DAT Survey of the Natural Sciences Quiz: Acid Base Chemistry
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Acid Base ChemistryQuestion 1 of 20

What is the pH of a 0.0050 M solution of perchloric acid (HClO₄)?

1.30
2.30
5.00
11.70
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DAT Survey of the Natural Sciences Quiz

DAT Survey of the Natural Sciences Quiz: Acid Base Chemistry

Practice Acid Base Chemistry in DAT Survey of the Natural Sciences with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Acid Base Chemistry, giving you a quick way to practice the rules, question types, and explanations that matter most for DAT Survey of the Natural Sciences.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

What is the pH of a 0.0050 M solution of perchloric acid (HClO₄)?

  1. 1.30
  2. 2.30 (correct answer)
  3. 5.00
  4. 11.70
Explanation: When you encounter a pH problem involving a strong acid like perchloric acid (HClO₄), remember that strong acids completely dissociate in water, making the calculation straightforward. Since HClO₄ is a strong acid, it dissociates completely: HClO₄ → H⁺ + ClO₄⁻. This means that in a 0.0050 M solution of HClO₄, the concentration of H⁺ ions equals the original acid concentration: [H⁺] = 0.0050 M. To find pH, use the formula: pH=−log⁡[H+]\text{pH} = -\log[\text{H}^+] Substituting our value: pH=−log⁡(0.0050)=−log⁡(5.0×10−3)=−(log⁡5.0+log⁡10−3)=−(0.70+(−3))=−(0.70−3)=−(−2.30)=2.30\text{pH} = -\log(0.0050) = -\log(5.0 \times 10^{-3}) = -(\log 5.0 + \log 10^{-3}) = -(0.70 + (-3)) = -(0.70 - 3) = -(-2.30) = 2.30 Therefore, answer B (2.30) is correct. Answer A (1.30) would correspond to a stronger acid concentration of about 0.050 M—you may have miscalculated the log or confused the molarity. Answer C (5.00) represents a neutral solution misconception; this would be appropriate for pure water, not an acid solution. Answer D (11.70) is actually the pOH of this solution, not the pH—this is a common trap where students calculate the wrong value or confuse acidic vs. basic solutions. For strong acid pH problems on the DAT, remember this simple two-step process: the H⁺ concentration equals the acid concentration, then apply the negative log. Always double-check that your pH makes sense—strong acids should give pH values well below 7.

Question 2

An aqueous solution of which of the following salts will be basic?

  1. NH₄Cl
  2. KCN (correct answer)
  3. NaNO₃
  4. LiBr
Explanation: When you encounter questions about salt solutions and pH, you need to analyze how each ion affects water's acidity or basicity through hydrolysis reactions. KCN (choice B) creates a basic solution because it contains the cyanide ion (CN⁻), which is the conjugate base of the weak acid HCN. When CN⁻ dissolves in water, it accepts protons from water molecules: CN−+H2O⇌HCN+OH−\text{CN}^- + \text{H}_2\text{O} \rightleftharpoons \text{HCN} + \text{OH}^-. This produces hydroxide ions, making the solution basic. The K⁺ ion doesn't affect pH since it's from the strong base KOH. Choice A (NH₄Cl) produces an acidic solution because NH₄⁺ is the conjugate acid of the weak base NH₃. It donates protons to water, forming H₃O⁺ ions. The Cl⁻ ion is neutral since it comes from strong acid HCl. Choice C (NaNO₃) creates a neutral solution because both Na⁺ (from strong base NaOH) and NO₃⁻ (from strong acid HNO₃) are spectator ions that don't undergo hydrolysis. Choice D (LiBr) also produces a neutral solution since Li⁺ comes from strong base LiOH and Br⁻ comes from strong acid HBr—neither ion affects the water's pH. Strategy tip: Remember that salts formed from weak acids produce basic solutions (their anions accept protons), while salts from weak bases produce acidic solutions (their cations donate protons). Salts from strong acids and strong bases remain neutral.

Question 3

If 150 mL of 0.20 M HCl is mixed with 150 mL of 0.10 M NaOH, what will be the pH of the resulting solution?

  1. 1.30 (correct answer)
  2. 2.00
  3. 7.00
  4. 12.70
Explanation: When you encounter acid-base mixing problems, you need to determine whether you have excess acid, excess base, or a neutral solution, then calculate the resulting pH accordingly. First, calculate the moles of each reactant. For HCl: 0.150 L×0.20 M=0.030 mol0.150 \text{ L} \times 0.20 \text{ M} = 0.030 \text{ mol}. For NaOH: 0.150 L×0.10 M=0.015 mol0.150 \text{ L} \times 0.10 \text{ M} = 0.015 \text{ mol}. Since HCl and NaOH react in a 1:1 ratio (HCl+NaOH→NaCl+H2O\text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O}), the NaOH will be completely consumed, leaving excess HCl. The excess HCl is: 0.030−0.015=0.015 mol0.030 - 0.015 = 0.015 \text{ mol}. The total solution volume is 150+150=300 mL=0.300 L150 + 150 = 300 \text{ mL} = 0.300 \text{ L}. Therefore, the concentration of excess H⁺ is: 0.015 mol0.300 L=0.050 M\frac{0.015 \text{ mol}}{0.300 \text{ L}} = 0.050 \text{ M}. The pH is: pH=−log⁡(0.050)=1.30\text{pH} = -\log(0.050) = 1.30, confirming answer A is correct. Answer B (2.00) would result from incorrectly calculating −log⁡(0.010)-\log(0.010), perhaps from an arithmetic error in determining excess acid. Answer C (7.00) represents the common misconception that mixing any acid and base always gives a neutral solution—this only happens when moles are equal. Answer D (12.70) would be the pOH of this solution, not the pH, representing confusion between these related but opposite scales. Remember: always calculate moles first, determine the limiting reagent, then find the concentration of excess H⁺ or OH⁻ in the final volume. Equal volumes doesn't mean equal moles unless the molarities are also equal.

Question 4

What volume of 0.250 M HCl is required to completely neutralize 50.0 mL of 0.100 M Ba(OH)₂?

  1. 10.0 mL
  2. 20.0 mL
  3. 40.0 mL (correct answer)
  4. 80.0 mL
Explanation: When you encounter acid-base neutralization problems, you're working with stoichiometry where the key is balancing the moles of H⁺ and OH⁻ ions. Start by writing the balanced chemical equation: Ba(OH)2+2HCl→BaCl2+2H2O\text{Ba(OH)}_2 + 2\text{HCl} \rightarrow \text{BaCl}_2 + 2\text{H}_2\text{O} Notice that one mole of Ba(OH)₂ produces 2 moles of OH⁻ ions, while one mole of HCl produces 1 mole of H⁺ ions. This 1:2 stoichiometric ratio is crucial. First, calculate moles of Ba(OH)₂: 0.0500 L×0.100 M=0.00500 mol Ba(OH)20.0500 \text{ L} \times 0.100 \text{ M} = 0.00500 \text{ mol Ba(OH)}_2 Since each Ba(OH)₂ provides 2 OH⁻ ions, you have 0.00500×2=0.0100 mol OH−0.00500 \times 2 = 0.0100 \text{ mol OH}^- For complete neutralization, you need 0.0100 mol H⁺, which means 0.0100 mol HCl. Using the HCl concentration: Volume=0.0100 mol0.250 M=0.0400 L=40.0 mL\text{Volume} = \frac{0.0100 \text{ mol}}{0.250 \text{ M}} = 0.0400 \text{ L} = 40.0 \text{ mL} The answer is C) 40.0 mL. Choice A) 10.0 mL would only provide 0.00250 mol H⁺, neutralizing just half the base. Choice B) 20.0 mL gives 0.00500 mol H⁺, which incorrectly assumes a 1:1 ratio between Ba(OH)₂ and HCl. Choice D) 80.0 mL provides twice the needed H⁺, representing an error in the stoichiometric calculation. Always identify the stoichiometric coefficients from the balanced equation first—polyprotic bases like Ba(OH)₂ are designed to test whether you account for multiple OH⁻ ions per formula unit.

Question 5

Identify the Lewis acid in the following reaction: Ag+(aq)+2NH3(aq)→[Ag(NH3)2]+(aq)\text{Ag}^+(aq) + 2\text{NH}_3(aq) \rightarrow [\text{Ag(NH}_3)_2]^+(aq)

  1. Ag⁺ (correct answer)
  2. NH₃
  3. [Ag(NH₃)₂]⁺
  4. No Lewis acid is present
Explanation: When you encounter a reaction involving coordination complexes, you need to identify Lewis acids and bases. A Lewis acid accepts electron pairs, while a Lewis base donates electron pairs. In this reaction, Ag+(aq)+2NH3(aq)→[Ag(NH3)2]+(aq)\text{Ag}^+(aq) + 2\text{NH}_3(aq) \rightarrow [\text{Ag(NH}_3)_2]^+(aq), the silver ion (Ag+\text{Ag}^+) has empty orbitals that can accept electron pairs. Ammonia (NH3\text{NH}_3) has a lone pair of electrons on nitrogen that it can donate. When the complex forms, ammonia's lone pairs are donated to silver's empty orbitals, creating coordinate covalent bonds. Choice A is correct because Ag+\text{Ag}^+ accepts electron pairs from the ammonia molecules, making it the Lewis acid by definition. Choice B is wrong because NH3\text{NH}_3 donates its lone pair electrons to form the coordinate bonds, making it the Lewis base, not the acid. Choice C is incorrect because [Ag(NH3)2]+[\text{Ag(NH}_3)_2]^+ is the product of the reaction, not a reactant that could function as either acid or base in this context. Choice D is wrong because Lewis acid-base reactions are exactly what's happening here—this is a classic example of complex ion formation through Lewis acid-base chemistry. Remember this pattern: in coordination chemistry, the central metal ion (especially cations with empty orbitals) typically acts as the Lewis acid, while ligands with lone pairs act as Lewis bases. This relationship drives the formation of most coordination complexes you'll see on the DAT.

Question 6

An acid-base indicator, HIn, has a pKa of 5.0. In its acidic form (HIn), it is yellow, and in its basic form (In⁻), it is blue. This indicator would be most suitable for a titration with an equivalence point pH of approximately:

  1. 3.0
  2. 5.0 (correct answer)
  3. 7.0
  4. 9.0
Explanation: When you encounter acid-base indicator questions, you need to understand that indicators work best when their pKa matches the pH at the equivalence point of the titration. This is because indicators undergo their most dramatic color change when the solution pH equals the indicator's pKa value. At pH = pKa, exactly 50% of the indicator exists in its acidic form (HIn) and 50% in its basic form (In⁻), according to the Henderson-Hasselbalch equation. This creates the most noticeable color transition, making it easiest to detect the endpoint. Since this indicator has a pKa of 5.0, it will show its sharpest color change from yellow to blue right around pH 5.0. Looking at the choices: Answer A (pH 3.0) is too acidic - at this pH, the indicator would remain predominantly yellow (HIn form) throughout the equivalence point region, making the color change difficult to detect. Answer C (pH 7.0) and Answer D (pH 9.0) are both too basic - at these pH values, the indicator would already be mostly blue (In⁻ form) before reaching equivalence, again making the transition less visible. Answer B (pH 5.0) is correct because it matches the indicator's pKa, ensuring the color change occurs precisely at the equivalence point. Study tip: Remember the "pKa = pH" rule for indicator selection. Always choose an indicator whose pKa is closest to the expected equivalence point pH. This ensures you'll see the clearest color change exactly when the titration reaches completion.

Question 7

For phosphoric acid (H3PO4H_3PO_4), the acid dissociation constants are Ka1=7.5×10−3K_{a1} = 7.5 \times 10^{-3}, Ka2=6.2×10−8K_{a2} = 6.2 \times 10^{-8}, and Ka3=4.2×10−13K_{a3} = 4.2 \times 10^{-13}. In a solution with a pH of 5.0, which species would be present in the highest concentration?

  1. H3PO4H_3PO_4
  2. PO43−PO_4^{3-}
  3. HPO42−HPO_4^{2-}
  4. H2PO4−H_2PO_4^- (correct answer)
Explanation: When you encounter polyprotic acid problems, you need to determine which species predominates at a given pH by comparing the pH to the pKa values. For phosphoric acid, calculate the pKa values: pKa₁ = 2.12, pKa₂ = 7.21, and pKa₃ = 12.38. At pH 5.0, you're between pKa₁ (2.12) and pKa₂ (7.21). The key principle is that when pH < pKa, the protonated form dominates, and when pH > pKa, the deprotonated form dominates. Since pH 5.0 > pKa₁, the first deprotonation is essentially complete, so H3PO4H_3PO_4 has been converted to H2PO4−H_2PO_4^-. However, since pH 5.0 < pKa₂, the second deprotonation hasn't occurred significantly yet. This means H2PO4−H_2PO_4^- is the predominant species at pH 5.0, making D correct. Choice A (H3PO4H_3PO_4) is wrong because at pH 5.0, which is well above pKa₁, this fully protonated form has been almost entirely converted to its conjugate base. Choice B (PO43−PO_4^{3-}) is incorrect because this triply deprotonated form only becomes significant at very high pH values (well above pKa₃ = 12.38). Choice C (HPO42−HPO_4^{2-}) is wrong because this doubly deprotonated form doesn't become predominant until the pH approaches and exceeds pKa₂ (7.21). Remember: for polyprotic acids, identify which pKa values bracket your given pH, then determine which species is favored in that pH range. The species will be the one that has lost the appropriate number of protons for that pH region.

Question 8

The acid dissociation constant (KaK_a) for hydrocyanic acid (HCN) is 4.9×10−104.9 \times 10^{-10}. What is the base dissociation constant (KbK_b) for the cyanide ion (CN⁻) at 25°C? (Kw=1.0×10−14K_w = 1.0 \times 10^{-14})

  1. 2.0×10−112.0 \times 10^{-11}
  2. 2.0×10−52.0 \times 10^{-5} (correct answer)
  3. 4.9×10−44.9 \times 10^{-4}
  4. 4.9×10−104.9 \times 10^{-10}
Explanation: When you encounter questions about acid-base conjugate pairs, remember that acids and their conjugate bases are fundamentally linked through the water dissociation constant. This relationship allows you to calculate one equilibrium constant when you know the other. For any conjugate acid-base pair, the relationship is: Ka×Kb=KwK_a \times K_b = K_w. Here, HCN is the weak acid and CN⁻ is its conjugate base. To find KbK_b for CN⁻, you rearrange the equation: Kb=KwKaK_b = \frac{K_w}{K_a}. Substituting the given values: Kb=1.0×10−144.9×10−10=2.04×10−5K_b = \frac{1.0 \times 10^{-14}}{4.9 \times 10^{-10}} = 2.04 \times 10^{-5}, which rounds to 2.0×10−52.0 \times 10^{-5}. This confirms answer B is correct. Looking at the wrong answers: A (2.0×10−112.0 \times 10^{-11}) results from incorrectly multiplying KaK_a and KwK_w instead of dividing. C (4.9×10−44.9 \times 10^{-4}) comes from a calculation error, likely misplacing decimal points during division. D (4.9×10−104.9 \times 10^{-10}) simply repeats the given KaK_a value, showing confusion between the acid and base constants. Remember this key relationship: Ka×Kb=Kw=1.0×10−14K_a \times K_b = K_w = 1.0 \times 10^{-14} at 25°C. When you know one constant for a conjugate pair, you can always find the other by division. Strong acids have weak conjugate bases (small KbK_b), while weak acids like HCN have relatively stronger conjugate bases (larger KbK_b).

Question 9

Consider the following acids and their Ka values: HCN,Ka=4.9×10−10\text{HCN}, K_a = 4.9 \times 10^{-10}; CH3COOH,Ka=1.8×10−5\text{CH}_3\text{COOH}, K_a = 1.8 \times 10^{-5}; HNO2,Ka=4.5×10−4\text{HNO}_2, K_a = 4.5 \times 10^{-4}. Which of the following correctly ranks their conjugate bases (CN⁻, CH₃COO⁻, NO₂⁻) from most basic to least basic?

  1. CN⁻ > CH₃COO⁻ > NO₂⁻ (correct answer)
  2. NO₂⁻ > CH₃COO⁻ > CN⁻
  3. CH₃COO⁻ > NO₂⁻ > CN⁻
  4. CN⁻ > NO₂⁻ > CH₃COO⁻
Explanation: When you encounter acid-base equilibrium questions involving Ka values, remember that acid strength and conjugate base strength are inversely related. The weaker the acid, the stronger its conjugate base. To solve this, you need to identify which acid is weakest by comparing Ka values. A smaller Ka indicates a weaker acid because it dissociates less in solution. Looking at the given values: HCN has Ka=4.9×10−10K_a = 4.9 \times 10^{-10}, CH₃COOH has Ka=1.8×10−5K_a = 1.8 \times 10^{-5}, and HNO₂ has Ka=4.5×10−4K_a = 4.5 \times 10^{-4}. Since 10−10<10−5<10−410^{-10} < 10^{-5} < 10^{-4}, the acid strength order is: HCN (weakest) < CH₃COOH < HNO₂ (strongest). Because conjugate base strength is inversely related to acid strength, the conjugate base ranking from most basic to least basic is: CN⁻ > CH₃COO⁻ > NO₂⁻. Choice A correctly shows this relationship. Choice B reverses the entire order, suggesting that the strongest acid has the strongest conjugate base, which violates fundamental acid-base theory. Choice C places CH₃COO⁻ as most basic, incorrectly suggesting that CH₃COOH is the weakest acid. Choice D incorrectly positions NO₂⁻ as more basic than CH₃COO⁻, which would mean HNO₂ is weaker than CH₃COOH. Remember this key relationship: when comparing conjugate bases, always look for the acid with the smallest Ka value first—its conjugate base will be the strongest base in the group.

Question 10

A buffer solution is prepared by mixing 50 mL of 0.20 M acetic acid (CH3COOH\text{CH}_3\text{COOH}, Ka=1.8×10−5K_a = 1.8 \times 10^{-5}) with 50 mL of 0.10 M sodium acetate (CH3COONa\text{CH}_3\text{COONa}). What is the pH of this buffer? (pKa ≈ 4.74)

  1. 4.44 (correct answer)
  2. 4.74
  3. 5.04
  4. 5.34
Explanation: When you encounter buffer problems, you're dealing with a system that resists pH changes by containing both a weak acid and its conjugate base. The key tool here is the Henderson-Hasselbalch equation: pH=pKa+log⁡([A−][HA])\text{pH} = \text{pK}_a + \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right). First, calculate the final concentrations after mixing. When you combine 50 mL of 0.20 M acetic acid with 50 mL of 0.10 M sodium acetate, the total volume becomes 100 mL. The moles don't change, but concentrations are halved: acetic acid becomes 0.10 M and acetate becomes 0.050 M in the final solution. Now apply Henderson-Hasselbalch: pH=4.74+log⁡(0.0500.10)=4.74+log⁡(0.50)=4.74+(−0.30)=4.44\text{pH} = 4.74 + \log\left(\frac{0.050}{0.10}\right) = 4.74 + \log(0.50) = 4.74 + (-0.30) = 4.44. This confirms answer A is correct. Looking at the wrong answers: B (4.74) assumes equal concentrations of acid and base, ignoring that you actually have twice as much acid as base. C (5.04) likely results from incorrectly adding 0.30 instead of subtracting it from the pKa. D (5.34) might come from using the wrong ratio or calculation errors. Study tip: For buffer problems, always remember that Henderson-Hasselbalch requires you to use final concentrations after mixing, not initial ones. When the weak acid concentration exceeds the conjugate base concentration, the pH will be below the pKa—this relationship helps you check if your answer makes sense.

Question 11

In the reaction HSO4−(aq)+OH−(aq)⇌SO42−(aq)+H2O(l)\text{HSO}_4^-(aq) + \text{OH}^-(aq) \rightleftharpoons \text{SO}_4^{2-}(aq) + \text{H}_2\text{O}(l), which pair represents a Brønsted-Lowry acid and its conjugate base?

  1. OH−\text{OH}^- and SO42−\text{SO}_4^{2-}
  2. OH−\text{OH}^- and H2O\text{H}_2\text{O}
  3. HSO4−\text{HSO}_4^- and H2O\text{H}_2\text{O}
  4. HSO4−\text{HSO}_4^- and SO42−\text{SO}_4^{2-} (correct answer)
Explanation: When you encounter acid-base reactions, the key is identifying which species donates a proton (H⁺) and which accepts it. In Brønsted-Lowry theory, an acid and its conjugate base differ by exactly one proton. Looking at this reaction, HSO4−\text{HSO}_4^- donates a proton to OH−\text{OH}^-, making HSO4−\text{HSO}_4^- the acid. When it loses that proton, it becomes SO42−\text{SO}_4^{2-}, which is its conjugate base. The correct answer is D because HSO4−\text{HSO}_4^- and SO42−\text{SO}_4^{2-} form a conjugate acid-base pair. Choice A pairs OH−\text{OH}^- and SO42−\text{SO}_4^{2-}, but these aren't related by a single proton transfer - they have completely different chemical structures. Choice B suggests OH−\text{OH}^- and H2O\text{H}_2\text{O} are acid and conjugate base, but this has the relationship backward. OH−\text{OH}^- is actually the conjugate base of water, not the acid. Choice C pairs HSO4−\text{HSO}_4^- and H2O\text{H}_2\text{O}, but water is the conjugate acid of OH−\text{OH}^-, not the conjugate base of HSO4−\text{HSO}_4^-. To master these problems, always trace the proton movement. Find what loses H⁺ (the acid) and what it becomes after losing that proton (the conjugate base). These two species will have identical formulas except for one fewer hydrogen and one less positive charge in the conjugate base.

Question 12

When 25.0 mL of 0.10 M hydrofluoric acid (HF, a weak acid) is titrated with 0.10 M sodium hydroxide (NaOH, a strong base), what is the expected pH at the equivalence point?

  1. Equal to 7.0
  2. Less than 7.0
  3. Greater than 7.0 (correct answer)
  4. Equal to the pKa of HF
Explanation: When you encounter a weak acid-strong base titration, the key is recognizing that the equivalence point occurs when all the weak acid has been converted to its conjugate base, creating a basic solution. At the equivalence point, the HF has been completely neutralized by NaOH, forming sodium fluoride (NaF) and water: HF+NaOH→NaF+H2O\text{HF} + \text{NaOH} \rightarrow \text{NaF} + \text{H}_2\text{O}. The resulting solution contains F⁻ ions, which are the conjugate base of the weak acid HF. Since F⁻ is a weak base, it will hydrolyze water: F−+H2O⇌HF+OH−\text{F}^- + \text{H}_2\text{O} \rightleftharpoons \text{HF} + \text{OH}^-. This produces hydroxide ions, making the solution basic with a pH greater than 7.0. Looking at the incorrect answers: (A) suggests pH equals 7.0, which would only occur if you had a strong acid-strong base titration where neither ion hydrolyzes. (B) claims pH less than 7.0, which would happen in a strong acid-weak base titration where the conjugate acid of the weak base creates an acidic solution. (D) suggests the pH equals the pKa of HF, but the pKa represents the pH at the half-equivalence point (when half the acid is neutralized), not the equivalence point. Remember this pattern: weak acid + strong base always gives a basic equivalence point (pH > 7), while strong acid + weak base gives an acidic equivalence point (pH < 7). The "stronger" component determines the final pH character at equivalence.

Question 13

During the titration of a weak monoprotic acid with a strong base, the pH is found to be equal to the pKa of the acid. This observation occurs at which point in the titration?

  1. At the half-equivalence point (correct answer)
  2. At the starting point before any base is added
  3. At the equivalence point
  4. After the equivalence point has been passed
Explanation: When you encounter titration questions involving weak acids and strong bases, focus on what's happening to the acid-base equilibrium at different points during the titration. At the half-equivalence point, exactly half of the weak acid has been neutralized by the strong base. This creates a special situation where you have equal concentrations of the weak acid (HA) and its conjugate base (A⁻) in solution. Using the Henderson-Hasselbalch equation: pH=pKa+log⁡[A−][HA]pH = pK_a + \log\frac{[A^-]}{[HA]}. When [A⁻] = [HA], the log term equals zero, so pH = pK_a. This is why answer A is correct. Let's examine why the other options are wrong. Answer B suggests this occurs before any base is added, but initially you only have weak acid present, so pH would be much lower than pK_a. Answer C points to the equivalence point, where all acid has been converted to conjugate base. Here, the pH is determined by the hydrolysis of the conjugate base and will be greater than pK_a. Answer D suggests this happens after the equivalence point, but at this stage you have excess strong base, making the solution even more basic than at the equivalence point. Remember this key relationship: pH = pK_a always occurs at the half-equivalence point in weak acid-strong base titrations. This is also where the buffering capacity is maximized, making it a critical concept for both titration curves and buffer problems on the DAT.

Question 14

Which of the following binary hydrides is the most acidic?

  1. H₂O
  2. H₂Te (correct answer)
  3. H₂Se
  4. H₂S
Explanation: When you encounter questions about the acidity of binary hydrides, you need to consider how readily these compounds donate protons (H⁺) in solution. For hydrides in the same group of the periodic table, acidity increases as you move down the group due to weakening bonds and increasing atomic size. Looking at these Group 16 hydrides, the key factor is bond strength. As you move from oxygen to tellurium, the atoms become larger and the H-X bonds become longer and weaker. Weaker bonds break more easily, making it easier for the compound to release H⁺ ions and act as an acid. Choice B (H₂Te) is correct because tellurium is the largest atom here, creating the weakest H-Te bonds. This makes H₂Te the most willing to donate protons, hence the most acidic. Choice A (H₂O) is wrong because oxygen forms the strongest bonds with hydrogen due to its small size and high electronegativity. Water is actually amphoteric (can act as both acid and base) but is the weakest acid in this series. Choice C (H₂Se) is wrong because while selenium is larger than sulfur, making H₂Se more acidic than H₂S, it's still smaller than tellurium, so its bonds are stronger than those in H₂Te. Choice D (H₂S) is wrong for similar reasons - sulfur is smaller than both selenium and tellurium, creating stronger H-S bonds that are less likely to break. Remember: For hydrides in the same group, acidity increases as you go down the periodic table due to decreasing bond strength.

Question 15

A 10.0 mL sample of a concentrated HCl solution with a pH of 1.0 is diluted with pure water to a final volume of 1.00 L. What is the pH of the diluted solution?

  1. 1.0
  2. 2.0
  3. 3.0 (correct answer)
  4. 4.0
Explanation: When you encounter pH dilution problems, you're working with the relationship between hydrogen ion concentration and pH, combined with the dilution formula. Start by finding the initial [H+][H^+] concentration. Since pH = -log[H+][H^+], a pH of 1.0 means [H+]=10−1.0=0.10[H^+] = 10^{-1.0} = 0.10 M in the original 10.0 mL sample. Next, apply the dilution formula: M1V1=M2V2M_1V_1 = M_2V_2. Here, (0.10 M)(10.0 mL)=M2(1000 mL)(0.10 \text{ M})(10.0 \text{ mL}) = M_2(1000 \text{ mL}). Solving for M2M_2: M2=1.01000=0.001M_2 = \frac{1.0}{1000} = 0.001 M or 1.0×10−31.0 \times 10^{-3} M. Finally, convert back to pH: pH = -log(1.0×10−3)=3.0(1.0 \times 10^{-3}) = 3.0. Looking at the wrong answers: Choice A (1.0) assumes no dilution occurred, ignoring the 100-fold volume increase. Choice B (2.0) might result from incorrectly thinking that a 10-fold dilution changes pH by 1 unit, but the dilution here is 100-fold. Choice D (4.0) represents an error in calculating the dilution factor or pH conversion. The key insight is that each 10-fold dilution increases pH by exactly 1 unit for strong acids like HCl. Since you diluted 100-fold (10.0 mL to 1000 mL), the pH increases by 2 units: from 1.0 to 3.0. Study tip: For strong acid dilutions, remember that pH change equals the log of the dilution factor. A 100-fold dilution means pH increases by log(100) = 2 units.

Question 16

What is the effect on the pH and the percent ionization of a 0.1 M solution of hydrofluoric acid (HF) when solid sodium fluoride (NaF) is added?

  1. pH increases, percent ionization increases
  2. pH increases, percent ionization decreases (correct answer)
  3. pH decreases, percent ionization decreases
  4. pH decreases, percent ionization increases
Explanation: This question tests your understanding of the common ion effect and how it impacts weak acid equilibrium. When you see a problem about adding a salt that shares an ion with a weak acid, think about Le Châtelier's principle and how the equilibrium will shift. HF is a weak acid that establishes this equilibrium: HF+H2O⇌H3O++F−HF + H_2O \rightleftharpoons H_3O^+ + F^- When you add solid NaF, it completely dissociates to produce Na⁺ and F⁻ ions. The added F⁻ ions increase the concentration of fluoride in solution, which is a product of the HF ionization reaction. According to Le Châtelier's principle, the equilibrium shifts left to counteract this change, meaning less HF ionizes. With less ionization, fewer H₃O⁺ ions are produced, so the solution becomes less acidic and the pH increases. Simultaneously, since a smaller fraction of the original HF molecules ionize, the percent ionization decreases. This confirms answer B is correct. Let's examine why the other options are wrong: Answer A incorrectly suggests that percent ionization increases when it actually decreases due to the common ion effect. Answer C wrongly claims pH decreases - but adding the conjugate base of a weak acid makes the solution less acidic, not more acidic. Answer D makes both errors, suggesting pH decreases (wrong direction) and percent ionization increases (opposite of what the common ion effect causes). Remember: adding the conjugate base of a weak acid always suppresses ionization and increases pH. This common ion effect pattern appears frequently on chemistry exams.

Question 17

A 20.0 mL sample of 0.20 M HBr is titrated with 0.20 M KOH. What is the pH of the solution after 10.0 mL of KOH has been added?

  1. 1.18 (correct answer)
  2. 1.30
  3. 7.00
  4. 12.82
Explanation: When you encounter a titration problem, you need to determine what species remain in solution after the reaction and calculate the resulting pH based on those species. First, calculate the moles of each reactant: HBr = 0.020 L × 0.20 M = 0.004 mol, and KOH = 0.010 L × 0.20 M = 0.002 mol. Since HBr and KOH react in a 1:1 ratio (HBr + KOH → KBr + H₂O), the limiting reagent is KOH. After the reaction, 0.002 mol of HBr remains unreacted (0.004 - 0.002 = 0.002 mol). The total volume is now 30.0 mL (20.0 + 10.0), so the concentration of excess HBr is 0.002 mol0.030 L=0.0667 M\frac{0.002 \text{ mol}}{0.030 \text{ L}} = 0.0667 \text{ M} Since HBr is a strong acid that completely dissociates, [H⁺] = 0.0667 M. Therefore: pH=−log⁡(0.0667)=1.18\text{pH} = -\log(0.0667) = 1.18 Choice A (1.18) is correct. Choice B (1.30) might result from calculation errors or rounding mistakes. Choice C (7.00) represents the equivalence point pH, but you're only halfway to equivalence since you've added 10.0 mL of the 20.0 mL needed to neutralize all the acid. Choice D (12.82) would be the pH if you had excess base, which doesn't occur until after the equivalence point. For titration problems, always identify whether you're before, at, or after the equivalence point by comparing moles of acid and base added. The species that determines pH changes dramatically at each stage.

Question 18

Which of the following reactions best illustrates water acting as a Brønsted-Lowry base?

  1. H2CO3(aq)+H2O(l)⇌HCO3−(aq)+H3O+(aq)\text{H}_2\text{CO}_3(aq) + \text{H}_2\text{O}(l) \rightleftharpoons \text{HCO}_3^-(aq) + \text{H}_3\text{O}^+(aq) (correct answer)
  2. NH3(aq)+H2O(l)⇌NH4+(aq)+OH−(aq)\text{NH}_3(aq) + \text{H}_2\text{O}(l) \rightleftharpoons \text{NH}_4^+(aq) + \text{OH}^-(aq)
  3. 2H2O(l)⇌H3O+(aq)+OH−(aq)2\text{H}_2\text{O}(l) \rightleftharpoons \text{H}_3\text{O}^+(aq) + \text{OH}^-(aq)
  4. HCl(g)+NH3(g)→NH4Cl(s)\text{HCl}(g) + \text{NH}_3(g) \rightarrow \text{NH}_4\text{Cl}(s)
Explanation: When you encounter Brønsted-Lowry acid-base questions, focus on proton (H⁺) transfer. A Brønsted-Lowry base accepts a proton, while an acid donates one. In option A, carbonic acid (H2CO3\text{H}_2\text{CO}_3) donates a proton to water. Water accepts this proton and becomes H3O+\text{H}_3\text{O}^+, demonstrating its role as a proton acceptor (base). This is exactly what the question asks for. Option B shows ammonia (NH3\text{NH}_3) accepting a proton from water. Here, water acts as the proton donor (acid), not the base. Water loses a proton to become OH−\text{OH}^-, while ammonia gains one to become NH4+\text{NH}_4^+. Option C represents water's autoionization, where one water molecule acts as an acid and another as a base simultaneously. Since the question asks for water specifically acting as a base, this doesn't clearly illustrate the concept. Option D involves no water molecules at all—it's a gas-phase reaction between HCl and ammonia forming solid ammonium chloride. This eliminates it immediately. The key insight is tracking proton movement: in A, the proton flows from H2CO3\text{H}_2\text{CO}_3 to H2O\text{H}_2\text{O}, making water the proton acceptor (base). Study tip: Always identify the proton donor and acceptor by looking for where H⁺ goes. The species gaining the proton is the Brønsted-Lowry base. Draw arrows showing proton transfer if it helps visualize the process.

Question 19

What is the approximate pH of a 0.10 M solution of formic acid (HCOOH), given that its acid dissociation constant (KaK_a) is 1.6×10−41.6 \times 10^{-4}?

  1. 1.00
  2. 2.40 (correct answer)
  3. 3.80
  4. 4.60
Explanation: When you encounter a weak acid pH calculation, you're dealing with partial dissociation where you need to use the acid dissociation constant (KaK_a) and an ICE table approach. For formic acid (HCOOH), set up the equilibrium: HCOOH ⇌ H⁺ + HCOO⁻. Since Ka=1.6×10−4K_a = 1.6 \times 10^{-4} is relatively large for a weak acid, you can't ignore the amount that dissociates. Using the quadratic formula or successive approximations: Ka=[H+][HCOO−][HCOOH]=x20.10−x=1.6×10−4K_a = \frac{[H^+][HCOO^-]}{[HCOOH]} = \frac{x^2}{0.10-x} = 1.6 \times 10^{-4} Solving this gives x = [H⁺] ≈ 0.0038 M, so pH = -log(0.0038) = 2.40. Choice A (1.00) represents the pH if this were a strong acid completely dissociating: pH = -log(0.10) = 1.00. This ignores that formic acid is weak and only partially ionizes. Choice C (3.80) might result from incorrectly assuming the acid is much weaker than it actually is, or from calculation errors in the equilibrium expression. Choice D (4.60) could come from mistakenly calculating -log(KaK_a) instead of solving the equilibrium problem, or from assuming the acid is extremely weak and using inappropriate approximations. Study tip: For weak acid pH problems, always check if KaK_a is large enough (> 10⁻⁵) that you need the full quadratic treatment rather than simplifying assumptions. Also remember that weak acid pH values fall between what you'd get for a strong acid and pure water.

Question 20

Each of the following can act as a Lewis acid EXCEPT one. Which one is the EXCEPTION?

  1. BF₃
  2. AlCl₃
  3. NH₃ (correct answer)
  4. Fe³⁺
Explanation: Lewis acids are electron pair acceptors - they can receive electrons from other molecules or ions. When you encounter Lewis acid/base questions, focus on electron availability and bonding capacity rather than just proton transfer. Let's examine each molecule's ability to accept electron pairs. BF₃ (choice A) is a classic Lewis acid because boron has only six valence electrons around it, leaving room to accept an electron pair to complete its octet. AlCl₃ (choice B) works similarly - aluminum can expand its valence shell and readily accepts electron pairs, making it a strong Lewis acid. Fe³⁺ (choice D) is a metal cation with empty d orbitals that can accept electron pairs from ligands, which is fundamental to coordination chemistry. NH₃ (choice C) is fundamentally different. Nitrogen has a complete octet and possesses a lone pair of electrons on the nitrogen atom. Rather than accepting electrons, NH₃ donates its lone pair to other species, making it a Lewis base, not a Lewis acid. This is why ammonia forms coordinate covalent bonds by donating electrons to electron-deficient species. The key distinction is that choices A, B, and D all have electron deficiencies or vacant orbitals that can accommodate additional electron pairs, while choice C has excess electrons available for donation. Remember this pattern: Lewis acids typically involve atoms with incomplete octets, expandable valence shells, or positive charges that create electron deficiency. Species with lone pairs like NH₃, H₂O, and halide ions are usually Lewis bases.