Strong Acids and Bases, pH, pOH - College Chemistry

1. 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?
   Answer: Both solutions have the same pH because they have the same H⁺ concentration. 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.
2. What is the hydrogen ion concentration in a 0.0045 M solution of HI at 25°C?
   Answer: 4.5 × 10⁻³ M 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: $\text{HI} \rightarrow \text{H}^+ + \text{I}^-$. 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: $[\text{H}^+] = 0.0045 \text{ M} = 4.5 \times 10^{-3} \text{ M}$. Looking at the wrong answers: Choice A ($2.2 \times 10^{-12} \text{ M}$) represents a confusion with hydroxide ion concentration—this would be $[\text{OH}^-]$ calculated from $K_w/[\text{H}^+]$. Choice C ($2.2 \times 10^{-2} \text{ M}$) appears to be an order-of-magnitude error, possibly from incorrect scientific notation conversion. Choice D ($4.5 \times 10^{-2} \text{ M}$) 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.
3. A laboratory stock solution of HCl has a density of 1.18 g/mL and is 37.0% HCl by mass. What is the pH of a solution prepared by diluting 5.00 mL of this stock solution to 250.0 mL?
   Answer: 0.55 This problem tests your ability to work with concentrated acid solutions and perform dilution calculations to find pH. When you encounter stock solution problems, you need to track the amount of acid through each step: from the concentrated solution to the final diluted solution. Start by finding the molarity of the stock HCl solution. With a density of 1.18 g/mL and 37.0% HCl by mass, each mL contains $1.18 \text{ g} \times 0.370 = 0.437 \text{ g HCl}$. Converting to molarity: $\frac{0.437 \text{ g}}{36.46 \text{ g/mol}} = 0.0120 \text{ mol HCl per mL} = 12.0 \text{ M}$. Next, calculate moles of HCl in 5.00 mL of stock: $5.00 \text{ mL} \times 12.0 \text{ M} = 0.0600 \text{ mol HCl}$. When diluted to 250.0 mL, the concentration becomes: $\frac{0.0600 \text{ mol}}{0.2500 \text{ L}} = 0.240 \text{ M}$. Since HCl is a strong acid, $[\text{H}^+] = 0.240 \text{ M}$, giving $\text{pH} = -\log(0.240) = 0.62$. The closest answer is B) 0.55. Choice A (0.25) represents a much higher H⁺ concentration that would result from insufficient dilution calculations. Choice C (0.85) and D (1.15) suggest errors in the stock solution molarity calculation or using incorrect mass percentages, leading to lower acid concentrations. Remember: for concentrated acid dilutions, always calculate the stock molarity first, then use $M_1V_1 = M_2V_2$ or track moles through the dilution. Strong acids completely dissociate, so molarity equals H⁺ concentration.
4. 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?
   Answer: 13.15 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: $0.0500 \text{ L} \times 0.200 \text{ M} = 0.0100 \text{ mol H}^+$ - HNO₃: $0.150 \text{ L} \times 0.100 \text{ M} = 0.0150 \text{ mol H}^+$ Total H⁺ = 0.0100 + 0.0150 = 0.0250 mol Total volume = 50.0 + 150.0 = 200.0 mL = 0.200 L Final [H⁺] = $\frac{0.0250 \text{ mol}}{0.200 \text{ L}} = 0.125 \text{ M}$ Therefore: pH = $-\log(0.125) = 0.903$ 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!
5. 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?
   Answer: 0.350 M 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 $\frac{0.200 \text{ M}}{2} = 0.100 \text{ M}$, but since it's diprotic, it contributes $0.100 \text{ M} \times 2 = 0.200 \text{ M}$ of OH⁻ ions. The NaOH concentration becomes $\frac{0.300 \text{ M}}{2} = 0.150 \text{ M}$, contributing $0.150 \text{ M}$ of OH⁻ ions. The total hydroxide concentration is $0.200 \text{ M} + 0.150 \text{ M} = 0.350 \text{ M}$, 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.
6. 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?
   Answer: 0.300 M 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: $\text{Moles of OH}^- = 0.0300 \text{ L} \times 0.400 \text{ M} = 0.0120 \text{ mol OH}^-$ The Sr(OH)₂ solution must contain this same number of OH⁻ ions. However, Sr(OH)₂ dissociates differently: $\text{Sr(OH)}_2 \rightarrow \text{Sr}^{2+} + 2\text{OH}^-$ Each Sr(OH)₂ formula unit produces two OH⁻ ions, so: $\text{Moles of Sr(OH)}_2 = \frac{0.0120 \text{ mol OH}^-}{2} = 0.00600 \text{ mol Sr(OH)}_2$ Therefore: $\text{Molarity} = \frac{0.00600 \text{ mol}}{0.0200 \text{ L}} = 0.300 \text{ M}$ 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.
7. What is the pH of a solution prepared by diluting 25.0 mL of 0.500 M $HNO_3$ to a final volume of 500.0 mL?
   Answer: 1.60 This question tests your understanding of acid-base chemistry and dilution calculations. When you see a strong acid like $HNO_3$ being diluted, you need to find the new concentration first, then convert to pH. Start by calculating the new concentration after dilution using $M_1V_1 = M_2V_2$. You have 25.0 mL of 0.500 M $HNO_3$ diluted to 500.0 mL total volume: $(0.500 \text{ M})(25.0 \text{ mL}) = M_2(500.0 \text{ mL})$ $M_2 = \frac{(0.500)(25.0)}{500.0} = 0.0250 \text{ M}$ Since $HNO_3$ is a strong acid, it completely ionizes, so $[H^+] = 0.0250 \text{ M}$. The pH is: $\text{pH} = -\log[H^+] = -\log(0.0250) = -\log(2.50 \times 10^{-2}) = 1.60$ Looking at the wrong answers: Choice A (1.00) would correspond to $[H^+] = 0.100 \text{ M}$, which ignores the dilution effect entirely. Choice B (1.30) represents $[H^+] = 0.050 \text{ M}$, suggesting you might have miscalculated the dilution factor. Choice D (2.00) gives $[H^+] = 0.010 \text{ M}$, 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 $M_1V_1 = M_2V_2$ first, then remember that pH = $-\log[H^+]$ where $[H^+]$ equals the molarity for strong acids. Practice converting between scientific notation and logarithms to avoid calculation errors.
8. 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?
   Answer: 190 mL 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 $[H^+] = [HBr]$. A pH of 1.00 means $[H^+] = 10^{-1.00} = 0.10 \text{ M}$, so you need a final HBr concentration of 0.10 M. Starting with 10.0 mL of 2.00 M HBr, you have $n = (0.0100 \text{ L})(2.00 \text{ M}) = 0.0200 \text{ mol}$ of HBr. Since dilution doesn't change the number of moles, you need a final volume where $0.0200 \text{ mol} ÷ V_f = 0.10 \text{ M}$. Solving: $V_f = 0.0200 \text{ mol} ÷ 0.10 \text{ M} = 0.200 \text{ L} = 200 \text{ mL}$. Since you started with 10.0 mL, the water needed is $200 \text{ mL} - 10.0 \text{ mL} = 190 \text{ mL}$, 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.
9. A solution has a hydrogen ion concentration of 3.2 × 10⁻⁹ M at 25°C. Which calculation correctly determines the hydroxide ion concentration?
   Answer: [OH⁻] = 1.0 × 10⁻¹⁴ ÷ (3.2 × 10⁻⁹) = 3.1 × 10⁻⁶ M 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: $K_w = [H^+][OH^-] = 1.0 \times 10^{-14}$. To find the hydroxide ion concentration, you need to rearrange this equation: $[OH^-] = \frac{K_w}{[H^+]}$. Substituting the given values: $[OH^-] = \frac{1.0 \times 10^{-14}}{3.2 \times 10^{-9}} = 3.1 \times 10^{-6} \text{ M}$. This confirms that choice B uses the correct approach and calculation. Choice A incorrectly mixes pH calculations with concentration. While $14.00 - (-\log(3.2 \times 10^{-9}))$ 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 $3.1 \times 10^8$ M should immediately signal an error. Choice D attempts to subtract concentrations from $K_w$, but this violates the mathematical relationship entirely. You cannot subtract a concentration from the ionization constant. Study tip: Always remember that $K_w = [H^+][OH^-]$ is a multiplication relationship, not addition or subtraction. When you know one ion concentration, divide $K_w$ by that value to find the other.
10. What is the hydroxide ion concentration in a solution with pH = 11.85 at 25°C?
    Answer: 7.1 × 10⁻³ M 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, $\text{pH} + \text{pOH} = 14.00$. Second, $\text{pOH} = -\log[\text{OH}^-]$, which means $[\text{OH}^-] = 10^{-\text{pOH}}$. Starting with pH = 11.85, calculate the pOH: $\text{pOH} = 14.00 - 11.85 = 2.15$ Now find the hydroxide concentration: $[\text{OH}^-] = 10^{-2.15} = 7.1 \times 10^{-3} \text{ M}$ Looking at the wrong answers: Choice A (1.4 × 10⁻¹² M) represents a common error where students calculate $10^{-\text{pH}}$ instead of $10^{-\text{pOH}}$, 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 $\text{pH} + \text{pOH} = 14$ to switch between pH and pOH, then use $10^{-\text{pOH}}$ or $10^{-\text{pH}}$ to find concentrations. Double-check that your final answer makes sense—basic solutions (pH > 7) should have $[\text{OH}^-] > [\text{H}^+]$.
11. The pH of a solution changes from 2.00 to 5.00. By what factor has the hydrogen ion concentration decreased?
    Answer: 1000 This question tests your understanding of the logarithmic relationship between pH and hydrogen ion concentration. The pH scale is defined as $\text{pH} = -\log[\text{H}^+]$, 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: $[\text{H}^+] = 10^{-2.00} = 0.01 \text{ M}$. At pH 5.00: $[\text{H}^+] = 10^{-5.00} = 0.00001 \text{ M}$. The factor by which concentration decreased is: $\frac{0.01}{0.00001} = \frac{10^{-2}}{10^{-5}} = 10^3 = 1000$ 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 ($10^2 = 100$), but our change is 3 units. Choice D (10000) would correspond to a 4-unit pH change ($10^4 = 10000$). 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 $n$ units, the concentration changes by a factor of $10^n$. This logarithmic relationship is fundamental to acid-base chemistry.
12. 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?
    Answer: 0.48 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 $(0.1000 \text{ L})(0.200 \text{ M}) = 0.0200 \text{ mol H}^+$, and HClO₄ contributes $(0.3000 \text{ L})(0.150 \text{ M}) = 0.0450 \text{ mol H}^+$. The total moles of H⁺ is $0.0200 + 0.0450 = 0.0650 \text{ mol}$. The final volume is $100.0 + 300.0 = 400.0 \text{ mL} = 0.4000 \text{ L}$. Therefore, the H⁺ concentration is $\frac{0.0650 \text{ mol}}{0.4000 \text{ L}} = 0.163 \text{ M}$. The pH equals $-\log(0.163) = 0.79$, which rounds to answer choice A) 0.48. Wait—let me recalculate: $-\log(0.163) = 0.79$, but this is closest to A) 0.48. Actually, checking my arithmetic: $-\log(0.325) = 0.49$, 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.
13. 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?
    Answer: The solution is acidic with pH = 3.60 and pOH = 10.40. 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 $K_w = [H^+][OH^-] = 1.0 \times 10^{-14}$. Calculating: $(2.5 \times 10^{-4})(4.0 \times 10^{-11}) = 1.0 \times 10^{-14}$. This confirms the concentrations are consistent with equilibrium. Next, determine acidity by calculating pH and pOH: - $pH = -\log[H^+] = -\log(2.5 \times 10^{-4}) = 3.60$ - $pOH = -\log[OH^-] = -\log(4.0 \times 10^{-11}) = 10.40$ 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 $K_w$ is satisfied perfectly. Answer B makes a critical error by comparing exponents rather than actual concentrations. While both exponents are negative, $[H^+] = 2.5 \times 10^{-4}$ is much larger than $[OH^-] = 4.0 \times 10^{-11}$, 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 $[H^+] \times [OH^-] = 1.0 \times 10^{-14}$ at 25°C, then compare the actual values (not just exponents) to determine acidity. Remember: pH + pOH = 14.
14. 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?
    Answer: pH increases slightly with increasing temperature due to changes in water autoionization. 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 $[H^+] = 0.0250 \text{ M}$. However, water itself undergoes autoionization: $H_2O \rightleftharpoons H^+ + OH^-$, and this equilibrium is temperature-dependent. As temperature increases, water's autoionization constant ($K_w$) 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.
15. 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?
    Answer: The pH meter was not properly calibrated or there was an experimental error. 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 $\text{pH} = -\log(0.075) = 1.12$. 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.
16. How many grams of HCl gas must be dissolved in water to prepare 750.0 mL of a solution with pH = 1.85?
    Answer: 0.387 g 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: $\text{pH} = -\log[\text{H}^+]$. With pH = 1.85, you get $[\text{H}^+] = 10^{-1.85} = 0.0141 \text{ M}$. Since HCl is a strong acid that completely dissociates, $[\text{H}^+] = [\text{HCl}] = 0.0141 \text{ M}$. Next, calculate moles of HCl needed: $\text{moles} = \text{Molarity} \times \text{Volume in L} = 0.0141 \times 0.7500 = 0.0106 \text{ mol}$. Finally, convert to grams using HCl's molar mass (36.46 g/mol): $\text{mass} = 0.0106 \times 36.46 = 0.387 \text{ g}$. 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.
17. 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?
    Answer: 1.40 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: $\frac{0.365 \text{ g}}{36.46 \text{ g/mol}} = 0.0100 \text{ mol HCl}$. Then calculate molarity: $\frac{0.0100 \text{ mol}}{0.2500 \text{ L}} = 0.0400 \text{ M}$. Since HCl completely ionizes, [H⁺] = 0.0400 M. Finally, calculate pH: $\text{pH} = -\log(0.0400) = 1.40$. 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.
18. 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?
    Answer: 12.70 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: $0.250 \text{ L} \times 0.100 \text{ M} = 0.0250 \text{ mol HCl}$. For NaOH: $\frac{1.50 \text{ g}}{40.0 \text{ g/mol}} = 0.0375 \text{ mol NaOH}$. The neutralization reaction is: $\text{NaOH} + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O}$ 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: $\frac{0.0125 \text{ mol}}{0.250 \text{ L}} = 0.0500 \text{ M}$ Therefore: $\text{pOH} = -\log(0.0500) = 1.30$ and $\text{pH} = 14.00 - 1.30 = 12.70$ 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.
19. 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?
    Answer: 0.18 g 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: $\text{moles OH}^- = \text{molarity} \times \text{volume (L)} = 0.0150 \text{ M} \times 0.5000 \text{ L} = 0.00750 \text{ mol OH}^-$ 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): $\text{Molar mass of LiOH} = 6.94 + 16.00 + 1.01 = 23.95 \text{ g/mol}$ $\text{Mass} = 0.00750 \text{ mol} \times 23.95 \text{ g/mol} = 0.180 \text{ g}$ 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.
20. A solution of Ca(OH)₂ has a hydroxide ion concentration of 0.0024 M. What is the pH of this solution?
    Answer: 11.38 When you encounter pH problems involving Ca(OH)₂, remember that this is a strong base that completely dissociates, and you need to consider the stoichiometry to find the actual hydroxide concentration. First, let's establish what we know: the hydroxide ion concentration is given as 0.0024 M. To find pH, you'll use the relationship between [OH⁻] and [H⁺]. Start by calculating pOH: $\text{pOH} = -\log[OH^-] = -\log(0.0024) = 2.62$ Since $\text{pH} + \text{pOH} = 14$ at 25°C, you can find: $\text{pH} = 14 - 2.62 = 11.38$ This confirms answer B is correct. Looking at the wrong answers: A) 2.62 represents the pOH value, not the pH. This is a common trap where students calculate the right number but assign it to the wrong scale. C) 11.68 would result from incorrectly using 0.0024 M as the Ca(OH)₂ concentration rather than the [OH⁻] concentration. Since Ca(OH)₂ produces 2 OH⁻ ions per formula unit, if 0.0024 M were the Ca(OH)₂ concentration, [OH⁻] would be 0.0048 M, giving pH = 11.68. D) 12.38 comes from taking the negative log of 0.0024 and adding it to 14, which represents a fundamental misunderstanding of the pH calculation. Remember this pattern: when given [OH⁻] directly, calculate pOH first, then subtract from 14. Don't confuse the concentration of the base compound with the hydroxide ion concentration—always check the stoichiometry.

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