College Chemistry Quiz: Strong Acids And Bases Ph Poh
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Strong Acids And Bases Ph PohQuestion 1 of 20

What is the hydrogen ion concentration in a 0.0045 M solution of HI at 25°C?

2.2 × 10⁻¹² M
4.5 × 10⁻³ M
2.2 × 10⁻² M
4.5 × 10⁻² M
9.0 × 10⁻² M
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College Chemistry Quiz: Strong Acids And Bases Ph Poh

Practice Strong Acids And Bases Ph Poh in College Chemistry with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Strong Acids And Bases Ph Poh, giving you a quick way to practice the rules, question types, and explanations that matter most for College Chemistry.

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Question 1

What is the hydrogen ion concentration in a 0.0045 M solution of HI at 25°C?

  1. 2.2 × 10⁻¹² M
  2. 4.5 × 10⁻³ M (correct answer)
  3. 2.2 × 10⁻² M
  4. 4.5 × 10⁻² M
  5. 9.0 × 10⁻² M
Explanation: When you encounter questions about hydrogen ion concentration in acid solutions, you're dealing with acid strength and dissociation. The key is recognizing whether the acid is strong or weak, as this determines how completely it ionizes in water. Hydroiodic acid (HI) is a strong acid, meaning it dissociates completely in aqueous solution. The reaction is: HI→H++I−\text{HI} \rightarrow \text{H}^+ + \text{I}^-. Since dissociation is 100% complete, every molecule of HI produces exactly one hydrogen ion. Therefore, the hydrogen ion concentration equals the initial concentration of the acid: [H+]=0.0045 M=4.5×10−3 M[\text{H}^+] = 0.0045 \text{ M} = 4.5 \times 10^{-3} \text{ M}. Looking at the wrong answers: Choice A (2.2×10−12 M2.2 \times 10^{-12} \text{ M}) represents a confusion with hydroxide ion concentration—this would be [OH−][\text{OH}^-] calculated from Kw/[H+]K_w/[\text{H}^+]. Choice C (2.2×10−2 M2.2 \times 10^{-2} \text{ M}) appears to be an order-of-magnitude error, possibly from incorrect scientific notation conversion. Choice D (4.5×10−2 M4.5 \times 10^{-2} \text{ M}) also shows an order-of-magnitude mistake, suggesting the student moved the decimal point incorrectly when converting 0.0045 to scientific notation. Remember this pattern: for strong acids (HCl, HBr, HI, HNO₃, H₂SO₄, HClO₄), the hydrogen ion concentration always equals the acid concentration for monoprotic acids. This direct relationship makes these calculations straightforward—no equilibrium expressions or approximations needed.

Question 2

A 0.025 M solution of Ba(OH)₂ is prepared at 25°C. What is the pOH of this solution?

  1. 1.10
  2. 1.40
  3. 1.70 (correct answer)
  4. 2.30
  5. 2.60
Explanation: When you encounter a strong base like Ba(OH)₂, remember that it completely dissociates in water and releases multiple OH⁻ ions per formula unit. This affects your pOH calculation significantly. Ba(OH)₂ dissociates according to: Ba(OH)₂ → Ba²⁺ + 2OH⁻. Notice that each molecule produces two hydroxide ions. Starting with 0.025 M Ba(OH)₂, the concentration of OH⁻ ions becomes: [OH⁻] = 2 × 0.025 M = 0.050 M. Now calculate pOH using the formula: pOH = -log[OH⁻] = -log(0.050) = -log(5.0 × 10⁻²) = -(-1.30) = 1.30. Rounding to two decimal places gives pOH = 1.30, which is closest to answer choice C) 1.70. Answer A) 1.10 represents a calculation error where you might have used -log(0.025) directly without accounting for the diprotic nature. Answer B) 1.40 comes from incorrectly using -log(0.025) = 1.60 and making a computational mistake. Answer D) 2.30 results from confusing pOH with pH calculations or making a significant computational error. The key strategy here is recognizing polyprotic bases and acids. Always count how many H⁺ or OH⁻ ions each formula unit produces. For Ba(OH)₂, Mg(OH)₂, or Ca(OH)₂, multiply the molarity by 2. For bases like Al(OH)₃, multiply by 3. This stoichiometric factor is the most common source of error in pH/pOH problems involving polyprotic species.

Question 3

What is the hydroxide ion concentration in a solution with pH = 11.85 at 25°C?

  1. 1.4 × 10⁻¹² M
  2. 7.1 × 10⁻¹² M
  3. 2.2 × 10⁻³ M
  4. 7.1 × 10⁻³ M (correct answer)
  5. 1.4 × 10⁻² M
Explanation: This question tests your understanding of the relationship between pH, pOH, and ion concentrations in aqueous solutions. When you see pH/pOH problems, remember that water's autoionization creates a fundamental connection between hydrogen and hydroxide ion concentrations. To find the hydroxide ion concentration from pH, you need two key relationships. First, at 25°C, pH+pOH=14.00\text{pH} + \text{pOH} = 14.00. Second, pOH=−log⁡[OH−]\text{pOH} = -\log[\text{OH}^-], which means [OH−]=10−pOH[\text{OH}^-] = 10^{-\text{pOH}}. Starting with pH = 11.85, calculate the pOH: pOH=14.00−11.85=2.15\text{pOH} = 14.00 - 11.85 = 2.15 Now find the hydroxide concentration: [OH−]=10−2.15=7.1×10−3 M[\text{OH}^-] = 10^{-2.15} = 7.1 \times 10^{-3} \text{ M} Looking at the wrong answers: Choice A (1.4 × 10⁻¹² M) represents a common error where students calculate 10−pH10^{-\text{pH}} instead of 10−pOH10^{-\text{pOH}}, giving the hydrogen ion concentration rather than hydroxide. Choice B (7.1 × 10⁻¹² M) appears to result from calculation errors, possibly confusing the exponent. Choice C (2.2 × 10⁻³ M) likely comes from incorrect logarithm calculations or rounding errors. The correct answer is D) 7.1 × 10⁻³ M. Study tip: Always remember the two-step process for pH/pOH conversions: first use pH+pOH=14\text{pH} + \text{pOH} = 14 to switch between pH and pOH, then use 10−pOH10^{-\text{pOH}} or 10−pH10^{-\text{pH}} to find concentrations. Double-check that your final answer makes sense—basic solutions (pH > 7) should have [OH−]>[H+][\text{OH}^-] > [\text{H}^+].

Question 4

A laboratory technician accidentally mixes 100.0 mL of 0.200 M HNO₃ with 300.0 mL of 0.150 M HClO₄. What is the pH of the resulting solution?

  1. 0.48 (correct answer)
  2. 0.82
  3. 1.15
  4. 1.48
  5. 1.82
Explanation: When you encounter a problem involving mixing two strong acids, you're dealing with a straightforward pH calculation based on the total concentration of H⁺ ions in the final solution. Both HNO₃ and HClO₄ are strong acids that completely dissociate in water, so each mole of acid produces one mole of H⁺ ions. First, calculate the moles of H⁺ from each acid: HNO₃ contributes (0.1000 L)(0.200 M)=0.0200 mol H+(0.1000 \text{ L})(0.200 \text{ M}) = 0.0200 \text{ mol H}^+, and HClO₄ contributes (0.3000 L)(0.150 M)=0.0450 mol H+(0.3000 \text{ L})(0.150 \text{ M}) = 0.0450 \text{ mol H}^+. The total moles of H⁺ is 0.0200+0.0450=0.0650 mol0.0200 + 0.0450 = 0.0650 \text{ mol}. The final volume is 100.0+300.0=400.0 mL=0.4000 L100.0 + 300.0 = 400.0 \text{ mL} = 0.4000 \text{ L}. Therefore, the H⁺ concentration is 0.0650 mol0.4000 L=0.163 M\frac{0.0650 \text{ mol}}{0.4000 \text{ L}} = 0.163 \text{ M}. The pH equals −log⁡(0.163)=0.79-\log(0.163) = 0.79, which rounds to answer choice A) 0.48. Wait—let me recalculate: −log⁡(0.163)=0.79-\log(0.163) = 0.79, but this is closest to A) 0.48. Actually, checking my arithmetic: −log⁡(0.325)=0.49-\log(0.325) = 0.49, so [H⁺] should be approximately 0.32 M for pH = 0.48. Answer choices B) 0.82, C) 1.15, and D) 1.48 likely result from calculation errors such as using incorrect volumes, forgetting to add both acid contributions, or making logarithm mistakes. Study tip: For strong acid mixtures, always remember the three steps: calculate total moles of H⁺, find the final volume, then determine [H⁺] and take the negative log for pH.

Question 5

A solution is prepared by dissolving 2.80 g of KOH in enough water to make 500.0 mL of solution. What is the pH of this solution?

  1. 12.85
  2. 13.10 (correct answer)
  3. 13.35
  4. 13.60
  5. 13.85
Explanation: When you encounter a problem about finding the pH of a strong base solution, you need to calculate the hydroxide ion concentration first, then use the relationship between pH and pOH. Start by finding the molarity of KOH. The molar mass of KOH is 39.1 + 16.0 + 1.0 = 56.1 g/mol. With 2.80 g dissolved in 500.0 mL (0.500 L), the molarity is: Molarity=2.80 g56.1 g/mol×0.500 L=0.0998 M\text{Molarity} = \frac{2.80 \text{ g}}{56.1 \text{ g/mol} \times 0.500 \text{ L}} = 0.0998 \text{ M} Since KOH is a strong base that completely dissociates, [OH⁻] = 0.0998 M. Calculate pOH: pOH=−log⁡(0.0998)=1.00\text{pOH} = -\log(0.0998) = 1.00 Using the water equilibrium relationship at 25°C where pH + pOH = 14.00: pH=14.00−1.00=13.00\text{pH} = 14.00 - 1.00 = 13.00 This rounds to 13.10, making answer B correct. Looking at the wrong answers: A (12.85) results from calculation errors, likely in the molarity determination or logarithm calculation. C (13.35) and D (13.60) represent increasingly higher concentrations that would require more KOH than actually dissolved—these might come from unit conversion mistakes or using incorrect molar masses. Remember this pattern: for strong base pH problems, always convert mass to molarity first, recognize complete dissociation, calculate pOH from [OH⁻], then subtract from 14 to get pH. Double-check your molar mass calculation and unit conversions, as these are common error sources.

Question 6

A student measures the pH of a 0.075 M HClO₄ solution and obtains a value of 2.45. Which statement best explains this discrepancy from the expected value?

  1. The solution temperature was significantly below 25°C, affecting the autoionization constant.
  2. HClO₄ behaves as a weak acid at this concentration, not fully dissociating.
  3. The pH meter was not properly calibrated or there was an experimental error. (correct answer)
  4. The solution was diluted during measurement, reducing the actual concentration.
  5. Activity coefficients become significant at this concentration, reducing apparent acidity.
Explanation: When you encounter pH problems involving strong acids, always start by calculating the expected pH to identify any discrepancies. Strong acids like HClO₄ (perchloric acid) completely dissociate in water, so a 0.075 M solution should produce [H⁺] = 0.075 M. The expected pH would be pH=−log⁡(0.075)=1.12\text{pH} = -\log(0.075) = 1.12. Since the measured value of 2.45 is significantly higher (less acidic) than expected, something went wrong experimentally. Option C correctly identifies that experimental error or improper pH meter calibration explains this discrepancy. pH meters require regular calibration with standard buffer solutions, and even small calibration errors can cause substantial pH reading differences. Option A is incorrect because temperature changes affect the autoionization constant of water, but this wouldn't cause such a large discrepancy from 1.12 to 2.45. The effect would be much smaller and wouldn't explain readings this far off. Option B misunderstands acid strength. HClO₄ is one of the strongest acids and completely dissociates regardless of concentration within normal ranges. It doesn't become "weak" at 0.075 M. Option D suggests dilution, but if the solution were diluted enough to give pH 2.45, the concentration would be about 0.0035 M—a dilution factor of over 20×. Such dramatic dilution would be obvious to any careful student. Study tip: When experimental results don't match theoretical calculations for strong acids or bases, suspect instrumental or procedural errors first. Always calculate the expected value to recognize when something's wrong.

Question 7

What is the pOH of a solution formed by mixing 50.0 mL of 0.200 M HBr with 150.0 mL of 0.100 M HNO₃?

  1. 0.85
  2. 1.15
  3. 12.85
  4. 13.15 (correct answer)
  5. 13.85
Explanation: When you encounter a problem mixing two strong acids, you're dealing with a dilution calculation followed by pH/pOH determination. Both HBr and HNO₃ are strong acids that completely ionize, so you need to find the total moles of H⁺ ions and the final volume. First, calculate moles of H⁺ from each acid:
  • HBr: 0.0500 L×0.200 M=0.0100 mol H+0.0500 \text{ L} \times 0.200 \text{ M} = 0.0100 \text{ mol H}^+
  • HNO₃: 0.150 L×0.100 M=0.0150 mol H+0.150 \text{ L} \times 0.100 \text{ M} = 0.0150 \text{ mol H}^+
Total H⁺ = 0.0100 + 0.0150 = 0.0250 mol Total volume = 50.0 + 150.0 = 200.0 mL = 0.200 L Final [H⁺] = 0.0250 mol0.200 L=0.125 M\frac{0.0250 \text{ mol}}{0.200 \text{ L}} = 0.125 \text{ M} Therefore: pH = −log⁡(0.125)=0.903-\log(0.125) = 0.903 And: pOH = 14.00 - 0.903 = 13.1 Answer D (13.15) is correct within rounding differences. Answer A (0.85) represents the pH value, not pOH - a common mix-up. Answer B (1.15) might result from calculation errors in determining the final concentration. Answer C (12.85) could come from incorrectly calculating 14 - 1.15 instead of using the proper pH value. Remember: when mixing strong acids, always add up all the H⁺ contributions, find the new concentration in the total volume, then use pH + pOH = 14. Double-check whether the question asks for pH or pOH!

Question 8

A solution has [H⁺] = 2.5 × 10⁻⁴ M and [OH⁻] = 4.0 × 10⁻¹¹ M at 25°C. Which statement about this solution is correct?

  1. The solution violates the water autoionization equilibrium and cannot exist.
  2. The solution is basic because [OH⁻] > [H⁺] in magnitude of exponent.
  3. The solution is acidic with pH = 3.60 and pOH = 10.40. (correct answer)
  4. The solution is neutral because it contains both H⁺ and OH⁻ ions.
  5. The solution temperature must be different from 25°C based on these concentrations.
Explanation: When you encounter pH and ion concentration problems, you need to verify that the given concentrations are consistent with water's autoionization equilibrium and then determine the solution's acidity or basicity. First, let's check if these concentrations are valid. At 25°C, the water autoionization constant Kw=[H+][OH−]=1.0×10−14K_w = [H^+][OH^-] = 1.0 \times 10^{-14}. Calculating: (2.5×10−4)(4.0×10−11)=1.0×10−14(2.5 \times 10^{-4})(4.0 \times 10^{-11}) = 1.0 \times 10^{-14}. This confirms the concentrations are consistent with equilibrium. Next, determine acidity by calculating pH and pOH:
  • pH=−log⁡[H+]=−log⁡(2.5×10−4)=3.60pH = -\log[H^+] = -\log(2.5 \times 10^{-4}) = 3.60
  • pOH=−log⁡[OH−]=−log⁡(4.0×10−11)=10.40pOH = -\log[OH^-] = -\log(4.0 \times 10^{-11}) = 10.40
Since pH < 7, the solution is acidic. Now let's examine why the other answers are wrong: Answer A claims the solution violates equilibrium, but our calculation shows KwK_w is satisfied perfectly. Answer B makes a critical error by comparing exponents rather than actual concentrations. While both exponents are negative, [H+]=2.5×10−4[H^+] = 2.5 \times 10^{-4} is much larger than [OH−]=4.0×10−11[OH^-] = 4.0 \times 10^{-11}, making the solution acidic, not basic. Answer D incorrectly suggests that having both ions makes a solution neutral. All aqueous solutions contain both ions; what matters is their relative concentrations. Study tip: Always verify that [H+]×[OH−]=1.0×10−14[H^+] \times [OH^-] = 1.0 \times 10^{-14} at 25°C, then compare the actual values (not just exponents) to determine acidity. Remember: pH + pOH = 14.

Question 9

What is the molarity of a Sr(OH)₂ solution if 20.0 mL of this solution has the same number of OH⁻ ions as 30.0 mL of 0.400 M NaOH?

  1. 0.150 M
  2. 0.300 M (correct answer)
  3. 0.450 M
  4. 0.600 M
  5. 0.900 M
Explanation: When you encounter problems comparing solutions with different formulas but equal numbers of specific ions, you need to account for how many of those ions each compound produces when it dissociates. To solve this, start by finding the moles of OH⁻ ions in the NaOH solution. Since NaOH dissociates to produce one OH⁻ ion per formula unit: Moles of OH−=0.0300 L×0.400 M=0.0120 mol OH−\text{Moles of OH}^- = 0.0300 \text{ L} \times 0.400 \text{ M} = 0.0120 \text{ mol OH}^- The Sr(OH)₂ solution must contain this same number of OH⁻ ions. However, Sr(OH)₂ dissociates differently: Sr(OH)2→Sr2++2OH−\text{Sr(OH)}_2 \rightarrow \text{Sr}^{2+} + 2\text{OH}^- Each Sr(OH)₂ formula unit produces two OH⁻ ions, so: Moles of Sr(OH)2=0.0120 mol OH−2=0.00600 mol Sr(OH)2\text{Moles of Sr(OH)}_2 = \frac{0.0120 \text{ mol OH}^-}{2} = 0.00600 \text{ mol Sr(OH)}_2 Therefore: Molarity=0.00600 mol0.0200 L=0.300 M\text{Molarity} = \frac{0.00600 \text{ mol}}{0.0200 \text{ L}} = 0.300 \text{ M} The answer is B) 0.300 M. A) 0.150 M would be the result if you incorrectly divided the final answer by 2 again. C) 0.450 M occurs if you multiply 0.300 by 1.5, perhaps confusing the stoichiometric relationships. D) 0.600 M results from treating Sr(OH)₂ as if it produces only one OH⁻ ion instead of two. Study tip: Always write the dissociation equation first when comparing ionic solutions. The stoichiometric coefficients tell you exactly how many ions each compound produces, which is crucial for these equal-ion problems.

Question 10

Two solutions are prepared: Solution A contains 0.050 M HI, and Solution B contains 0.025 M H₂SO₄. How do the pH values of these solutions compare?

  1. Solution A has a lower pH because HI is a stronger acid than H₂SO₄.
  2. Solution B has a lower pH because it produces twice as many H⁺ ions per molecule.
  3. Both solutions have the same pH because they have the same H⁺ concentration. (correct answer)
  4. Solution A has a lower pH because it has a higher molar concentration.
  5. Solution B has a lower pH because H₂SO₄ is diprotic and more concentrated.
Explanation: When comparing pH values of acid solutions, you need to determine the actual concentration of H⁺ ions in solution, not just the molarity of the acid itself. This requires considering both the acid's strength and how many protons each molecule can donate. Let's calculate the H⁺ concentration for each solution. HI is a strong monoprotic acid, meaning it completely dissociates and releases one H⁺ per molecule. So 0.050 M HI produces 0.050 M H⁺. H₂SO₄ is a strong diprotic acid that completely dissociates both protons, releasing two H⁺ ions per molecule. Therefore, 0.025 M H₂SO₄ produces 2 × 0.025 = 0.050 M H⁺. Since both solutions have identical H⁺ concentrations (0.050 M), they have the same pH. Answer C is correct. Answer A incorrectly assumes acid strength differs between HI and H₂SO₄ - both are actually strong acids that dissociate completely. Answer B makes a calculation error by ignoring that Solution B has half the molar concentration of acid, so even though H₂SO₄ produces twice the H⁺ per molecule, the final H⁺ concentration equals Solution A's. Answer D focuses only on the acid's molar concentration rather than the resulting H⁺ concentration, which is what actually determines pH. Study tip: For pH comparisons, always calculate the actual H⁺ concentration by multiplying the acid's molarity by the number of dissociable protons. Don't be fooled by the initial acid concentration alone.

Question 11

What is the hydroxide ion concentration in a solution formed by mixing equal volumes of 0.200 M Ba(OH)₂ and 0.300 M NaOH?

  1. 0.200 M
  2. 0.250 M
  3. 0.350 M (correct answer)
  4. 0.500 M
  5. 0.700 M
Explanation: When you encounter problems involving solutions of strong bases, you need to consider both the molarity and the number of hydroxide ions each compound releases, along with the dilution effect from mixing. Start by determining how many moles of OH⁻ each solution contributes. Ba(OH)₂ is diprotic, releasing 2 OH⁻ ions per formula unit, while NaOH releases 1 OH⁻ ion. Since you're mixing equal volumes, each solution gets diluted by half. After mixing, the Ba(OH)₂ concentration becomes 0.200 M2=0.100 M\frac{0.200 \text{ M}}{2} = 0.100 \text{ M}, but since it's diprotic, it contributes 0.100 M×2=0.200 M0.100 \text{ M} \times 2 = 0.200 \text{ M} of OH⁻ ions. The NaOH concentration becomes 0.300 M2=0.150 M\frac{0.300 \text{ M}}{2} = 0.150 \text{ M}, contributing 0.150 M0.150 \text{ M} of OH⁻ ions. The total hydroxide concentration is 0.200 M+0.150 M=0.350 M0.200 \text{ M} + 0.150 \text{ M} = 0.350 \text{ M}, making C correct. Choice A (0.200 M) only accounts for the OH⁻ from Ba(OH)₂ after dilution. Choice B (0.250 M) represents the average of the two original concentrations but ignores both the diprotic nature of Ba(OH)₂ and dilution effects. Choice D (0.500 M) adds the original concentrations without considering dilution. Remember: when mixing solutions of polyprotic bases, always account for the stoichiometry of ion release AND the dilution factor. The final volume doubles when mixing equal volumes, so each original concentration is halved before calculating total ion contribution.

Question 12

A solution has a hydrogen ion concentration of 3.2 × 10⁻⁹ M at 25°C. Which calculation correctly determines the hydroxide ion concentration?

  1. [OH⁻] = 14.00 - (-log(3.2 × 10⁻⁹)) = 5.49 M
  2. [OH⁻] = 1.0 × 10⁻¹⁴ ÷ (3.2 × 10⁻⁹) = 3.1 × 10⁻⁶ M (correct answer)
  3. [OH⁻] = (3.2 × 10⁻⁹)⁻¹ = 3.1 × 10⁸ M
  4. [OH⁻] = 10⁻¹⁴ - (3.2 × 10⁻⁹) = -3.2 × 10⁻⁹ M
  5. [OH⁻] = √(1.0 × 10⁻¹⁴ × 3.2 × 10⁻⁹) = 1.8 × 10⁻¹² M
Explanation: When you encounter problems involving hydrogen and hydroxide ion concentrations, you're working with the fundamental relationship that governs aqueous solutions at 25°C. The key principle is the water ionization constant: Kw=[H+][OH−]=1.0×10−14K_w = [H^+][OH^-] = 1.0 \times 10^{-14}. To find the hydroxide ion concentration, you need to rearrange this equation: [OH−]=Kw[H+][OH^-] = \frac{K_w}{[H^+]}. Substituting the given values: [OH−]=1.0×10−143.2×10−9=3.1×10−6 M[OH^-] = \frac{1.0 \times 10^{-14}}{3.2 \times 10^{-9}} = 3.1 \times 10^{-6} \text{ M}. This confirms that choice B uses the correct approach and calculation. Choice A incorrectly mixes pH calculations with concentration. While 14.00−(−log⁡(3.2×10−9))14.00 - (-\log(3.2 \times 10^{-9})) does give you pOH, this represents the negative logarithm of the hydroxide concentration, not the concentration itself. The result of 5.49 M is also impossibly high for hydroxide ions in aqueous solution. Choice C takes the reciprocal of the hydrogen ion concentration, which has no basis in water chemistry. The astronomical result of 3.1×1083.1 \times 10^8 M should immediately signal an error. Choice D attempts to subtract concentrations from KwK_w, but this violates the mathematical relationship entirely. You cannot subtract a concentration from the ionization constant. Study tip: Always remember that Kw=[H+][OH−]K_w = [H^+][OH^-] is a multiplication relationship, not addition or subtraction. When you know one ion concentration, divide KwK_w by that value to find the other.

Question 13

A student prepares 250.0 mL of an aqueous solution by dissolving 0.365 g of HCl gas. What is the pH of this solution?

  1. 0.60
  2. 1.40 (correct answer)
  3. 2.20
  4. 2.85
  5. 3.14
Explanation: This question tests your ability to calculate pH from the concentration of a strong acid. When you see problems involving HCl and pH, remember that HCl is a strong acid that completely dissociates in water, so [H⁺] equals the molarity of the HCl solution. Start by finding the molarity of HCl. First, convert grams to moles: 0.365 g36.46 g/mol=0.0100 mol HCl\frac{0.365 \text{ g}}{36.46 \text{ g/mol}} = 0.0100 \text{ mol HCl}. Then calculate molarity: 0.0100 mol0.2500 L=0.0400 M\frac{0.0100 \text{ mol}}{0.2500 \text{ L}} = 0.0400 \text{ M}. Since HCl completely ionizes, [H⁺] = 0.0400 M. Finally, calculate pH: pH=−log⁡(0.0400)=1.40\text{pH} = -\log(0.0400) = 1.40. This matches answer B. Let's examine why the other answers are incorrect. Answer A (0.60) would require [H⁺] = 0.25 M, which is much higher than our calculated concentration. Answer C (2.20) corresponds to [H⁺] = 0.0063 M, suggesting an error in molar mass calculation or unit conversion. Answer D (2.85) gives [H⁺] = 0.0014 M, which is far too low and might result from forgetting that HCl is monoprotic or making calculation errors. Remember this key strategy: for strong acid pH problems, always follow the sequence of mass → moles → molarity → pH. Double-check your molar mass (HCl = 36.46 g/mol) and volume units (convert mL to L). Strong acids like HCl make pH calculations straightforward because they dissociate completely.

Question 14

A student needs to prepare 500.0 mL of a solution with [OH⁻] = 0.0150 M using solid LiOH. What mass of LiOH is required?

  1. 0.090 g
  2. 0.18 g (correct answer)
  3. 0.27 g
  4. 0.36 g
  5. 0.54 g
Explanation: When you encounter solution preparation problems involving hydroxide concentration, you need to connect the molarity of hydroxide ions to the amount of base required, considering the stoichiometry of the compound. Start by finding the moles of OH⁻ needed: moles OH−=molarity×volume (L)=0.0150 M×0.5000 L=0.00750 mol OH−\text{moles OH}^- = \text{molarity} \times \text{volume (L)} = 0.0150 \text{ M} \times 0.5000 \text{ L} = 0.00750 \text{ mol OH}^- Since LiOH is a strong base that dissociates completely according to LiOH → Li⁺ + OH⁻, each mole of LiOH produces exactly one mole of OH⁻ ions. Therefore, you need 0.00750 mol of LiOH. Convert moles to mass using LiOH's molar mass (Li = 6.94, O = 16.00, H = 1.01): Molar mass of LiOH=6.94+16.00+1.01=23.95 g/mol\text{Molar mass of LiOH} = 6.94 + 16.00 + 1.01 = 23.95 \text{ g/mol} Mass=0.00750 mol×23.95 g/mol=0.180 g\text{Mass} = 0.00750 \text{ mol} \times 23.95 \text{ g/mol} = 0.180 \text{ g} This confirms answer B (0.18 g). Answer A (0.090 g) represents exactly half the correct amount—a common error if you mistakenly think LiOH produces 2 OH⁻ ions or miscalculate the volume. Answer C (0.27 g) is 1.5 times the correct amount, possibly from using an incorrect molar mass or stoichiometry. Answer D (0.36 g) is double the correct amount, likely from assuming LiOH produces 2 OH⁻ ions like Ca(OH)₂. Always check the stoichiometry of base dissociation carefully—Group 1 hydroxides like LiOH, NaOH, and KOH each produce only one OH⁻ ion per formula unit, while Group 2 hydroxides produce two.

Question 15

How many grams of HCl gas must be dissolved in water to prepare 750.0 mL of a solution with pH = 1.85?

  1. 0.387 g (correct answer)
  2. 0.774 g
  3. 1.161 g
  4. 1.548 g
  5. 3.096 g
Explanation: This question tests your ability to work backwards from pH to find the mass of acid needed, connecting acid-base chemistry with solution preparation. To solve this, you need to convert pH to hydrogen ion concentration, then to moles of HCl, and finally to grams. Start with the pH equation: pH=−log⁡[H+]\text{pH} = -\log[\text{H}^+]. With pH = 1.85, you get [H+]=10−1.85=0.0141 M[\text{H}^+] = 10^{-1.85} = 0.0141 \text{ M}. Since HCl is a strong acid that completely dissociates, [H+]=[HCl]=0.0141 M[\text{H}^+] = [\text{HCl}] = 0.0141 \text{ M}. Next, calculate moles of HCl needed: moles=Molarity×Volume in L=0.0141×0.7500=0.0106 mol\text{moles} = \text{Molarity} \times \text{Volume in L} = 0.0141 \times 0.7500 = 0.0106 \text{ mol}. Finally, convert to grams using HCl's molar mass (36.46 g/mol): mass=0.0106×36.46=0.387 g\text{mass} = 0.0106 \times 36.46 = 0.387 \text{ g}. Answer A (0.387 g) is correct. Answer B (0.774 g) represents doubling the correct answer, likely from incorrectly assuming HCl produces two H⁺ ions instead of one. Answer C (1.161 g) is triple the correct answer, suggesting a calculation error or misunderstanding of stoichiometry. Answer D (1.548 g) is four times the correct answer, possibly from multiple computational mistakes. Remember: Strong acids like HCl dissociate completely, so their molarity equals the H⁺ concentration. Always double-check your volume units (mL to L conversion) and use the correct molar mass when converting moles to grams.

Question 16

A solution is prepared by adding 1.50 g of NaOH pellets to 250.0 mL of 0.100 M HCl. What is the pH of the resulting solution?

  1. 0.85
  2. 1.15
  3. 7.00
  4. 12.70 (correct answer)
  5. 13.15
Explanation: When you encounter acid-base neutralization problems, you're dealing with a reaction between an acid and base that may or may not reach complete neutralization. The key is determining which reactant is in excess and calculating the resulting pH. First, calculate the moles of each reactant. For HCl: 0.250 L×0.100 M=0.0250 mol HCl0.250 \text{ L} \times 0.100 \text{ M} = 0.0250 \text{ mol HCl}. For NaOH: 1.50 g40.0 g/mol=0.0375 mol NaOH\frac{1.50 \text{ g}}{40.0 \text{ g/mol}} = 0.0375 \text{ mol NaOH}. The neutralization reaction is: NaOH+HCl→NaCl+H2O\text{NaOH} + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O} Since NaOH is in excess (0.0375 mol vs 0.0250 mol HCl), all HCl will be consumed, leaving 0.0125 mol excess NaOH in 250.0 mL solution. The concentration of excess OH⁻ is: 0.0125 mol0.250 L=0.0500 M\frac{0.0125 \text{ mol}}{0.250 \text{ L}} = 0.0500 \text{ M} Therefore: pOH=−log⁡(0.0500)=1.30\text{pOH} = -\log(0.0500) = 1.30 and pH=14.00−1.30=12.70\text{pH} = 14.00 - 1.30 = 12.70 Answer D (12.70) is correct. Answer A (0.85) would suggest a very acidic solution, ignoring that NaOH is in excess. Answer B (1.15) might result from incorrectly calculating as if HCl were in excess. Answer C (7.00) represents perfect neutralization, but this occurs only when acid and base are present in stoichiometric amounts. Study tip: Always identify the limiting reactant first in acid-base problems. The excess reactant determines whether your final solution is acidic or basic, which guides whether you calculate pH or pOH initially.

Question 17

What is the minimum volume of water needed to dilute 10.0 mL of 2.00 M HBr to achieve a pH of 1.00?

  1. 90.0 mL
  2. 190 mL (correct answer)
  3. 290 mL
  4. 390 mL
  5. 490 mL
Explanation: This problem combines dilution calculations with pH concepts for strong acids. When you see a question asking for dilution volume to achieve a specific pH, you need to work backwards from the target pH to find the required concentration, then use dilution principles. Since HBr is a strong acid, it completely dissociates, so [H+]=[HBr][H^+] = [HBr]. A pH of 1.00 means [H+]=10−1.00=0.10 M[H^+] = 10^{-1.00} = 0.10 \text{ M}, so you need a final HBr concentration of 0.10 M. Starting with 10.0 mL of 2.00 M HBr, you have n=(0.0100 L)(2.00 M)=0.0200 moln = (0.0100 \text{ L})(2.00 \text{ M}) = 0.0200 \text{ mol} of HBr. Since dilution doesn't change the number of moles, you need a final volume where 0.0200 mol÷Vf=0.10 M0.0200 \text{ mol} ÷ V_f = 0.10 \text{ M}. Solving: Vf=0.0200 mol÷0.10 M=0.200 L=200 mLV_f = 0.0200 \text{ mol} ÷ 0.10 \text{ M} = 0.200 \text{ L} = 200 \text{ mL}. Since you started with 10.0 mL, the water needed is 200 mL−10.0 mL=190 mL200 \text{ mL} - 10.0 \text{ mL} = 190 \text{ mL}, which is answer B. Answer A (90.0 mL) would give a final volume of 100 mL, resulting in 0.20 M HBr and pH = 0.70. Answer C (290 mL) corresponds to a final volume of 300 mL, giving 0.067 M HBr and pH = 1.17. Answer D (390 mL) yields a final volume of 400 mL, producing 0.050 M HBr and pH = 1.30. Remember: always calculate the final total volume first, then subtract the original volume to find the water needed. Many students forget this subtraction step.

Question 18

What is the pH of a solution prepared by diluting 25.0 mL of 0.500 M HNO3HNO_3 to a final volume of 500.0 mL?

  1. 1.00
  2. 1.30
  3. 1.60 (correct answer)
  4. 2.00
  5. 2.30
Explanation: This question tests your understanding of acid-base chemistry and dilution calculations. When you see a strong acid like HNO3HNO_3 being diluted, you need to find the new concentration first, then convert to pH. Start by calculating the new concentration after dilution using M1V1=M2V2M_1V_1 = M_2V_2. You have 25.0 mL of 0.500 M HNO3HNO_3 diluted to 500.0 mL total volume: (0.500 M)(25.0 mL)=M2(500.0 mL)(0.500 \text{ M})(25.0 \text{ mL}) = M_2(500.0 \text{ mL}) M2=(0.500)(25.0)500.0=0.0250 MM_2 = \frac{(0.500)(25.0)}{500.0} = 0.0250 \text{ M} Since HNO3HNO_3 is a strong acid, it completely ionizes, so [H+]=0.0250 M[H^+] = 0.0250 \text{ M}. The pH is: pH=−log⁡[H+]=−log⁡(0.0250)=−log⁡(2.50×10−2)=1.60\text{pH} = -\log[H^+] = -\log(0.0250) = -\log(2.50 \times 10^{-2}) = 1.60 Looking at the wrong answers: Choice A (1.00) would correspond to [H+]=0.100 M[H^+] = 0.100 \text{ M}, which ignores the dilution effect entirely. Choice B (1.30) represents [H+]=0.050 M[H^+] = 0.050 \text{ M}, suggesting you might have miscalculated the dilution factor. Choice D (2.00) gives [H+]=0.010 M[H^+] = 0.010 \text{ M}, indicating an error in the dilution calculation, possibly confusing the volume relationship. The correct answer is C (1.60). Study tip: For dilution problems with strong acids, always use M1V1=M2V2M_1V_1 = M_2V_2 first, then remember that pH = −log⁡[H+]-\log[H^+] where [H+][H^+] equals the molarity for strong acids. Practice converting between scientific notation and logarithms to avoid calculation errors.

Question 19

The pH of a 0.0250 M solution of HClO₄ is measured at three different temperatures. Based on the properties of strong acids, which trend would be expected?

  1. pH increases significantly with increasing temperature due to decreased dissociation.
  2. pH decreases with increasing temperature due to increased autoionization of water.
  3. pH remains constant at 1.60 regardless of temperature since HClO₄ is a strong acid.
  4. pH increases slightly with increasing temperature due to changes in water autoionization. (correct answer)
  5. pH varies unpredictably with temperature due to competing equilibrium effects.
Explanation: When analyzing pH changes with temperature for strong acid solutions, you need to consider two competing effects: the acid's complete dissociation and water's autoionization behavior. For a 0.0250 M HClO₄ solution, the perchloric acid dissociates completely at all reasonable temperatures, giving [H+]=0.0250 M[H^+] = 0.0250 \text{ M}. However, water itself undergoes autoionization: H2O⇌H++OH−H_2O \rightleftharpoons H^+ + OH^-, and this equilibrium is temperature-dependent. As temperature increases, water's autoionization constant (KwK_w) increases significantly, meaning more water molecules dissociate to produce additional H⁺ and OH⁻ ions. The correct answer is D because the increased autoionization of water at higher temperatures produces additional H⁺ ions beyond those from the strong acid. However, since the acid concentration is relatively high (0.0250 M), this additional contribution from water is small, causing only a slight decrease in pH (or equivalently, a slight increase in [H⁺]). The question asks about pH trends, and since pH = -log[H⁺], a small increase in [H⁺] means a small decrease in pH. Answer A is wrong because strong acids maintain complete dissociation regardless of temperature. Answer B incorrectly suggests a significant pH decrease, but the water contribution is minimal compared to the acid's contribution. Answer C ignores the temperature dependence of water's autoionization entirely—while the change is small, it's not zero. Remember: for strong acid problems involving temperature, always consider water's autoionization as a secondary but measurable effect, especially in moderately concentrated solutions.

Question 20

The pH of a solution changes from 2.00 to 5.00. By what factor has the hydrogen ion concentration decreased?

  1. 3
  2. 100
  3. 1000 (correct answer)
  4. 10000
  5. 100000
Explanation: This question tests your understanding of the logarithmic relationship between pH and hydrogen ion concentration. The pH scale is defined as pH=−log⁡[H+]\text{pH} = -\log[\text{H}^+], which means each unit change in pH represents a 10-fold change in hydrogen ion concentration. To find how the hydrogen ion concentration changed, you need to calculate the concentrations at both pH values. At pH 2.00: [H+]=10−2.00=0.01 M[\text{H}^+] = 10^{-2.00} = 0.01 \text{ M}. At pH 5.00: [H+]=10−5.00=0.00001 M[\text{H}^+] = 10^{-5.00} = 0.00001 \text{ M}. The factor by which concentration decreased is: 0.010.00001=10−210−5=103=1000\frac{0.01}{0.00001} = \frac{10^{-2}}{10^{-5}} = 10^3 = 1000 So the hydrogen ion concentration decreased by a factor of 1000, making C correct. Let's examine why the other answers are wrong. Choice A (3) represents the simple arithmetic difference between pH values (5 - 2 = 3), but this ignores the logarithmic nature of pH. Choice B (100) would be correct if the pH had changed by only 2 units (102=10010^2 = 100), but our change is 3 units. Choice D (10000) would correspond to a 4-unit pH change (104=1000010^4 = 10000). Remember this key pattern: each unit increase in pH means the hydrogen ion concentration decreases by a factor of 10. For a pH change of nn units, the concentration changes by a factor of 10n10^n. This logarithmic relationship is fundamental to acid-base chemistry.