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 0.10 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
View Answer
Answer Explanation:
To find the concentration of the formic acid, we use the titration formula at the equivalence point: MacidVacid = MbaseVbase.
Cacid × Vacid = Cbase × Vbase
Cacid × 20.0 mL = 0.10 M × 10.0 mL
Cacid = (0.10 M × 10.0 mL) / 20.0 mL = 0.050 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) HCOONa
- (B) HC2H3O2
- (C) NH3
- (D) H2O
View Answer
Correct Answer: (A)
Question 5:
The equation CH3NH2(aq) + H2O(l) ⇌ CH3NH3+(aq) + OH-(aq) represents the reaction between the weak base methylamine and water. Which of the following best represents the concentrations of the various species at equilibrium?
Options:
- (A) [CH3NH2] > [CH3NH3+] = [OH-]
- (B) [OH-] > [CH3NH2] > [CH3NH3+]
- (C) [CH3NH3+] = [OH-] > [CH3NH2]
- (D) [CH3NH2] = [CH3NH3+] = [OH-]
View Answer
Answer Explanation: Because methylamine is a weak base, it only reacts with water to a small extent. This means at equilibrium, most of the methylamine remains unreacted. The products, the conjugate acid CH3NH3+ and hydroxide OH-, are formed in equal, small amounts.
Correct Answer: (A)
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), Ka = 5.9 × 10-2
- (B) Arsenic acid (H3AsO4), Ka = 5.6 × 10-3
- (C) Acetic acid (HC2H3O2), Ka = 1.8 × 10-5
- (D) Hypochlorous acid (HOCl), Ka = 3.0 × 10-8
View Answer
Answer Explanation: To create an effective buffer, you should choose an acid whose pKa is as close as possible to the target pH.
pKa = -log(Ka)
For acetic acid: pKa = -log(1.8 × 10-5) = 4.74
This value is the closest to a pH of 5.
Correct Answer: (C)
Question 8:
A solution of sulfurous acid (H2SO3) is present in an aqueous solution. Which of the following represents the relative concentrations of the different species 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:
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 Ka of acetic acid is 1.8 × 10-5.
Options:
- (A) 4.44
- (B) 4.14
- (C) 4.74
- (D) 5.00
View Answer
Answer Explanation:
Use the Henderson-Hasselbalch equation: pH = pKa + log([conjugate base]/[acid])
First, calculate the pKa: pKa = -log(1.8 × 10-5) = 4.74
Now, substitute the concentrations: pH = 4.74 + log(0.25 / 0.50) = 4.74 + log(0.50) = 4.74 - 0.30 = 4.44
Correct Answer: (A)
Question 10:
What is the pH of a 0.010 M solution of formic acid (HCOOH)? The Ka of formic acid is 1.8 × 10-4.
Options:
- (A) 3.18
- (B) 2.74
- (C) 2.90
- (D) 3.74
View Answer
Answer Explanation:
Use the Ka expression: Ka = [H+][A-] / [HA] = x2 / (0.010 - x) = 1.8 × 10-4
Because dissociation is significant, use the quadratic formula: x2 + (1.8 × 10-4)x - (1.8 × 10-6) = 0
Solving for x gives: x = [H+] = 1.255 × 10-3 M
The pH is: pH = -log(1.255 × 10-3) = 2.90
Correct Answer: (C)
Question 11:
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)
View Answer
Correct Answer: (B) Phenolphthalein changes color in the range of 8.3 to 10.0, which is perfect for detecting an endpoint around pH 9.5.
Question 12:
An aqueous solution of NH4Cl is mildly acidic. Which of the following statements explains this observation?
Options:
- (A) The NH4+ ion reacts with water to produce H3O+, lowering the pH.
- (B) The Cl- ion reacts with water to produce HCl, lowering the pH.
- (C) Both NH4+ and Cl- ions hydrolyze water to produce an acidic solution.
- (D) NH4Cl is the salt of a strong base and strong acid, which makes its solution acidic.
View Answer
Correct Answer: (A)
Ammonium chloride is acidic in water because the NH4+ (the conjugate acid of a weak base) donates a proton to water:
NH4+ + H2O → NH3 + H3O+
The Cl- ion, being the conjugate base of a strong acid (HCl), does not hydrolyze water.
Question 13:
An aqueous solution of sodium acetate (NaC2H3O2) is basic. What is the best explanation for this?
Options:
- (A) The acetate ion (C2H3O2-) reacts with water to produce OH-, raising the pH.
- (B) The Na+ ion reacts with water to produce NaOH, raising the pH.
- (C) Sodium acetate is the salt of a strong base and strong acid, making the solution basic.
- (D) Both Na+ and C2H3O2- ions hydrolyze to produce a basic solution.
View Answer
Correct Answer: (A)
Sodium acetate yields a basic solution because the acetate ion (the conjugate base of acetic acid) hydrolyzes water:
C2H3O2- + H2O → HC2H3O2 + OH-
The Na+ ion (from a strong base, NaOH) does not react with water.
Question 14:
Which of the following salts, when dissolved in water, produces a neutral (pH ≈ 7) solution?
Options:
- (A) NaCl
- (B) NH4Cl
- (C) Na2CO3
- (D) CH3COONa
View Answer
Correct Answer: (A)
Sodium chloride is a neutral salt because it is the product of a strong acid (HCl) and a strong base (NaOH).
Neither Na+ nor Cl- hydrolyzes water, so the solution remains neutral.
(By contrast, NH4Cl is acidic and both Na2CO3 and CH3COONa are basic in solution.)
Question 15:
Which of the following pairs of substances, when mixed in solution, would produce a buffer solution?
Options:
- (A) HCl and NaCl (equal volumes)
- (B) CH3COOH and CH3COONa (equal concentrations)
- (C) HNO3 and NaNO3 (equal moles)
- (D) NaOH and KCl (equal moles)
View Answer
Correct Answer: (B)
A buffer can be made from a weak acid and its conjugate base.
Mixing acetic acid with sodium acetate provides CH3COOH and CH3COO- in the same solution.
None of the other combinations provide a weak acid/base conjugate pair (for example, HCl/NaCl is a strong acid and neutral salt, not a buffer).
Question 16:
A buffer solution contains appreciable amounts of acetic acid (CH3COOH) and sodium acetate (CH3COONa). Which equation best represents what happens when a small amount of NaOH is added to this buffer?
Options:
- (A) CH3COOH + OH- → CH3COO- + H2O
- (B) CH3COO- + OH- → CH3COOH + O2-
- (C) CH3COOH + H+ → CH3COOH2+
- (D) CH3COO- + H+ → CH3COOH
View Answer
Correct Answer: (A)
When a strong base (OH-) is added to an acetic acid/acetate buffer, the acetic acid (weak acid component) neutralizes the base:
CH3COOH + OH- → CH3COO- + H2O
This reaction consumes the added OH- and generates more acetate, resisting a significant pH change.
Question 17:
During the titration of a weak acid with a strong base, the half-equivalence point is reached when half of the acid has been neutralized. Which of the following is true at the half-equivalence point?
Options:
- (A) pH = pKa of the acid (and [HA] = [A-]).
- (B) pH = 7.00 for any weak acid titration.
- (C) The moles of base added are equal to the initial moles of acid.
- (D) The buffering capacity of the solution is effectively zero.
View Answer
Correct Answer: (A)
At the half-equivalence point, exactly half of the weak acid has been converted to its conjugate base.
This means [HA] = [A-] in solution, making it an ideal buffer.
Under these conditions, the Henderson-Hasselbalch equation gives pH = pKa.
(Note: pH = 7 at half-neutralization only if the acid’s pKa is 7, which is not generally true.)
Question 18:
In a buffer solution, if the pH is greater than the pKa of the weak acid, which of the following statements is true about the concentrations of the conjugate base [A-] and the acid [HA]?
Options:
- (A) [A-] > [HA]
- (B) [A-] = [HA]
- (C) [A-] < [HA]
- (D) No relationship can be determined from the pH and pKa.
View Answer
Correct Answer: (A)
If pH > pKa, then the buffer solution has a higher concentration of the conjugate base than the acid.
This is because pH = pKa + log([A-]/[HA]).
When pH is above pKa, the log term must be positive, implying [A-] > [HA].
Question 19:
What is the percent dissociation of 0.020 M formic acid (HCOOH)? (Ka = 1.8 × 10-4)
Options:
- (A) ~3.0%
- (B) ~6.0%
- (C) ~9.5%
- (D) ~19%
View Answer
Correct Answer: (C)
For a weak monoprotic acid, an excellent approximation is:
% dissociation ≈ √(Ka/C) × 100%
Compute: Ka/C = (1.8 × 10-4) / 0.020 = 9.0 × 10-3
√(9.0 × 10-3) ≈ 0.0949 → 9.49%
Thus, the percent dissociation is approximately 9.5%.
Question 20:
Estimate the percent dissociation of 0.010 M hydrofluoric acid (HF). (Ka = 6.8 × 10-4)
Options:
- (A) ~8%
- (B) ~14%
- (C) ~26%
- (D) ~2.6%
View Answer
Correct Answer: (C)
Use % dissociation ≈ √(Ka/C) × 100%.
Ka/C = (6.8 × 10-4) / (1.0 × 10-2) = 6.8 × 10-2 = 0.068
√(0.068) ≈ 0.261 → 26.1%
So the percent dissociation is about 26% (quadratic would give a very similar value).
Question 21:
A 0.10 M solution of formic acid (HCOOH) is prepared. Which change will decrease its percent dissociation the most?
Options:
- (A) Diluting the solution to 0.010 M
- (B) Adding solid NaCl
- (C) Adding 0.10 M NaHCOO (sodium formate)
- (D) Warming the solution by 5 °C
View Answer
Correct Answer: (C)
Adding the conjugate base (HCOO-) introduces a strong common ion effect, shifting the equilibrium toward the undissociated acid and sharply reducing percent dissociation.
Dilution (A) increases percent dissociation (∝ 1/√C). Adding inert salt (B) has negligible effect. A small temperature change (D) is minor compared to the common ion effect.
Question 22:
A 0.10 M acetic acid (CH3COOH) solution has a percent dissociation of about 1.3%. If this solution is diluted tenfold to 0.010 M (same Ka), the percent dissociation will:
Options:
- (A) Stay the same
- (B) Approximately double
- (C) Increase by a factor of about √10 (≈ 3.16)
- (D) Decrease by a factor of about √10 (≈ 0.32)
View Answer
Correct Answer: (C)
For weak acids: % dissociation ≈ √(Ka/C) × 100% ∝ 1/√C.
Reducing concentration by 10× increases percent dissociation by √10 ≈ 3.16.
Thus, ~1.3% → ~4.1% (consistent with the relation).
Question 23:
What is the approximate percent dissociation of 0.050 M benzoic acid (C6H5COOH)? (Ka = 6.3 × 10-5)
Options:
- (A) ~1.1%
- (B) ~2.2%
- (C) ~3.6%
- (D) ~7.2%
View Answer
Correct Answer: (C)
% dissociation ≈ √(Ka/C) × 100%.
Ka/C = (6.3 × 10-5) / 0.050 = 1.26 × 10-3
√(1.26 × 10-3) ≈ 0.0355 → 3.55%
Therefore, the percent dissociation is approximately 3.6%.
Question 24:
What is the pH of a 0.10 M NH4Cl solution at 25 °C? (Kb for NH3 = 1.8 × 10-5)
Options:
- (A) 5.13
- (B) 6.50
- (C) 7.00
- (D) 8.87
View Answer
Correct Answer: (A)
NH4+ is a weak acid: Ka = Kw/Kb = 1.0 × 10-14 / 1.8 × 10-5 = 5.6 × 10-10.
[H+] ≈ √(KaC) = √(5.6 × 10-10 × 0.10) = 7.5 × 10-6; pH ≈ 5.13.
Question 25:
What is the pH of a 0.10 M CH3COONa solution at 25 °C? (Ka for CH3COOH = 1.8 × 10-5)
Options:
- (A) 5.13
- (B) 8.87
- (C) 7.00
- (D) 6.50
View Answer
Correct Answer: (B)
CH3COO- is a weak base: Kb = Kw/Ka = 1.0 × 10-14 / 1.8 × 10-5 = 5.6 × 10-10.
[OH-] ≈ √(KbC) = 7.5 × 10-6; pOH ≈ 5.13; pH ≈ 8.87.
Question 26:
A weak acid HA (pKa = 3.17) is titrated with a strong base. What is the pH at the half-equivalence point?
Options:
- (A) 3.17
- (B) 7.00
- (C) 2.00
- (D) 4.74
View Answer
Correct Answer: (A)
At half-equivalence, [HA] = [A-] so pH = pKa = 3.17.
Question 27:
Mix 100.0 mL of 0.200 M CH3COOH with 50.0 mL of 0.100 M NaOH. What is the resulting pH? (pKa of CH3COOH = 4.74)
Options:
- (A) 4.26
- (B) 4.74
- (C) 5.21
- (D) 3.74
View Answer
Correct Answer: (A)
n(HA) = 0.100 L × 0.200 M = 0.0200 mol; n(OH-) = 0.0500 L × 0.100 M = 0.00500 mol.
After reaction: n(HA) = 0.0150 mol; n(A-) = 0.00500 mol. Ratio = 0.00500/0.0150 = 1/3.
pH = pKa + log([A-]/[HA]) = 4.74 + log(1/3) = 4.74 - 0.477 ≈ 4.26.
Question 28:
Which indicator is most suitable for a titration of a weak base with a strong acid if the expected endpoint pH is ~5.0?
Options:
- (A) Phenolphthalein (8.3–10.0)
- (B) Methyl orange (3.1–4.4)
- (C) Methyl red (4.4–6.2)
- (D) Bromothymol blue (6.0–7.6)
View Answer
Correct Answer: (C)
Endpoint near pH 5 lies within methyl red’s transition range (4.4–6.2).
Question 29:
A buffer contains 0.20 M HA and 0.40 M A- (pKa = 4.74). What is the pH?
Options:
- (A) 4.44
- (B) 5.04
- (C) 4.74
- (D) 5.44
View Answer
Correct Answer: (B)
pH = 4.74 + log(0.40/0.20) = 4.74 + log(2) = 4.74 + 0.30 = 5.04.
Question 30:
Which change increases the percent dissociation of a weak acid HA the most?
Options:
- (A) Adding its conjugate base salt (A-)
- (B) Diluting the solution
- (C) Adding a neutral salt (e.g., NaCl)
- (D) Slightly increasing temperature (assume endothermic dissociation is negligible)
View Answer
Correct Answer: (B)
For weak acids, % dissociation ∝ 1/√C; dilution increases percent dissociation. Adding A- causes the common ion effect and decreases dissociation.
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 Ka of acetic acid is 1.8 × 10-5.
Options:
- (A) 4.44
- (B) 4.14
- (C) 4.74
- (D) 5.00
View Answer
Answer Explanation: Use the Henderson-Hasselbalch equation for buffer solutions: pH = pKa + log([conjugate base]/[acid]) First, calculate the pKa: pKa = −log(1.8 × 10−5) = 4.74 Now, substitute the concentrations into the equation: pH = 4.74 + log(0.25/0.50) = 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 Ka of formic acid is 1.8 × 10-4.
Options:
- (A) 3.18
- (B) 2.74
- (C) 2.90
- (D) 3.74
View Answer
Answer Explanation: Use the Ka expression for formic acid: Ka = [H+][A-]/[HA] = x2/(0.010 − x) = 1.8 × 10−4 In this case, the dissociation is significant (>5%), so we should use the quadratic formula to solve for x = [H+]: x2 + (1.8 × 10−4)x − (1.8 × 10−6) = 0 Solving for x gives: x = 1.255 × 10−3 M The pH is: pH = −log(1.255 × 10−3) = 2.90 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 Ka of lactic acid is 1.4 × 10-4.
Options:
- (A) 3.08
- (B) 3.24
- (C) 3.55
- (D) 3.00
View Answer
Answer Explanation: Use the Henderson-Hasselbalch equation: pH = pKa + log([conjugate base]/[acid]) Calculate pKa: pKa = −log(1.4 × 10−4) = 3.85 Substitute the concentrations into the equation: pH = 3.85 + log(0.05/0.10) = 3.85 + log(0.50) = 3.85 − 0.30 = 3.55 Correct Answer: (C)
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 Ka of carbonic acid is 4.3 × 10-7.
Options:
- (A) 6.12
- (B) 6.27
- (C) 6.50
- (D) 6.74
View Answer
Answer Explanation: Use the Henderson-Hasselbalch equation: pH = pKa + log([conjugate base]/[acid]) Calculate pKa: pKa = −log(4.3 × 10−7) = 6.37 Substitute the concentrations into the equation: pH = 6.37 + log(0.40/0.50) = 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 Ka of carbonic acid is 4.3 × 10-7.
Options:
- (A) 6.54
- (B) 6.37
- (C) 6.67
- (D) 6.20
View Answer
Answer Explanation: Using the Henderson-Hasselbalch equation: pH = pKa + log([conjugate base]/[acid]) Calculate pKa: pKa = − log(4.3 × 10−7) = 6.37 Substitute the concentrations into the equation: pH = 6.37 + log(0.10/0.05) = 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
Correct Answer: (A) 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
Correct Answer: (B) This pair forms a buffer solution as NH₃ is a weak base and NH₄Cl contains its conjugate acid.
Question 39: Which of the following is a weak acid?
- (A) HNO₃
- (B) CH₃COOH
- (C) H₂SO₄
- (D) HI
View Answer
Correct Answer: (B) 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
Correct Answer: (D) 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
Correct Answer: (B) 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
Correct Answer: (B) 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
Correct Answer: (B) 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
Correct Answer: (B) 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
Correct Answer: (A) 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) NH₄NO₃
- (C) KNO₃
- (D) CaCl₂
View Answer
Correct Answer: (B) Ammonium nitrate (NH₄NO₃) produces an acidic solution due to the presence of NH₄⁺, which is a weak acid.
Question 47: Select the weak acid among the following options.
- (A) HNO₃
- (B) HCl
- (C) H₂SO₄
- (D) H₂CO₃
View Answer
Correct Answer: (D) 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
Correct Answer: (A) 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
Correct Answer: (A) 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
Correct Answer: (A) 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
Correct Answer: (C) 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
Correct Answer: (D) 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
Correct Answer: (A) 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
Correct Answer: (A) 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
Correct Answer: (C) 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
Correct Answer: (B) 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
Correct Answer: (B) 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
Correct Answer: (C) 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
Correct Answer: (B) 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
Correct Answer: (C) 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
Correct Answer: (B) 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
Correct Answer: (C) This buffer is made from a weak base (NH₃) and its conjugate acid, so its buffering range is in the basic pH region (around 9.25), which is higher than the acidic buffers listed.
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
Correct Answer: (C) At the equivalence point, the weak acid has been converted to its conjugate base, which makes the solution basic (pH > 7).
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
Correct Answer: (D) For a weak base-strong acid titration, the pH at the equivalence point is acidic (pH < 7). An indicator like methyl orange is suitable because its color change range is in the acidic region.
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
Correct Answer: (A) When the concentrations of the weak acid and its conjugate base are equal, the pH of the buffer 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
Correct Answer: (B) Acetate (CH₃COO⁻) is the conjugate base of a weak acid and will hydrolyze water 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
Correct Answer: (B) Phenolphthalein changes color in the range of 8.3 to 10.0, which is perfect for detecting an endpoint around pH 9.5.
Question 68:
An aqueous solution of NH4Cl is mildly acidic. Which of the following statements explains this observation?
Options:
- (A) The NH4+ ion reacts with water to produce H3O+, lowering the pH.
- (B) The Cl- ion reacts with water to produce HCl, lowering the pH.
- (C) Both NH4+ and Cl- ions hydrolyze water to produce an acidic solution.
- (D) NH4Cl is the salt of a strong base and strong acid, which makes its solution acidic.
View Answer
Correct Answer: (A)
Ammonium chloride is acidic in water because the NH4+ (the conjugate acid of a weak base) donates a proton to water:
NH4+ + H2O → NH3 + H3O+
The Cl- ion, being the conjugate base of a strong acid (HCl), does not hydrolyze water.
Question 69:
An aqueous solution of sodium acetate (NaC2H3O2) is basic. What is the best explanation for this?
Options:
- (A) The acetate ion (C2H3O2-) reacts with water to produce OH-, raising the pH.
- (B) The Na+ ion reacts with water to produce NaOH, raising the pH.
- (C) Sodium acetate is the salt of a strong base and strong acid, making the solution basic.
- (D) Both Na+ and C2H3O2- ions hydrolyze to produce a basic solution.
View Answer
Correct Answer: (A)
Sodium acetate yields a basic solution because the acetate ion (the conjugate base of acetic acid) hydrolyzes water:
C2H3O2- + H2O → HC2H3O2 + OH-
The Na+ ion (from a strong base, NaOH) does not react with water.
Question 70:
Which of the following salts, when dissolved in water, produces a neutral (pH ≈ 7) solution?
Options:
- (A) NaCl
- (B) NH4Cl
- (C) Na2CO3
- (D) CH3COONa
View Answer
Correct Answer: (A)
Sodium chloride is a neutral salt because it is the product of a strong acid (HCl) and a strong base (NaOH).
Neither Na+ nor Cl- hydrolyzes water, so the solution remains neutral.
(By contrast, NH4Cl is acidic and both Na2CO3 and CH3COONa are basic in solution.)
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