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 0.05 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)
- (B) Phosphoric acid (H₃AsO₄)
- (C) Acetic acid (HC₂H₃O₂)
- (D) Hypochlorous acid (HOCl)
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
Options:
- (A)
- (B)
- (C)
- (D)
View Answer
Correct Answer: (A)
Question 11:
The value of
Options:
- (A)
- (B)
- (C)
- (D)
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
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
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
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
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
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
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
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
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
Options:
- (A)
- (B)
- (C)
- (D)
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
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
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
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
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
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
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
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)
Question 37: Which of the following is a strong acid?
- (A) HCl
- (B) HF
- (C) H₂CO₃
- (D) H₃PO₄
View Answer
(A) HCl - Hydrochloric acid (HCl) is a strong acid that completely dissociates in water.
Question 38: Which of the following pairs can form a buffer solution?
- (A) HCl and NaCl
- (B) NH₃ and NH₄Cl
- (C) H₂SO₄ and Na₂SO₄
- (D) NaOH and KOH
View Answer
(B) NH₃ and NH₄Cl - This pair forms a buffer solution as NH₃ is a weak base and NH₄Cl is its conjugate acid salt.
Question 39: Which of the following is a weak acid?
- (A) HNO₃
- (B) CH₃COOH
- (C) H₂SO₄
- (D) HI
View Answer
(B) CH₃COOH - Acetic acid (CH₃COOH) is a weak acid because it only partially dissociates in water.
Question 40: Which of the following salts will produce a basic solution when dissolved in water?
- (A) NaCl
- (B) NH₄Cl
- (C) KNO₃
- (D) Na₂CO₃
View Answer
(D) Na₂CO₃ - Sodium carbonate (Na₂CO₃) is a salt that will produce a basic solution because the carbonate ion (CO₃²⁻) is a weak base.
Question 41: Identify the strong base among the following compounds.
- (A) NH₃
- (B) NaOH
- (C) Al(OH)₃
- (D) Mg(OH)₂
View Answer
(B) NaOH - Sodium hydroxide (NaOH) is a strong base that fully dissociates in water.
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?
- (A) 7.00
- (B) 8.72
- (C) 4.74
- (D) 3.00
View Answer
(B) 8.72 - At the equivalence point, the solution contains the conjugate base (CH₃COO⁻), which makes the solution basic.
Question 43: Which of the following is a weak base?
- (A) KOH
- (B) NH₃
- (C) Ca(OH)₂
- (D) LiOH
View Answer
(B) NH₃ - Ammonia (NH₃) is a weak base because it only partially dissociates in water.
Question 44: Which of the following solutions would best resist a pH change when a small amount of HCl is added?
- (A) HCl and NaCl
- (B) CH₃COOH and CH₃COONa
- (C) NaOH and KOH
- (D) H₂SO₄ and Na₂SO₄
View Answer
(B) CH₃COOH and CH₃COONa - This is 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?
- (A) HClO₄
- (B) H₃PO₄
- (C) H₂CO₃
- (D) CH₃COOH
View Answer
(A) HClO₄ - Perchloric acid (HClO₄) is a strong acid and completely dissociates in water.
Question 46: Which of the following salts will produce an acidic solution when dissolved in water?
- (A) NaCl
- (B) NH₄Cl
- (C) KNO₃
- (D) Na₂CO₃
View Answer
(B) NH₄Cl - Ammonium chloride (NH₄Cl) produces an acidic solution as the NH₄⁺ ion is a weak acid.
Question 47: Select the weak acid among the following options.
- (A) HNO₃
- (B) HCl
- (C) H₂SO₄
- (D) H₂CO₃
View Answer
(D) H₂CO₃ - Carbonic acid (H₂CO₃) is a weak acid because it only partially dissociates in water.
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?
- (A) 4.74
- (B) 7.00
- (C) 5.00
- (D) 3.00
View Answer
(A) 4.74 - At the halfway point, the pH is equal to the pKa of the acid. The pKa of acetic acid is 4.74.
Question 49: Which of the following is a strong base?
- (A) NaOH
- (B) NH₄OH
- (C) Al(OH)₃
- (D) CH₃NH₂
View Answer
(A) NaOH - Sodium hydroxide (NaOH) is a strong base and dissociates completely in water.
Question 50: A buffer solution has a pH of 5.0. Which of the following acids would be best for creating this buffer?
- (A) Acetic acid, pKa = 4.74
- (B) Formic acid, pKa = 3.75
- (C) Hydrochloric acid, pKa < 0
- (D) Nitric acid, pKa < 0
View Answer
(A) Acetic acid - The pKa of acetic acid (4.74) is close to the desired pH of 5.0, making it suitable for the buffer.
Question 51: Identify the weak base in this list.
- (A) KOH
- (B) Ca(OH)₂
- (C) NH₃
- (D) Ba(OH)₂
View Answer
(C) NH₃ - Ammonia (NH₃) is a weak base as it only partially dissociates in water.
Question 52: Which salt, when dissolved in water, would likely produce a solution with a pH greater than 7?
- (A) NaCl
- (B) NH₄Cl
- (C) KNO₃
- (D) Na₂CO₃
View Answer
(D) Na₂CO₃ - Sodium carbonate (Na₂CO₃) forms a basic solution as the carbonate ion (CO₃²⁻) is a weak base.
Question 53: Which of the following is a strong acid?
- (A) HClO₃
- (B) H₂CO₃
- (C) HF
- (D) CH₃COOH
View Answer
(A) HClO₃ - Chloric acid (HClO₃) is a strong acid and dissociates completely in water.
Question 54: Which of the following is a strong acid?
- (A) HBr
- (B) CH₃COOH
- (C) H₂CO₃
- (D) H₃PO₄
View Answer
(A) HBr - Hydrobromic acid (HBr) is a strong acid that fully dissociates in water.
Question 55: Which indicator would be most suitable for a titration between a strong acid and a strong base?
- (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)
View Answer
(C) Bromothymol blue - Bromothymol blue is suitable for strong acid-strong base titrations as it changes color around the neutral pH range.
Question 56: Which of the following salts will produce an acidic solution when dissolved in water?
- (A) Na₂SO₄
- (B) NH₄NO₃
- (C) KCl
- (D) CaCl₂
View Answer
(B) NH₄NO₃ - Ammonium nitrate (NH₄NO₃) produces an acidic solution due to the presence of NH₄⁺, which is a weak acid.
Question 57: Which of the following would not form a buffer solution?
- (A) CH₃COOH and CH₃COONa
- (B) HCl and NaCl
- (C) NH₃ and NH₄Cl
- (D) H₂CO₃ and NaHCO₃
View Answer
(B) HCl and NaCl - This combination does not form a buffer as HCl is a strong acid and does not have a weak conjugate base in this pair.
Question 58: In the titration of a weak acid with a strong base, which indicator would be most appropriate?
- (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)
View Answer
(C) Phenolphthalein - For weak acid-strong base titrations, phenolphthalein is suitable as the pH at the equivalence point is usually above 7.
Question 59: Which of the following acids is considered weak?
- (A) HNO₃
- (B) HF
- (C) HCl
- (D) H₂SO₄
View Answer
(B) HF - Hydrofluoric acid (HF) is a weak acid as it only partially dissociates in water.
Question 60: Which of the following indicators would change color at a pH of approximately 5.5?
- (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)
View Answer
(C) Methyl red - 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?
- (A) NH₃
- (B) KOH
- (C) CH₃NH₂
- (D) Al(OH)₃
View Answer
(B) KOH - Potassium hydroxide (KOH) is a strong base that fully dissociates in water.
Question 62: Which of the following buffer solutions would have the highest pH?
- (A) CH₃COOH and CH₃COONa
- (B) H₂CO₃ and NaHCO₃
- (C) NH₄Cl and NH₃
- (D) H₃PO₄ and NaH₂PO₄
View Answer
(C) NH₄Cl and NH₃ - This buffer has a higher pH due to the presence of NH₃, a weak base.
Question 63: In a titration of a strong base with a weak acid, the pH at the equivalence point is likely to be:
- (A) 7.0
- (B) Less than 7.0
- (C) Greater than 7.0
- (D) Depends on the initial concentration
View Answer
(C) Greater than 7.0 - In a strong base-weak acid titration, the solution is 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?
- (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)
View Answer
(D) Methyl orange - For a weak base-strong acid titration, an acidic indicator like methyl orange is suitable as the pH at the endpoint is below 7.
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?
- (A) 4.74
- (B) 5.00
- (C) 4.50
- (D) 7.00
View Answer
(A) 4.74 - Since the concentrations of the acid and conjugate base are equal, the pH is equal to the pKa.
Question 66: When sodium acetate (CH₃COONa) dissolves in water, the resulting solution will be:
- (A) Acidic
- (B) Basic
- (C) Neutral
- (D) Depends on concentration
View Answer
(B) Basic - Acetate (CH₃COO⁻) is the conjugate base of a weak acid and will hydrolyze to form a basic solution.
Question 67: Which of the following indicators would be best for a titration where the endpoint pH is expected to be around 9.5?
- (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)
View Answer
(B) Phenolphthalein - Phenolphthalein changes color in the range suitable for an endpoint near pH 9.5.
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