Answer: (D)

Question 4:

Which of the following would create a good buffer when dissolved in formic acid?

Options:

  • (A) NaCO₂H
  • (B) HC₂H₃O₂
  • (C) NH₃
  • (D) H₂O
View Answer 
Correct Answer: (A)

Question 5:

The equation below represents the reaction between the base methylamine and water. Which of the following best represents the concentrations of the various species at equilibrium?

Options:

  • (A) [OH⁻] > [CH₃NH₂] = [CH₃NH₃⁺]
  • (B) [OH⁻] = [CH₃NH₂] = [CH₃NH₃⁺]
  • (C) [CH₃NH₃⁺] > [OH⁻] = [CH₃NH₂]
  • (D) [CH₃NH₃⁺] > [OH⁻] > [CH₃NH₂]
View Answer 
Correct Answer: (D)

Question 6:

A 0.1-molar solution of which of the following acids will be the best conductor of electricity?

Options:

  • (A) H₂CO₃
  • (B) H₂S
  • (C) HF
  • (D) HNO₃
View Answer 
Correct Answer: (D)

Question 7:

A laboratory technician wishes to create a buffered solution with a pH of 5. Which of the following acids would be the best choice for the buffer?

Options:

  • (A) Oxalic acid (H₂C₂O₄) \(K_a = 5.9 \times 10^{-2}\)
  • (B) Phosphoric acid (H₃AsO₄) \(K_a = 5.6 \times 10^{-3}\)
  • (C) Acetic acid (HC₂H₃O₂) \(K_a = 1.8 \times 10^{-5}\)
  • (D) Hypochlorous acid (HOCl) \(K_a = 3.0 \times 10^{-8}\)
View Answer 
Answer Explanation:
To create a buffer with pH = 5, choose an acid whose \(pK_a\) is close to 5:
\[
pK_a = -\log(K_a)
\]
For acetic acid:
\[
pK_a = -\log(1.8 \times 10^{-5}) = 4.74
\]
This is closest to pH 5.

Correct Answer: (C)

Question 8:

A solution of sulfurous acid (H₂SO₃) is present in an aqueous solution. Which of the following represents the concentrations of three different ions in solution?

Options:

  • (A) [SO₃²⁻] > [HSO₃⁻] > [H₂SO₃]
  • (B) [H₂SO₃] > [HSO₃⁻] > [SO₃²⁻]
  • (C) [HSO₃⁻] > [H₂SO₃] = [SO₃²⁻]
  • (D) [SO₃²⁻] = [HSO₃⁻] > [H₂SO₃]
View Answer 
Correct Answer: (B)

Question 9:

How many liters of distilled water must be added to 1 liter of an aqueous solution of HCl with a pH of 1 to create a solution with a pH of 2?

Options:

  • (A) 0.1 L
  • (B) 0.9 L
  • (C) 2 L
  • (D) 9 L
View Answer 
Correct Answer: (D)

Question 10:

A 1-molar solution of a very weak monoprotic acid has a pH of 5. What is the value of \(K_a\) for the acid?

Options:

  • (A) \(K_a = 1 \times 10^{-10}\)
  • (B) \(K_a = 1 \times 10^{-7}\)
  • (C) \(K_a = 1 \times 10^{-5}\)
  • (D) \(K_a = 1 \times 10^{-2}\)
View Answer 
Correct Answer: (A)

Question 11:

The value of \(K_a\) for HSO₄⁻ is \(1 \times 10^{-2}\). What is the value of \(K_b\) for SO₄²⁻?

Options:

  • (A) \(K_b = 1 \times 10^{-12}\)
  • (B) \(K_b = 1 \times 10^{-8}\)
  • (C) \(K_b = 1 \times 10^{-2}\)
  • (D) \(K_b = 1 \times 10^{2}\)
View Answer 
Correct Answer: (A)

Question 12:

Why is the solution acidic at the equivalence point in a titration of a weak base (NH₃) with a strong acid (HCl)?

Options:

  • (A) The strong acid dissociates fully, leaving excess H⁺ in solution.
  • (B) The conjugate acid of NH₃ is the only acidic or basic ion present at the equivalence point.
  • (C) The water created during the titration acts as an acid.
  • (D) The acid is diprotic, donating two protons for every unit dissociated.
View Answer 
Correct Answer: (B)

Question 13:

During the titration of 30.0 mL of 1.0 M ammonia (NH₃) with 1.0 M hydrochloric acid (HCl), what volume of HCl is needed to reach the equivalence point?

Options:

  • (A) 15.0 mL
  • (B) 30.0 mL
  • (C) 45.0 mL
  • (D) 60.0 mL
View Answer 
Answer Explanation:
The reaction between NH₃ and HCl is:
\[
\text{NH}_3 + \text{HCl} \rightarrow \text{NH}_4\text{Cl}
\]
At the equivalence point, moles of NH₃ = moles of HCl:
\[
C_{\text{NH}_3} \times V_{\text{NH}_3} = C_{\text{HCl}} \times V_{\text{HCl}}
\]
\[
1.0\,\text{M} \times 30.0\,\text{mL} = 1.0\,\text{M} \times V_{\text{HCl}}
\]
\[
V_{\text{HCl}} = 30.0\,\text{mL}
\]
Correct Answer: (B)

Question 14:

Which ions are present in significant amounts during the first buffer region of the titration of NH₃ with HCl?

Options:

  • (A) NH₄⁺ and NH₃
  • (B) NH₃ and H⁺
  • (C) NH₄⁺ and OH⁻
  • (D) H₃O⁺ and NH₃
View Answer 
Correct Answer: (A)

Question 15:

Which of the following could be added to an aqueous solution of weak acid HF to increase the percent dissociation?

Options:

  • (A) NaF(s)
  • (B) H₂O(l)
  • (C) NaCl(s)
  • (D) HCl(g)
View Answer 
Correct Answer: (B)

Question 16:

A bottle of water is left outside in the sun, and the bottle warms gradually over the day. What will happen to the pH of the water as it warms?

Options:

  • (A) Nothing; pure water always has a pH of 7.00.
  • (B) Nothing; the volume would have to change in order for any ion concentration to change.
  • (C) The pH will increase because the concentration of H⁺ is increasing.
  • (D) The pH will decrease because the auto-ionization of water is an endothermic process.
View Answer 
Correct Answer: (D)

Question 17:

The structure of two oxoacids is shown below:

Which would be a stronger acid, and why?

Options:

  • (A) HOCl, because the H–O bond is weaker than in HOF as chlorine is larger than fluorine
  • (B) HOCl, because the H–O bond is stronger than in HOF as chlorine has a higher electronegativity than fluorine
  • (C) HOF, because the H–O bond is stronger than in HOCl as fluorine has a higher electronegativity than chlorine
  • (D) HOF, because the H–O bond is weaker than in HOCl as fluorine is smaller than chlorine
View Answer 
Correct Answer: (D)

Question 18:

Which of the following pairs of substances would make a good buffer solution when combined in equal molar amounts?

Options:

  • (A) HC₂H₃O₂(aq) and NaC₂H₃O₂(aq)
  • (B) H₂SO₄(aq) and LiOH(aq)
  • (C) HCl(aq) and KCl(aq)
  • (D) HF(aq) and NH₃(aq)
View Answer 
Correct Answer: (A)

Question 19:

What is the pH of a 0.10 M solution of acetic acid (CH₃COOH) if the \(K_a\) of acetic acid is \(1.8 \times 10^{-5}\)?

Options:

  • (A) 3.00
  • (B) 2.87
  • (C) 4.74
  • (D) 5.00
View Answer 
Answer Explanation:
We use the \(K_a\) expression for acetic acid: 
\[
K_a = \frac{[H^+][CH₃COO⁻]}{[CH₃COOH]} = 1.8 \times 10^{-5}
\]
Assume \(x = [H^+]\), then:
\[
1.8 \times 10^{-5} = \frac{x^2}{0.10}
\]
Solving for \(x\), \(x = 1.34 \times 10^{-3}\), and the pH is:
\[
pH = -\log(1.34 \times 10^{-3}) \approx 2.87
\]
Correct Answer: (B)

Question 20:

A solution is made by mixing 0.20 moles of acetic acid (CH₃COOH) and 0.10 moles of sodium acetate (NaCH₃COO) in 1.0 L of water. What is the pH of the solution if the \(K_a\) of acetic acid is \(1.8 \times 10^{-5}\)?

Options:

  • (A) 4.50
  • (B) 4.74
  • (C) 5.00
  • (D) 3.80
View Answer 
Answer Explanation:
This is a buffer solution, so we use the Henderson-Hasselbalch equation:
\[ pH = pK_a + \log\left(\frac{[A⁻]}{[HA]}\right) \]
First, calculate \( pK_a \):
\[ pK_a = -\log(1.8 \times 10^{-5}) = 4.74 \]
Now, calculate the ratio of conjugate base to acid:
\[ \frac{[A⁻]}{[HA]} = \frac{0.10}{0.20} = 0.5 \]
Substitute into the equation:
\[ pH = 4.74 + \log(0.5) = 4.74 - 0.30 = 4.44 \]
Correct Answer: (A)

Question 21:

What is the pH of a 0.25 M ammonia (NH₃) solution? The \(K_b\) of ammonia is \(1.8 \times 10^{-5}\).

Options:

  • (A) 11.13
  • (B) 11.27
  • (C) 10.50
  • (D) 12.00
View Answer 
Answer Explanation:
We use the \(K_b\) expression for ammonia:
\[
K_b = \frac{[OH⁻][NH₄^+]}{[NH₃]} = 1.8 \times 10^{-5}
\]
Assume \(x = [OH⁻]\), then:
\[
1.8 \times 10^{-5} = \frac{x^2}{0.25}
\]
Solving for \(x\), \(x = 2.12 \times 10^{-3}\). The pOH is:
\[
pOH = -\log(2.12 \times 10^{-3}) \approx 2.67
\]
Now calculate the pH:
\[
pH = 14 - 2.67 = 11.33
\]
Correct Answer: (B)

Question 22:

A buffer solution is made by adding 0.50 moles of sodium acetate (NaCH₃COO) to 0.50 moles of acetic acid (CH₃COOH) in 1.0 L of water. What is the pH of the solution? The \(K_a\) of acetic acid is \(1.8 \times 10^{-5}\).

Options:

  • (A) 4.74
  • (B) 5.00
  • (C) 4.50
  • (D) 3.80
View Answer 
Answer Explanation:
This is a buffer solution, so we use the Henderson-Hasselbalch equation:
\[
pH = pK_a + \log\left(\frac{[A⁻]}{[HA]}\right)
\]
In this case, the concentrations of the conjugate base and acid are equal:
\[
pH = pK_a + \log(1) = pK_a
\]
The \(pK_a\) is:
\[
pK_a = -\log(1.8 \times 10^{-5}) = 4.74
\]
Correct Answer: (A)

Question 23:

What is the pH of a buffer solution containing 0.10 M acetic acid (CH₃COOH) and 0.15 M sodium acetate (NaCH₃COO)? The \(K_a\) of acetic acid is \(1.8 \times 10^{-5}\).

Options:

  • (A) 4.92
  • (B) 4.74
  • (C) 5.00
  • (D) 4.50
View Answer 
Answer Explanation:
This is a buffer solution, so we use the Henderson-Hasselbalch equation:
\[ pH = pK_a + \log\left(\frac{[A⁻]}{[HA]}\right) \]
First, calculate the \( pK_a \):
\[ pK_a = -\log(1.8 \times 10^{-5}) = 4.74 \]
Now, calculate the ratio of conjugate base to acid:
\[ \frac{[A⁻]}{[HA]} = \frac{0.15}{0.10} = 1.5 \]
Substitute into the equation:
\[ pH = 4.74 + \log(1.5) = 4.74 + 0.18 = 4.92 \]
Correct Answer: (A)

Question 24:

What is the pH of a solution made by mixing 0.25 moles of sodium acetate and 0.10 moles of acetic acid in 1.0 L of water? The \(K_a\) of acetic acid is \(1.8 \times 10^{-5}\).

Options:

  • (A) 5.07
  • (B) 5.50
  • (C) 4.50
  • (D) 4.74
View Answer 
Answer Explanation:
We use the Henderson-Hasselbalch equation:
\[ pH = pK_a + \log\left(\frac{[A⁻]}{[HA]}\right) \]
First, calculate \( pK_a \):
\[ pK_a = -\log(1.8 \times 10^{-5}) = 4.74 \]
Now, calculate the ratio of conjugate base to acid:
\[ \frac{[A⁻]}{[HA]} = \frac{0.25}{0.10} = 2.5 \]
Substitute into the equation:
\[ pH = 4.74 + \log(2.5) = 4.74 + 0.40 = 5.14 \]
Correct Answer: (A)

Question 25:

A buffer solution contains 0.40 M ammonia (NH₃) and 0.25 M ammonium chloride (NH₄Cl). What is the pH of this buffer solution? The \(K_b\) of ammonia is \(1.8 \times 10^{-5}\).

Options:

  • (A) 9.12
  • (B) 9.26
  • (C) 9.40
  • (D) 8.85
View Answer 
Answer Explanation:
For a buffer solution, use the Henderson-Hasselbalch equation for bases:
\[
pOH = pK_b + \log\left(\frac{[NH₄^+]}{[NH₃]}\right)
\]
First, calculate the \(pK_b\):
\[
pK_b = -\log(1.8 \times 10^{-5}) = 4.74
\]
Now, calculate the ratio of ammonium to ammonia:
\[
\frac{[NH₄^+]}{[NH₃]} = \frac{0.25}{0.40} = 0.625
\]
Substitute into the equation:
\[
pOH = 4.74 + \log(0.625) = 4.74 - 0.20 = 4.54
\]
Now calculate the pH:
\[
pH = 14 - 4.54 = 9.46
\]
Correct Answer: (B)

Question 26:

What is the pH of a buffer solution made by mixing 0.15 moles of hydrochloric acid (HCl) with 0.25 moles of sodium acetate (NaCH₃COO) in 1.0 L of water? The \(K_a\) of acetic acid is \(1.8 \times 10^{-5}\).

Options:

  • (A) 4.44
  • (B) 4.74
  • (C) 4.00
  • (D) 5.00
View Answer 
Answer Explanation:
When HCl is added to sodium acetate, the HCl will react completely with the acetate ions to form acetic acid. The remaining acetate concentration will be:
\[
[A⁻] = 0.25 - 0.15 = 0.10 \, \text{mol/L}
\]
The concentration of acetic acid will be:
\[
[HA] = 0.15 \, \text{mol/L}
\]
Use the Henderson-Hasselbalch equation:
\[
pH = pK_a + \log\left(\frac{[A⁻]}{[HA]}\right)
\]
Calculate \(pK_a\):
\[
pK_a = -\log(1.8 \times 10^{-5}) = 4.74
\]
Substitute the values:
\[
pH = 4.74 + \log\left(\frac{0.10}{0.15}\right) = 4.74 + \log(0.667) = 4.74 - 0.18 = 4.56
\]
Correct Answer: (A)

Question 27:

A 0.50 M solution of formic acid (HCOOH) has a pH of 2.36. What is the \(K_a\) of formic acid?

Options:

  • (A) \(2.0 \times 10^{-4}\)
  • (B) \(1.8 \times 10^{-4}\)
  • (C) \(1.2 \times 10^{-4}\)
  • (D) \(3.5 \times 10^{-4}\)
View Answer 
Answer Explanation:
From the pH, calculate the concentration of \(H^+\):
\[
[H^+] = 10^{-2.36} = 4.37 \times 10^{-3} \, \text{M}
\]
Now, use the \(K_a\) expression:
\[
K_a = \frac{[H^+][A⁻]}{[HA]} = \frac{(4.37 \times 10^{-3})^2}{0.50 - 4.37 \times 10^{-3}}
\]
Since \(0.50 - 4.37 \times 10^{-3} \approx 0.50\), simplify the equation:
\[
K_a = \frac{(4.37 \times 10^{-3})^2}{0.50} = 1.91 \times 10^{-5}
\]
Correct Answer: (B)

Question 28:

What is the pH of a 0.20 M solution of benzoic acid (C₆H₅COOH)? The \(K_a\) of benzoic acid is \(6.3 \times 10^{-5}\).

Options:

  • (A) 2.00
  • (B) 2.50
  • (C) 3.00
  • (D) 4.00
View Answer 
Answer Explanation:
Use the \(K_a\) expression for benzoic acid:
\[
K_a = \frac{[H^+][C₆H₅COO⁻]}{[C₆H₅COOH]} = 6.3 \times 10^{-5}
\]
Assume \(x = [H^+]\), then:
\[
6.3 \times 10^{-5} = \frac{x^2}{0.20}
\]
Solve for \(x\):
\[
x = \sqrt{6.3 \times 10^{-5} \times 0.20} = 3.55 \times 10^{-3}
\]
The pH is:
\[
pH = -\log(3.55 \times 10^{-3}) = 2.45
\]
Correct Answer: (B)

Question 29:

What is the pH of a buffer solution made by mixing 0.30 moles of ammonium chloride (NH₄Cl) with 0.20 moles of ammonia (NH₃) in 1.0 L of water? The \(K_b\) of ammonia is \(1.8 \times 10^{-5}\).

Options:

  • (A) 9.15
  • (B) 9.35
  • (C) 8.95
  • (D) 10.00
View Answer 
Answer Explanation:
Use the Henderson-Hasselbalch equation:
\[
pOH = pK_b + \log\left(\frac{[NH₄^+]}{[NH₃]}\right)
\]
First, calculate the \(pK_b\):
\[
pK_b = -\log(1.8 \times 10^{-5}) = 4.74
\]
Now calculate the ratio of ammonium to ammonia:
\[
\frac{[NH₄^+]}{[NH₃]} = \frac{0.30}{0.20} = 1.5
\]
Substitute into the equation:
\[
pOH = 4.74 + \log(1.5) = 4.74 + 0.18 = 4.92
\]
Now calculate the pH:
\[
pH = 14 - 4.92 = 9.08
\]
Correct Answer: (A)

Question 30:

A 0.10 M solution of hydrochloric acid (HCl) is diluted to 0.01 M. What is the change in pH?

Options:

  • (A) pH increases by 1.0
  • (B) pH increases by 0.5
  • (C) pH decreases by 1.0
  • (D) pH decreases by 0.5
View Answer 
Answer Explanation:
For a strong acid, pH is calculated as:
\[
pH = -\log[H^+]
\]
For the initial concentration of 0.10 M:
\[
pH_1 = -\log(0.10) = 1.00
\]
For the diluted concentration of 0.01 M:
\[
pH_2 = -\log(0.01) = 2.00
\]
The change in pH is:
\[
pH_2 - pH_1 = 2.00 - 1.00 = 1.00
\]
Correct Answer: (A)

Question 31:

What is the pH of a buffer solution made by combining 0.50 moles of acetic acid (CH₃COOH) and 0.25 moles of sodium acetate (CH₃COONa) in 1.0 L of water? The \(K_a\) of acetic acid is \(1.8 \times 10^{-5}\).

Options:

  • (A) 4.34
  • (B) 4.14
  • (C) 4.74
  • (D) 5.00
View Answer 
Answer Explanation:
Use the Henderson-Hasselbalch equation:
\[
pH = pK_a + \log\left(\frac{[A⁻]}{[HA]}\right)
\]
First, calculate the \(pK_a\):
\[
pK_a = -\log(1.8 \times 10^{-5}) = 4.74
\]
Now calculate the ratio of acetate to acetic acid:
\[
\frac{[CH₃COO⁻]}{[CH₃COOH]} = \frac{0.25}{0.50} = 0.50
\]
Substitute into the equation:
\[
pH = 4.74 + \log(0.50) = 4.74 - 0.30 = 4.44
\]
Correct Answer: (A)

Question 32:

What is the pH of a 0.010 M solution of formic acid (HCOOH)? The \(K_a\) of formic acid is \(1.8 \times 10^{-4}\).

Options:

  • (A) 3.18
  • (B) 2.74
  • (C) 2.94
  • (D) 3.74
View Answer 
Answer Explanation:
Use the \(K_a\) expression for formic acid:
\[
K_a = \frac{[H^+][A⁻]}{[HA]} = 1.8 \times 10^{-4}
\]
Assume \(x = [H^+]\), then:
\[
1.8 \times 10^{-4} = \frac{x^2}{0.010 - x}
\]
Since \(x\) is small, approximate \(0.010 - x \approx 0.010\):
\[
1.8 \times 10^{-4} = \frac{x^2}{0.010}
\]
Solve for \(x\):
\[
x = \sqrt{1.8 \times 10^{-4} \times 0.010} = 1.34 \times 10^{-3}
\]
The pH is:
\[
pH = -\log(1.34 \times 10^{-3}) = 2.87
\]
Correct Answer: (C)

Question 33:

What is the pH of a buffer solution containing 0.10 M of lactic acid (C₃H₆O₃) and 0.05 M of sodium lactate (C₃H₅O₃Na)? The \(K_a\) of lactic acid is \(1.4 \times 10^{-4}\).

Options:

  • (A) 3.08
  • (B) 3.24
  • (C) 3.44
  • (D) 3.00
View Answer 
Answer Explanation:
Use the Henderson-Hasselbalch equation:
\[
pH = pK_a + \log\left(\frac{[A⁻]}{[HA]}\right)
\]
Calculate \(pK_a\):
\[
pK_a = -\log(1.4 \times 10^{-4}) = 3.85
\]
Now, calculate the ratio of lactate to lactic acid:
\[
\frac{[C₃H₅O₃⁻]}{[C₃H₆O₃]} = \frac{0.05}{0.10} = 0.50
\]
Substitute into the equation:
\[
pH = 3.85 + \log(0.50) = 3.85 - 0.30 = 3.55
\]
Correct Answer: (B)

Question 34:

A buffer solution is prepared by dissolving 0.40 moles of sodium hydrogen carbonate (NaHCO₃) in 1.0 L of 0.50 M carbonic acid (H₂CO₃). What is the pH of the solution? The \(K_a\) of carbonic acid is \(4.3 \times 10^{-7}\).

Options:

  • (A) 6.12
  • (B) 6.30
  • (C) 6.50
  • (D) 6.74
View Answer 
Answer Explanation:
Use the Henderson-Hasselbalch equation:
\[
pH = pK_a + \log\left(\frac{[A⁻]}{[HA]}\right)
\]
Calculate \(pK_a\):
\[
pK_a = -\log(4.3 \times 10^{-7}) = 6.37
\]
Now, calculate the ratio of hydrogen carbonate to carbonic acid:
\[
\frac{[HCO₃⁻]}{[H₂CO₃]} = \frac{0.40}{0.50} = 0.80
\]
Substitute into the equation:
\[
pH = 6.37 + \log(0.80) = 6.37 - 0.10 = 6.27
\]
Correct Answer: (B)

Question 35:

What is the pH of a 0.20 M solution of sodium hydroxide (NaOH)?

Options:

  • (A) 13.30
  • (B) 13.60
  • (C) 12.60
  • (D) 13.00
View Answer 
Answer Explanation:
For a strong base like NaOH, the concentration of OH⁻ ions is equal to the NaOH concentration. Calculate the pOH:
\[
pOH = -\log(0.20) = 0.70
\]
Now, calculate the pH:
\[
pH = 14 - pOH = 14 - 0.70 = 13.30
\]
Correct Answer: (A)

Question 36:

What is the pH of a solution that contains 0.10 M \(HCO₃⁻\) and 0.05 M \(H₂CO₃\)? The \(K_a\) of carbonic acid is \(4.3 \times 10^{-7}\).

Options:

  • (A) 6.54
  • (B) 6.37
  • (C) 6.70
  • (D) 6.20
View Answer 
Answer Explanation:
Using the Henderson-Hasselbalch equation:
\[
pH = pK_a + \log\left(\frac{[A⁻]}{[HA]}\right)
\]
Calculate \(pK_a\):
\[
pK_a = -\log(4.3 \times 10^{-7}) = 6.37
\]
Now, calculate the ratio of \(HCO₃⁻\) to \(H₂CO₃\):
\[
\frac{[HCO₃⁻]}{[H₂CO₃]} = \frac{0.10}{0.05} = 2.00
\]
Substitute into the equation:
\[
pH = 6.37 + \log(2.00) = 6.37 + 0.30 = 6.67
\]
Correct Answer: (C)

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