Question 1:

A solution of 0.05 M sodium hydroxide (NaOH) is being titrated into 20.0 mL of formic acid (HCOOH) of unknown concentration. It takes 10.0 mL 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

Answer: (A)
Explanation:
\[ C_{\text{acid}} \times V_{\text{acid}} = C_{\text{base}} \times V_{\text{base}} \]
\[ C_{\text{acid}} \times 20.0\,\text{mL} = 0.05\,\text{M} \times 10.0\,\text{mL} \]
\[ C_{\text{acid}} = 0.050\,\text{M} \]
        

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) Neutral; equal moles of both acid and base are present.
• (D) Basic; the higher concentration of the base is the determining factor.

Answer: (B)
Explanation: At the equivalence point, the conjugate base of the weak acid dominates, making the solution basic.
        

Question 3:

If the formic acid were replaced with a strong acid such as HCl at the same concentration, 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.

Answer: (D)
Explanation: The volume required to reach equivalence depends only on moles, not on acid strength.
        

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

Answer: (A)
Explanation: Sodium formate (NaCO₂H) is the conjugate base of formic acid, forming a buffer system.
        

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]

Answer: (D)
Explanation: The conjugate acid dominates due to partial dissociation, leading to [CH₃NH₃⁺] > [OH⁻].
        

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

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) \(K_a = 3.0 \times 10^{-8}\)

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]

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

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) \(K_a = 1 \times 10^{-2}\)

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) \(K_b = 1 \times 10^{2}\)

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.

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

Answer: (B)
Explanation: 
The reaction is:
\[
\text{NH₃} + \text{HCl} \rightarrow \text{NH₄Cl}
\]
Moles of NH₃ = \( C \times V = 1.0 \, \text{M} \times 30.0 \, \text{mL} = 0.030 \, \text{mol} \). 
Equal moles of HCl are needed, so \( V_{\text{HCl}} = \frac{0.030 \, \text{mol}}{1.0 \, \text{M}} = 30.0 \, \text{mL} \).
        

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

Answer: (A)
Explanation: 
During the buffer region, the weak base (NH₃) and its conjugate acid (NH₄⁺) are present in significant amounts.
        

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)

Answer: (B)
Explanation: 
Adding more water dilutes the solution, decreasing \([HF]\) and shifting the equilibrium toward dissociation according to Le Chatelier's principle.
        

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.

Answer: (D)
Explanation: 
As temperature increases, the auto-ionization of water (\(2H₂O \leftrightarrow H₃O⁺ + OH⁻\)) increases, and \([H₃O⁺]\) increases, lowering the pH.
        

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

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)

Answer: (A)
Explanation: 
Acetic acid (HC₂H₃O₂) and its conjugate base sodium acetate (NaC₂H₃O₂) form a buffer solution that resists pH changes.
        

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

Answer: (B)
Explanation:
Using the formula:
\[
K_a = \frac{[H^+][A⁻]}{[HA]}
\]
Assume \([H^+] = x\):
\[
1.8 \times 10^{-5} = \frac{x^2}{0.10 - x}
\]
Since \(x\) is small, approximate \(0.10 - x \approx 0.10\):
\[
x = \sqrt{1.8 \times 10^{-5} \times 0.10} = 1.34 \times 10^{-3}
\]
\[
pH = -\log(1.34 \times 10^{-3}) \approx 2.87
\]
        

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

Answer: (A)
Explanation:
This is a buffer solution, so 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:
\[
\frac{[A⁻]}{[HA]} = \frac{0.10}{0.20} = 0.5
\]
Substitute into the equation:
\[
pH = 4.74 + \log(0.5) = 4.74 - 0.30 = 4.44
\]
        

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

Answer: (B)
Explanation:
Using the formula:
\[
K_b = \frac{[OH⁻][NH₄^+]}{[NH₃]}
\]
Assume \([OH⁻] = x\):
\[
1.8 \times 10^{-5} = \frac{x^2}{0.25}
\]
Solve for \(x\):
\[
x = \sqrt{1.8 \times 10^{-5} \times 0.25} = 2.12 \times 10^{-3}
\]
Calculate \(pOH\):
\[
pOH = -\log(2.12 \times 10^{-3}) \approx 2.67
\]
Finally, calculate \(pH\):
\[
pH = 14 - pOH = 14 - 2.67 = 11.33
\]
        

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

Answer: (A)
Explanation:
Since the concentrations of acid and base are equal:
\[
pH = pK_a = -\log(1.8 \times 10^{-5}) = 4.74
\]
        

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

Answer: (A)
Explanation:
Use the Henderson-Hasselbalch equation:
\[
pH = pK_a + \log\left(\frac{[A⁻]}{[HA]}\right)
\]
\[
pH = 4.74 + \log\left(\frac{0.15}{0.10}\right) = 4.74 + 0.18 = 4.92
\]
        

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

Answer: (A)
Explanation:
Use the Henderson-Hasselbalch equation:
\[
pH = pK_a + \log\left(\frac{[A⁻]}{[HA]}\right)
\]
\[
pH = 4.74 + \log\left(\frac{0.25}{0.10}\right) = 4.74 + 0.40 = 5.14
\]
        

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

Answer: (B)
Explanation:
For a buffer solution, use the Henderson-Hasselbalch equation for bases:
\[
pOH = pK_b + \log\left(\frac{[NH₄^+]}{[NH₃]}\right)
\]
First, calculate \(pK_b\):
\[
pK_b = -\log(1.8 \times 10^{-5}) = 4.74
\]
Now calculate the ratio:
\[
\frac{[NH₄^+]}{[NH₃]} = \frac{0.25}{0.40} = 0.625
\]
Substitute into the equation:
\[
pOH = 4.74 + \log(0.625) = 4.74 - 0.20 = 4.54
\]
Finally, calculate \(pH\):
\[
pH = 14 - 4.54 = 9.46
\]
        

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

Answer: (A)
Explanation:
When HCl reacts with sodium acetate, acetate ions form acetic acid:
\[
[A⁻] = 0.25 - 0.15 = 0.10
\]
\[
[HA] = 0.15
\]
Use the Henderson-Hasselbalch equation:
\[
pH = pK_a + \log\left(\frac{[A⁻]}{[HA]}\right)
\]
\[
pH = 4.74 + \log\left(\frac{0.10}{0.15}\right) = 4.74 - 0.18 = 4.56
\]
        

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) \(3.5 \times 10^{-4}\)

Answer: (B)
Explanation:
From the pH, calculate \([H^+]\):
\[
[H^+] = 10^{-2.36} = 4.37 \times 10^{-3}
\]
Using \(K_a = \frac{[H^+]^2}{[HA] - [H^+]}\):
\[
K_a = \frac{(4.37 \times 10^{-3})^2}{0.50 - 4.37 \times 10^{-3}} \approx \frac{1.91 \times 10^{-5}}{0.50} = 1.8 \times 10^{-4}
\]
        

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

Answer: (C)
Explanation:
Using \(K_a = \frac{[H^+]^2}{[HA]}\), assume \([H^+] = x\):
\[
6.3 \times 10^{-5} = \frac{x^2}{0.20}
\]
Solve for \(x\):
\[
x = \sqrt{6.3 \times 10^{-5} \times 0.20} = 3.55 \times 10^{-3}
\]
\[
pH = -\log(3.55 \times 10^{-3}) \approx 3.45
\]
        

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

Answer: (A)
Explanation:
Using the Henderson-Hasselbalch equation:
\[
pOH = pK_b + \log\left(\frac{[NH₄^+]}{[NH₃]}\right)
\]
\[
pK_b = -\log(1.8 \times 10^{-5}) = 4.74
\]
\[
pOH = 4.74 + \log\left(\frac{0.30}{0.20}\right) = 4.74 + 0.18 = 4.92
\]
\[
pH = 14 - 4.92 = 9.08
\]
        

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

Answer: (A)
Explanation:
Calculate initial \(pH_1\):
\[
pH_1 = -\log(0.10) = 1.00
\]
Calculate new \(pH_2\):
\[
pH_2 = -\log(0.01) = 2.00
\]
Change in pH:
\[
\Delta pH = pH_2 - pH_1 = 2.00 - 1.00 = 1.00
\]
        

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

Answer: (A)
Explanation:
Using 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:
\[
\frac{[CH₃COO⁻]}{[CH₃COOH]} = \frac{0.25}{0.50} = 0.50
\]
Substitute into the equation:
\[
pH = 4.74 + \log(0.50) = 4.74 - 0.30 = 4.44
\]
        

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

Answer: (C)
Explanation:
Using the \(K_a\) expression:
\[
K_a = \frac{[H^+][A⁻]}{[HA]} = 1.8 \times 10^{-4}
\]
Assume \([H^+] = x\), 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:
\[
pH = -\log(1.34 \times 10^{-3}) = 2.87
\]
        

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

Answer: (B)
Explanation:
Using 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:
\[
\frac{[C₃H₅O₃⁻]}{[C₃H₆O₃]} = \frac{0.05}{0.10} = 0.50
\]
Substitute into the equation:
\[
pH = 3.85 + \log(0.50) = 3.85 - 0.30 = 3.55
\]
        

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

Answer: (B)
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:
\[
\frac{[HCO₃⁻]}{[H₂CO₃]} = \frac{0.40}{0.50} = 0.80
\]
Substitute into the equation:
\[
pH = 6.37 + \log(0.80) = 6.37 - 0.10 = 6.27
\]
        

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

Answer: (A)
Explanation:
For a strong base like NaOH:
\[
[OH⁻] = 0.20
\]
Calculate \(pOH\):
\[
pOH = -\log(0.20) = 0.70
\]
Now calculate \(pH\):
\[
pH = 14 - pOH = 14 - 0.70 = 13.30
\]
        

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

Answer: (B)
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:
\[
\frac{[HCO₃⁻]}{[H₂CO₃]} = \frac{0.10}{0.05} = 2.00
\]
Substitute into the equation:
\[
pH = 6.37 + \log(2.00) = 6.37 + 0.30 = 6.67
\]
        

Question 37:

Which of the following is a strong acid?

Options:

• (A) HCl
• (B) HF
• (C) H₂CO₃
• (D) H₃PO₄

Answer: (A)
Explanation:
Hydrochloric acid (HCl) is a strong acid that fully dissociates in water, unlike HF, H₂CO₃, or H₃PO₄.
        

Question 38:

Which of the following pairs can form a buffer solution?

Options:

• (A) HCl and NaCl
• (B) NH₃ and NH₄Cl
• (C) H₂SO₄ and Na₂SO₄
• (D) NaOH and KOH

Answer: (B)
Explanation:
NH₃ (a weak base) and NH₄Cl (its conjugate acid salt) form a buffer solution that resists pH changes when small amounts of acid or base are added.
        

Question 39:

Which of the following is a weak acid?

Options:

• (A) HNO₃
• (B) CH₃COOH
• (C) H₂SO₄
• (D) HI

Answer: (B)
Explanation:
Acetic acid (CH₃COOH) is a weak acid because it only partially dissociates in water, unlike strong acids like HNO₃, H₂SO₄, or HI.
        

Question 40:

Which of the following salts will produce a basic solution when dissolved in water?

Options:

• (A) NaCl
• (B) NH₄Cl
• (C) KNO₃
• (D) Na₂CO₃

Answer: (D)
Explanation:
Sodium carbonate (Na₂CO₃) produces a basic solution because the carbonate ion (CO₃²⁻) is the conjugate base of a weak acid and can hydrolyze in water.
        

Question 41:

Identify the strong base among the following compounds:

Options:

• (A) NH₃
• (B) NaOH
• (C) Al(OH)₃
• (D) Mg(OH)₂

Answer: (B)
Explanation:
Sodium hydroxide (NaOH) is a strong base that fully dissociates in water, unlike weak bases like NH₃ or slightly soluble bases like Mg(OH)₂ and Al(OH)₃.
        

Question 42:

During the titration of 25.0 mL of 0.1 M acetic acid (CH₃COOH) with 0.1 M NaOH, what is the pH at the equivalence point?

Options:

• (A) 7.00
• (B) 8.72
• (C) 4.74
• (D) 3.00

Answer: (B)
Explanation:
At the equivalence point, the solution contains the conjugate base (CH₃COO⁻), which hydrolyzes in water to make the solution basic. The pH is higher than 7.
        

Question 43:

Which of the following is a weak base?

Options:

• (A) KOH
• (B) NH₃
• (C) Ca(OH)₂
• (D) Ba(OH)₂

Answer: (B)
Explanation:
Ammonia (NH₃) is a weak base because it partially dissociates in water, unlike strong bases such as KOH, Ca(OH)₂, or Ba(OH)₂.
        

Question 44:

Which of the following solutions would best resist a pH change when a small amount of HCl is added?

Options:

• (A) HCl and NaCl
• (B) CH₃COOH and CH₃COONa
• (C) NaOH and KOH
• (D) H₂SO₄ and Na₂SO₄

Answer: (B)
Explanation:
A mixture of acetic acid (CH₃COOH) and sodium acetate (CH₃COONa) forms a buffer solution that resists pH changes when a small amount of acid or base is added.
        

Question 45:

Which of the following acids is strong?

Options:

• (A) HClO₄
• (B) H₃PO₄
• (C) H₂CO₃
• (D) CH₃COOH

Answer: (A)
Explanation:
Perchloric acid (HClO₄) is a strong acid that completely dissociates in water, unlike H₃PO₄, H₂CO₃, or CH₃COOH.
        

Question 46:

Which of the following salts will produce an acidic solution when dissolved in water?

Options:

• (A) Na₂SO₄
• (B) NH₄NO₃
• (C) KCl
• (D) CaCl₂

Answer: (B)
Explanation:
Ammonium nitrate (NH₄NO₃) produces an acidic solution because the ammonium ion (NH₄⁺) is a weak acid.
        

Question 47:

Select the weak acid among the following options:

Options:

• (A) HNO₃
• (B) HCl
• (C) H₂CO₃
• (D) H₂SO₄

Answer: (C)
Explanation:
Carbonic acid (H₂CO₃) is a weak acid because it only partially dissociates in water, unlike the strong acids HNO₃, HCl, and H₂SO₄.
        

Question 48:

What is the pH at the halfway point in the titration of 50 mL of 0.1 M acetic acid with 0.1 M NaOH?

Options:

• (A) 4.74
• (B) 7.00
• (C) 5.00
• (D) 3.00

Answer: (A)
Explanation:
At the halfway point of a weak acid-strong base titration, pH = pKa. For acetic acid, pKa = 4.74.
        

Question 49:

Which of the following is a strong base?

Options:

• (A) NaOH
• (B) NH₄OH
• (C) Al(OH)₃
• (D) CH₃NH₂

Answer: (A)
Explanation:
Sodium hydroxide (NaOH) is a strong base that fully dissociates in water, unlike NH₄OH, Al(OH)₃, or CH₃NH₂.
        

Question 50:

A buffer solution has a pH of 5.0. Which of the following acids would be best for creating this buffer?

Options:

• (A) Acetic acid, pKa = 4.74
• (B) Formic acid, pKa = 3.75
• (C) Hydrochloric acid, pKa < 0
• (D) Nitric acid, pKa < 0

Answer: (A)
Explanation:
Acetic acid is ideal for a buffer with pH = 5.0, as its pKa (4.74) is close to the desired pH.
        

Question 51:

Identify the weak base in this list:

Options:

• (A) KOH
• (B) Ca(OH)₂
• (C) NH₃
• (D) Ba(OH)₂

Answer: (C)
Explanation:
Ammonia (NH₃) is a weak base as it only partially dissociates in water, unlike strong bases such as KOH, Ca(OH)₂, and Ba(OH)₂.
        

Question 52:

Which salt, when dissolved in water, would likely produce a solution with a pH greater than 7?

Options:

• (A) NaCl
• (B) NH₄Cl
• (C) KNO₃
• (D) Na₂CO₃

Answer: (D)
Explanation:
Sodium carbonate (Na₂CO₃) forms a basic solution because the carbonate ion (CO₃²⁻) is a weak base that hydrolyzes in water.
        

Question 53:

Which of the following is a strong acid?

Options:

• (A) HClO₃
• (B) H₂CO₃
• (C) HF
• (D) CH₃COOH

Answer: (A)
Explanation:
Chloric acid (HClO₃) is a strong acid that completely dissociates in water, unlike weak acids such as HF, H₂CO₃, and CH₃COOH.
        

Question 54:

Which of the following is a strong acid?

Options:

• (A) HBr
• (B) CH₃COOH
• (C) H₂CO₃
• (D) H₃PO₄

Answer: (A)
Explanation:
Hydrobromic acid (HBr) is a strong acid that fully dissociates in water. The others are weak acids.
        

Question 55:

Which indicator would be most suitable for a titration between a strong acid and a strong base?

Options:

• (A) Methyl orange (pH range: 3.1 – 4.4)
• (B) Phenolphthalein (pH range: 8.3 – 10.0)
• (C) Bromothymol blue (pH range: 6.0 – 7.6)
• (D) Thymol blue (pH range: 1.2 – 2.8)

Answer: (C)
Explanation:
Bromothymol blue is suitable for titrations involving strong acids and strong bases as it changes color near the neutral pH range (6.0 - 7.6).
        

Question 56:

Which of the following salts will produce an acidic solution when dissolved in water?

Options:

• (A) Na₂SO₄
• (B) NH₄NO₃
• (C) KCl
• (D) CaCl₂

Answer: (B)
Explanation:
Ammonium nitrate (NH₄NO₃) produces an acidic solution because NH₄⁺ is a weak acid.
        

Question 57:

Which of the following would not form a buffer solution?

Options:

• (A) CH₃COOH and CH₃COONa
• (B) HCl and NaCl
• (C) NH₃ and NH₄Cl
• (D) H₂CO₃ and NaHCO₃

Answer: (B)
Explanation:
HCl and NaCl do not form a buffer because HCl is a strong acid and lacks a weak conjugate base.
        

Question 58:

In the titration of a weak acid with a strong base, which indicator would be most appropriate?

Options:

• (A) Methyl red (pH range: 4.4 – 6.2)
• (B) Bromothymol blue (pH range: 6.0 – 7.6)
• (C) Phenolphthalein (pH range: 8.3 – 10.0)
• (D) Thymol blue (pH range: 1.2 – 2.8)

Answer: (C)
Explanation:
Phenolphthalein is suitable for weak acid-strong base titrations because the pH at the equivalence point is usually above 7.
        

Question 59:

Which of the following acids is considered weak?

Options:

• (A) HNO₃
• (B) HF
• (C) HCl
• (D) H₂SO₄

Answer: (B)
Explanation:
Hydrofluoric acid (HF) is a weak acid as it only partially dissociates in water, unlike strong acids like HNO₃, HCl, and H₂SO₄.
        

Question 60:

Which of the following indicators would change color at a pH of approximately 5.5?

Options:

• (A) Methyl orange (pH range: 3.1 – 4.4)
• (B) Bromothymol blue (pH range: 6.0 – 7.6)
• (C) Methyl red (pH range: 4.4 – 6.2)
• (D) Phenolphthalein (pH range: 8.3 – 10.0)

Answer: (C)
Explanation:
Methyl red changes color in the pH range 4.4 to 6.2, covering pH 5.5.
        

Question 61:

Which of the following is a strong base?

Options:

• (A) NH₃
• (B) KOH
• (C) CH₃NH₂
• (D) Al(OH)₃

Answer: (B)
Explanation:
Potassium hydroxide (KOH) is a strong base that fully dissociates in water, unlike weak bases such as NH₃ and CH₃NH₂.
        

Question 62:

Which of the following buffer solutions would have the highest pH?

Options:

• (A) CH₃COOH and CH₃COONa
• (B) H₂CO₃ and NaHCO₃
• (C) NH₄Cl and NH₃
• (D) H₃PO₄ and NaH₂PO₄

Answer: (C)
Explanation:
NH₄Cl and NH₃ form a basic buffer, and NH₃ is a weak base. This buffer has a higher pH compared to the others, which are either acidic or less basic.
        

Question 63:

In a titration of a strong base with a weak acid, the pH at the equivalence point is likely to be:

Options:

• (A) 7.0
• (B) Less than 7.0
• (C) Greater than 7.0
• (D) Depends on the initial concentration

Answer: (C)
Explanation:
In a strong base-weak acid titration, the conjugate base of the weak acid hydrolyzes, making the solution basic at the equivalence point.
        

Question 64:

Which indicator would be best for detecting the endpoint in a titration involving a weak base and a strong acid?

Options:

• (A) Phenolphthalein (pH range: 8.3 – 10.0)
• (B) Bromophenol blue (pH range: 3.0 – 4.6)
• (C) Bromothymol blue (pH range: 6.0 – 7.6)
• (D) Methyl orange (pH range: 3.1 – 4.4)

Answer: (D)
Explanation:
For a weak base-strong acid titration, the equivalence point pH is below 7. Methyl orange, with a color change in the acidic range, is suitable.
        

Question 65:

A buffer solution is made with 0.1 M acetic acid (CH₃COOH) and 0.1 M sodium acetate (CH₃COONa). What is the pH of the buffer if the pKa of acetic acid is 4.74?

Options:

• (A) 4.74
• (B) 5.00
• (C) 4.50
• (D) 7.00

Answer: (A)
Explanation:
Since the concentrations of the acid and conjugate base are equal, the pH is equal to the pKa of acetic acid, which is 4.74.
        

Question 66:

When sodium acetate (CH₃COONa) dissolves in water, the resulting solution will be:

Options:

• (A) Acidic
• (B) Basic
• (C) Neutral
• (D) Depends on concentration

Answer: (B)
Explanation:
Sodium acetate forms a basic solution because acetate (CH₃COO⁻) is the conjugate base of the weak acid acetic acid.
        

Question 67:

Which of the following indicators would be best for a titration where the endpoint pH is expected to be around 9.5?

Options:

• (A) Bromothymol blue (pH range: 6.0 – 7.6)
• (B) Phenolphthalein (pH range: 8.3 – 10.0)
• (C) Methyl orange (pH range: 3.1 – 4.4)
• (D) Litmus (pH range: 4.5 – 8.3)

Answer: (B)
Explanation:
Phenolphthalein is ideal for detecting endpoints around pH 9.5 as it changes color in the range of 8.3 to 10.0.
        

Question 68:

What is the pH of a 0.10 M solution of hydrochloric acid (HCl)?

Options:

• (A) 1.00
• (B) 2.00
• (C) 0.10
• (D) 7.00

Answer: (A)
Explanation:
For a strong acid like HCl, the pH is calculated as:
pH = -log[H⁺] = -log(0.10) = 1.00.
        

Question 69:

What is the pH of a buffer solution containing 0.15 M acetic acid (CH₃COOH) and 0.10 M sodium acetate (CH₃COONa)? The \(K_a\) of acetic acid is \(1.8 \times 10^{-5}\).

Options:

• (A) 4.62
• (B) 4.74
• (C) 4.56
• (D) 4.82

Answer: (C)
Explanation:
Using 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 sodium acetate to acetic acid:
\[ \frac{[CH₃COONa]}{[CH₃COOH]} = \frac{0.10}{0.15} = 0.67 \]
Substitute into the equation:
\[ pH = 4.74 + \log(0.67) = 4.74 - 0.18 = 4.56 \]
        

Question 70:

Which of the following salts will produce an acidic solution when dissolved in water?

Options:

• (A) NaCl
• (B) NH₄Cl
• (C) KNO₃
• (D) Na₂CO₃

Answer: (B)
Explanation:
NH₄Cl dissociates to form NH₄⁺ and Cl⁻. NH₄⁺ is the conjugate acid of NH₃, making the solution acidic.
        

Question 71:

What is the pH of a 0.05 M solution of sulfuric acid (H₂SO₄)? Assume complete dissociation of the first proton and negligible dissociation of the second proton.

Options:

• (A) 1.30
• (B) 1.00
• (C) 1.70
• (D) 2.00

Answer: (A)
Explanation:
For strong acids like H₂SO₄, assume full dissociation of the first proton:
\[ [H⁺] = 0.05 \, \text{M} \]
Calculate pH:
\[ pH = -\log(0.05) = 1.30 \]
        

Question 72:

What is the pH of a 0.01 M solution of nitric acid (HNO₃)?

Options:

• (A) 1.00
• (B) 2.00
• (C) 3.00
• (D) 4.00

Answer: (B)
Explanation:
For strong acids like HNO₃:
\[ [H⁺] = 0.01 \, \text{M} \]
Calculate pH:
\[ pH = -\log(0.01) = 2.00 \]
        

Question 73:

What is the pH of a buffer solution made by mixing 0.30 M NH₃ and 0.20 M NH₄Cl? The \(K_b\) of NH₃ is \(1.8 \times 10^{-5}\).

Options:

• (A) 9.12
• (B) 9.04
• (C) 9.26
• (D) 8.95

Answer: (C)
Explanation:
Use the Henderson-Hasselbalch equation for bases:
\[ pOH = pK_b + \log\left(\frac{[NH₄Cl]}{[NH₃]}\right) \]
Calculate \( pK_b \):
\[ pK_b = -\log(1.8 \times 10^{-5}) = 4.74 \]
Calculate the ratio:
\[ \frac{[NH₄Cl]}{[NH₃]} = \frac{0.20}{0.30} = 0.67 \]
Substitute:
\[ pOH = 4.74 + \log(0.67) = 4.74 - 0.18 = 4.56 \]
Convert to pH:
\[ pH = 14 - 4.56 = 9.44 \]
        

Question 74:

Which of the following solutions would act as the best buffer at a pH of 4.8?

Options:

• (A) HCl and NaCl
• (B) CH₃COOH and CH₃COONa
• (C) NH₄Cl and NH₃
• (D) NaOH and KOH

Answer: (B)
Explanation:
A buffer works best when the pH is close to the pKa of the weak acid. Acetic acid (CH₃COOH) has a pKa of 4.74, making it suitable for a buffer at pH 4.8.
        

Question 75:

A buffer solution contains 0.25 M NH₃ and 0.20 M NH₄Cl. What will happen to the pH if a small amount of HCl is added?

Options:

• (A) pH increases slightly
• (B) pH decreases slightly
• (C) pH remains the same
• (D) pH decreases significantly

Answer: (B)
Explanation:
Adding HCl to a buffer solution reacts with NH₃, forming NH₄⁺. This reduces the base concentration slightly, causing a small decrease in pH.