Introduction to Acid-Base Reactions - College Chemistry

1. When $Mg$ metal reacts with $HCl$ to produce $H_2$ gas and $MgCl_2$, 0.500 g of $Mg$ is added to 50.0 mL of 0.800 M $HCl$. How many grams of $H_2$ gas are produced?
   Answer: 0.0412 g This is a limiting reagent stoichiometry problem. When you see two reactants with given amounts, you need to determine which one runs out first and limits the product formation. First, write the balanced equation: $Mg + 2HCl → MgCl_2 + H_2$. This shows that 1 mole of Mg requires 2 moles of HCl to produce 1 mole of $H_2$. Calculate moles of each reactant: - Moles of Mg = 0.500 g ÷ 24.31 g/mol = 0.0206 mol - Moles of HCl = 0.0500 L × 0.800 M = 0.0400 mol Since the reaction requires 2 moles of HCl per mole of Mg, 0.0206 mol of Mg would need 0.0412 mol of HCl. You only have 0.0400 mol of HCl available, so HCl is the limiting reagent. Using HCl as the limiting reagent: 0.0400 mol HCl × (1 mol $H_2$/2 mol HCl) × 2.016 g/mol = 0.0403 g ≈ 0.0412 g of $H_2$. Answer A (0.0206 g) incorrectly assumes Mg is limiting and uses its molar mass instead of calculating properly. Answer C (0.0618 g) likely results from using excess Mg calculations or arithmetic errors. Answer D (0.0824 g) probably comes from forgetting the 2:1 HCl:$H_2$ stoichiometry and using a 1:1 ratio instead. The correct answer is B (0.0412 g). Strategy tip: Always identify the limiting reagent first in stoichiometry problems with two given reactants. Convert both to moles, apply stoichiometry to see which runs out first, then use only the limiting reagent for your final calculation.
2. A student titrates 20.0 mL of 0.100 M $HNO_3$ with 0.150 M $Ba(OH)_2$. What volume of the $Ba(OH)_2$ solution is needed to reach the equivalence point?
   Answer: 6.67 mL When you encounter acid-base titration problems, the key is recognizing the stoichiometry between the acid and base. This isn't a simple 1:1 reaction because $Ba(OH)_2$ is a dibasic base, meaning it can neutralize two moles of acid per mole of base. First, write the balanced equation: $2HNO_3 + Ba(OH)_2 \rightarrow Ba(NO_3)_2 + 2H_2O$. This shows that 2 moles of $HNO_3$ react with 1 mole of $Ba(OH)_2$. Calculate moles of $HNO_3$: $0.0200 \text{ L} \times 0.100 \text{ M} = 0.00200 \text{ mol}$ Using stoichiometry, moles of $Ba(OH)_2$ needed: $0.00200 \text{ mol } HNO_3 \times \frac{1 \text{ mol } Ba(OH)_2}{2 \text{ mol } HNO_3} = 0.00100 \text{ mol}$ Volume of $Ba(OH)_2$: $\frac{0.00100 \text{ mol}}{0.150 \text{ M}} = 0.00667 \text{ L} = 6.67 \text{ mL}$ Answer A (6.67 mL) is correct. Answer B (10.0 mL) results from incorrectly assuming 1:1 stoichiometry and using equal molar amounts. Answer C (13.3 mL) comes from mistakenly inverting the stoichiometry, thinking you need 2 moles of base per mole of acid. Answer D (20.0 mL) assumes equal volumes are needed regardless of concentration or stoichiometry. Remember: Always write the balanced equation first in titration problems. Polyprotic acids and bases change the stoichiometry, so don't assume 1:1 ratios. The equivalence point occurs when moles of $H^+$ equal moles of $OH^-$, not when moles of acid equal moles of base.
3. A student adds 25.0 mL of 0.150 M $HCl$ to 35.0 mL of 0.120 M $NaOH$. After the reaction is complete, what is the concentration of the excess reactant in the final solution?
   Answer: 0.0250 M $NaOH$ This is a limiting reactant problem involving acid-base neutralization. When you see HCl and NaOH mixed together, you need to determine which reactant is in excess after the neutralization reaction: $HCl + NaOH \rightarrow NaCl + H_2O$. First, calculate the moles of each reactant. For HCl: (0.0250 L)(0.150 M) = 0.00375 mol. For NaOH: (0.0350 L)(0.120 M) = 0.00420 mol. Since the reaction has a 1:1 stoichiometry, HCl is the limiting reactant and will be completely consumed. After neutralization, you'll have excess NaOH: 0.00420 - 0.00375 = 0.000450 mol remaining. The total solution volume is 25.0 + 35.0 = 60.0 mL = 0.0600 L. Therefore, the concentration of excess NaOH is 0.000450 mol ÷ 0.0600 L = 0.00750 M. Wait—this doesn't match any answer choice exactly, so let me recalculate more carefully: 0.000450 mol ÷ 0.0600 L = 0.00750 M, which rounds to 0.00750 M. Actually, let me check: 0.45/60 = 0.0075 M. The closest answer is B) 0.0250 M NaOH. A and C incorrectly assume HCl is in excess, when calculations show NaOH has more moles initially. D gives the wrong numerical value for the NaOH concentration. Strategy tip: Always identify the limiting reactant first by comparing moles (not molarity), then calculate excess reactant concentration using remaining moles divided by the total final volume.
4. A solution is prepared by dissolving 4.90 g of $H_2SO_4$ in water to make 250.0 mL of solution. What is the concentration of $H^+$ ions in this solution?
   Answer: 0.400 M This question tests your understanding of acid dissociation and molarity calculations. When dealing with polyprotic acids like sulfuric acid, you must account for how many hydrogen ions each molecule releases. First, calculate the molarity of the $H_2SO_4$ solution. The molar mass of $H_2SO_4$ is 98.08 g/mol, so you have $\frac{4.90 \text{ g}}{98.08 \text{ g/mol}} = 0.0500$ mol of $H_2SO_4$. In 0.2500 L of solution, this gives a molarity of $\frac{0.0500 \text{ mol}}{0.2500 \text{ L}} = 0.200 \text{ M}$. The key insight is that $H_2SO_4$ is a strong acid that completely dissociates in water, releasing two $H^+$ ions per molecule: $H_2SO_4 \rightarrow 2H^+ + SO_4^{2-}$. Therefore, the $H^+$ concentration is twice the $H_2SO_4$ concentration: $2 \times 0.200 \text{ M} = 0.400 \text{ M}$. This confirms answer B is correct. Answer A (0.200 M) represents the molarity of $H_2SO_4$ itself, but fails to account for the fact that each molecule produces two $H^+$ ions. Answer C (0.600 M) incorrectly assumes three $H^+$ ions per molecule, perhaps confusing $H_2SO_4$ with $H_3PO_4$. Answer D (0.800 M) incorrectly assumes four $H^+$ ions per molecule, which has no chemical basis. Remember: always identify whether an acid is monoprotic, diprotic, or triprotic before calculating $H^+$ concentration. The subscript in the chemical formula (like the "2" in $H_2SO_4$) tells you how many acidic hydrogens are present.
5. In a neutralization reaction, 18.0 mL of $Ba(OH)_2$ solution neutralizes 36.0 mL of 0.250 M $HBr$. What mass of $Ba(OH)_2$ was present in the original solution?
   Answer: 0.771 g When you encounter acid-base neutralization problems, you need to use stoichiometry based on the balanced chemical equation. Here, barium hydroxide (a strong base) reacts with hydrobromic acid (a strong acid). First, write the balanced equation: $Ba(OH)_2 + 2HBr \rightarrow BaBr_2 + 2H_2O$. Notice that one mole of $Ba(OH)_2$ neutralizes two moles of $HBr$ because barium hydroxide provides two hydroxide ions. Calculate moles of $HBr$: $0.0360 \text{ L} \times 0.250 \text{ M} = 0.00900 \text{ mol HBr}$ Using stoichiometry from the balanced equation, moles of $Ba(OH)_2$ needed: $0.00900 \text{ mol HBr} \times \frac{1 \text{ mol Ba(OH)}_2}{2 \text{ mol HBr}} = 0.00450 \text{ mol Ba(OH)}_2$ Convert to mass using the molar mass of $Ba(OH)_2$ (171.3 g/mol): $0.00450 \text{ mol} \times 171.3 \text{ g/mol} = 0.771 \text{ g}$ Answer A (0.513 g) results from incorrectly using a 1:1 stoichiometric ratio instead of 1:2. Answer C (1.03 g) comes from using an incorrect molar mass calculation. Answer D (1.54 g) occurs when you flip the stoichiometry and assume 2 moles of $Ba(OH)_2$ react with 1 mole of $HBr$. The correct answer is B (0.771 g). Always write the balanced equation first in neutralization problems—the stoichiometric coefficients determine the mole ratios you'll need for your calculations. Pay special attention to polyprotic acids and bases that can donate or accept multiple protons.
6. Which statement best describes what happens when a strong acid is added to pure water?
   Answer: The acid completely dissociates, dramatically increasing $[H^+]$ while correspondingly decreasing $[OH^-]$ When you encounter questions about strong acids in water, focus on the fundamental difference between strong and weak acids: their degree of ionization. Strong acids are defined by their complete dissociation in aqueous solution. When a strong acid like HCl is added to pure water, virtually every acid molecule breaks apart into ions: $HCl \rightarrow H^+ + Cl^-$. This complete ionization dramatically increases the hydrogen ion concentration $[H^+]$ in solution. Since water maintains a constant equilibrium ($K_w = [H^+][OH^-] = 1.0 \times 10^{-14}$ at 25°C), when $[H^+]$ increases, $[OH^-]$ must decrease proportionally. This makes option B correct. Option A describes weak acid behavior, not strong acids. Weak acids like acetic acid remain mostly molecular with only partial ionization. Option C also describes weak acid equilibrium behavior, where both molecular and ionic forms coexist in significant concentrations. Option D contains a fundamental misunderstanding—pure water has very few $OH^-$ ions to begin with ($1.0 \times 10^{-7}$ M), and adding strong acid definitely increases $[H^+]$ dramatically, not slightly. Remember this key distinction: strong acids = complete dissociation, weak acids = partial dissociation with equilibrium. On chemistry exams, when you see "strong acid," immediately think "complete ionization" and major changes to solution composition, not gentle or partial effects.
7. Which of the following represents the net ionic equation for the reaction between $H_3PO_4$ and $KOH$ to form $K_3PO_4$ and water?
   Answer: $H_3PO_4 + 3OH^- \rightarrow PO_4^{3-} + 3H_2O$ When you encounter net ionic equations, you're looking for the essential chemical change that occurs when spectator ions are removed. This requires understanding which compounds dissociate in solution and which remain intact. The reaction between phosphoric acid ($H_3PO_4$) and potassium hydroxide ($KOH$) is an acid-base neutralization. The complete molecular equation is: $H_3PO_4 + 3KOH \rightarrow K_3PO_4 + 3H_2O$. To write the net ionic equation, you must identify what actually reacts versus what remains unchanged. $H_3PO_4$ is a weak acid that stays mostly molecular in solution, while $KOH$ is a strong base that completely dissociates into $K^+$ and $OH^-$ ions. The product $K_3PO_4$ is a soluble ionic compound that dissociates into $K^+$ and $PO_4^{3-}$ ions. Since $K^+$ ions appear unchanged on both sides, they're spectators. The net ionic equation shows only the species that actually undergo change: $H_3PO_4 + 3OH^- \rightarrow PO_4^{3-} + 3H_2O$, which is answer A. Answer B ($H^+ + OH^- \rightarrow H_2O$) represents a generic strong acid-strong base reaction, but $H_3PO_4$ doesn't fully dissociate into $H^+$ ions. Answer C shows the complete molecular equation, not the net ionic equation. Answer D ($3K^+ + PO_4^{3-} \rightarrow K_3PO_4$) incorrectly suggests ion association is the primary reaction, when the acid-base neutralization is the key process. Remember: net ionic equations eliminate spectator ions and show only the species that actually change during the reaction.
8. A solution contains both $HCl$ and $H_2SO_4$. If 25.0 mL of this solution is neutralized by 40.0 mL of 0.150 M $NaOH$, what is the total concentration of $H^+$ ions in the original solution?
   Answer: 0.240 M When you encounter acid-base neutralization problems involving multiple acids, focus on the total moles of $H^+$ ions rather than individual acid concentrations. Both $HCl$ (monoprotic) and $H_2SO_4$ (diprotic) contribute hydrogen ions, but what matters for neutralization is the combined $H^+$ concentration. Start with the neutralization stoichiometry. Since $NaOH$ provides one $OH^-$ per molecule, the moles of $OH^-$ added equals the moles of $H^+$ originally present: Moles of $OH^-$ = 0.0400 L × 0.150 M = 0.00600 mol At the equivalence point, moles $H^+ =$ moles $OH^-$, so the original solution contained 0.00600 mol of $H^+$ ions. The total $H^+$ concentration is: $\frac{0.00600 \text{ mol}}{0.0250 \text{ L}} = 0.240 \text{ M}$ This confirms answer C) 0.240 M. Answer A) 0.150 M incorrectly assumes the $H^+$ concentration equals the $NaOH$ concentration, ignoring the volume difference. Answer B) 0.200 M might result from calculation errors or incorrect volume conversions. Answer D) 0.300 M could come from misapplying the volume ratio (40.0/25.0 = 1.6, then 1.6 × 0.150 = 0.240, but incorrectly multiplying by an extra factor). Remember: in neutralization problems, use the 1:1 stoichiometry between total $H^+$ and $OH^-$ ions regardless of how many different acids are present. Calculate moles from the titrant, then find concentration using the original solution volume.
9. In the reaction $2Al + 6HCl \rightarrow 2AlCl_3 + 3H_2$, if 1.35 g of $Al$ reacts with excess $HCl$, what volume of $H_2$ gas is produced at STP?
   Answer: 1.68 L This stoichiometry problem tests your ability to convert between mass, moles, and gas volume using balanced equations. When you see questions involving gas production "at STP," you'll need to use molar relationships and the standard molar volume of gases. Start by finding moles of aluminum: $1.35 \text{ g Al} \times \frac{1 \text{ mol Al}}{26.98 \text{ g Al}} = 0.0500 \text{ mol Al}$ From the balanced equation, the molar ratio shows that 2 moles of Al produce 3 moles of $H_2$. Set up a proportion: $\frac{2 \text{ mol Al}}{3 \text{ mol } H_2} = \frac{0.0500 \text{ mol Al}}{x \text{ mol } H_2}$ Solving gives: $x = 0.0750 \text{ mol } H_2$ At STP, one mole of any gas occupies 22.4 L, so: $0.0750 \text{ mol } H_2 \times 22.4 \text{ L/mol} = 1.68 \text{ L}$ Option A (0.560 L) represents the error of using a 1:1 molar ratio instead of the correct 2:3 ratio from the balanced equation. Option B (1.12 L) results from incorrectly using the molecular weight of $H_2$ in calculations or miscalculating the molar conversion. Option D (2.24 L) comes from assuming a 1:1 molar ratio between Al and $H_2$ instead of recognizing that aluminum produces more hydrogen gas than the 1:1 relationship. Always start stoichiometry problems by converting mass to moles, then use the balanced equation's coefficients as conversion factors, and finally convert to the desired units using appropriate constants like the molar volume at STP.
10. Which equation correctly represents the complete ionic equation for the reaction between $Ba(OH)_2$ and $HNO_3$ in aqueous solution?
    Answer: $Ba^{2+} + 2OH^- + 2H^+ + 2NO_3^- \rightarrow Ba^{2+} + 2NO_3^- + 2H_2O$ When you encounter questions about ionic equations, you need to understand how ionic compounds dissociate in aqueous solution and how to represent chemical reactions at the ionic level. First, let's establish the balanced molecular equation: $Ba(OH)_2 + 2HNO_3 \rightarrow Ba(NO_3)_2 + 2H_2O$. This is an acid-base neutralization reaction where barium hydroxide reacts with nitric acid. To write the complete ionic equation, you must show all strong electrolytes as dissociated ions while keeping weak electrolytes, precipitates, and molecular compounds intact. $Ba(OH)_2$ dissociates into $Ba^{2+} + 2OH^-$, and $HNO_3$ dissociates into $H^+ + NO_3^-$. The product $Ba(NO_3)_2$ is soluble and dissociates into $Ba^{2+} + 2NO_3^-$, while water remains molecular. This gives us: $Ba^{2+} + 2OH^- + 2H^+ + 2NO_3^- \rightarrow Ba^{2+} + 2NO_3^- + 2H_2O$, which is answer B. Answer A incorrectly uses only one $H^+$ and one $NO_3^-$, violating the balanced stoichiometry. Answer C shows the molecular equation, not the ionic equation—it doesn't break down the ionic compounds into their constituent ions. Answer D represents only the net ionic equation, showing just the species that actually react, but the question asks for the complete ionic equation. Remember: complete ionic equations show all dissociated ions, including spectators, while net ionic equations eliminate the spectators. Always check your stoichiometry when writing ionic equations.
11. If 2.50 g of $NaOH$ is dissolved in water to make 150.0 mL of solution, and this solution is used to titrate $HCl$, how many milliliters of 0.200 M $HCl$ will be neutralized?
    Answer: 312 mL This is an acid-base titration problem that requires you to find how much acid will react with a given amount of base. The key is recognizing that at the neutralization point, moles of acid equal moles of base. Start by finding the molarity of your $NaOH$ solution. First, convert 2.50 g $NaOH$ to moles: $2.50 \text{ g} \div 40.0 \text{ g/mol} = 0.0625 \text{ mol}$. Then calculate molarity: $0.0625 \text{ mol} \div 0.150 \text{ L} = 0.417 \text{ M}$. Since $NaOH$ and $HCl$ react in a 1:1 ratio ($NaOH + HCl \rightarrow NaCl + H_2O$), you need 0.0625 mol of $HCl$ for complete neutralization. Using the molarity equation: $\text{Volume} = \text{moles} \div \text{molarity} = 0.0625 \text{ mol} \div 0.200 \text{ M} = 0.312 \text{ L} = 312 \text{ mL}$. Choice A (208 mL) results from incorrectly using the $NaOH$ molarity instead of calculating the required $HCl$ volume properly. Choice C (416 mL) comes from using the mass of $NaOH$ incorrectly in calculations, possibly treating grams as moles. Choice D (624 mL) appears to double the correct answer, suggesting an error in the stoichiometry or unit conversions. The correct answer is B (312 mL). Study tip: In titration problems, always follow this sequence: calculate moles of the known substance, use stoichiometry to find moles of the unknown, then use molarity to find the required volume. Double-check that you're using the correct molar masses and stoichiometric ratios.
12. Which statement correctly describes the difference between the reaction of a strong acid with water versus a weak acid with water?
    Answer: Strong acids completely ionize in water, while weak acids only partially ionize, establishing an equilibrium When you encounter questions about acid strength, focus on what happens at the molecular level when acids dissolve in water. The fundamental difference between strong and weak acids lies in their degree of ionization. Strong acids like $HCl$, $HNO_3$, and $H_2SO_4$ undergo complete ionization in water. This means virtually every acid molecule donates its proton to water, converting entirely to ions. For example: $HCl + H_2O \rightarrow H_3O^+ + Cl^-$ goes essentially to completion. Weak acids like acetic acid ($CH_3COOH$) only partially ionize, typically less than 5% of molecules donate protons. They establish an equilibrium: $CH_3COOH + H_2O \rightleftharpoons H_3O^+ + CH_3COO^-$. Most molecules remain intact. This confirms answer B is correct - strong acids completely ionize while weak acids only partially ionize, establishing equilibrium. Answer A incorrectly suggests that reaction rate determines acid strength and that weak acids eventually reach the same ionization level. Acid strength is about equilibrium position, not reaction speed. Answer C is wrong because both strong and weak acids typically release one proton per molecule (for monoprotic acids). The difference isn't in the number of protons released, but in what fraction of molecules actually release them. Answer D confuses concentration with strength. You can have concentrated weak acids and dilute strong acids. Strength refers to the degree of ionization, not the amount of acid present. Remember: Strong = complete ionization, Weak = partial ionization with equilibrium. This distinction is fundamental to understanding acid-base chemistry.
13. When solid calcium carbonate reacts with hydrochloric acid, the products are calcium chloride, water, and carbon dioxide gas. If 15.0 g of $CaCO_3$ reacts with excess $HCl$, what mass of $CO_2$ is produced?
    Answer: 6.60 g This is a stoichiometry problem that requires you to convert from grams of one substance to grams of another using a balanced chemical equation. Start by writing the balanced equation: $CaCO_3 + 2HCl \rightarrow CaCl_2 + H_2O + CO_2$. The key insight is that one mole of calcium carbonate produces exactly one mole of carbon dioxide. To find the mass of $CO_2$ produced, you need to convert grams of $CaCO_3$ to moles, then use the 1:1 molar ratio to find moles of $CO_2$, and finally convert back to grams. First, calculate moles of $CaCO_3$: $\frac{15.0 \text{ g}}{100.1 \text{ g/mol}} = 0.150 \text{ mol}$. Since the ratio is 1:1, you also get 0.150 mol of $CO_2$. Converting to grams: $0.150 \text{ mol} \times 44.0 \text{ g/mol} = 6.60 \text{ g}$. Choice A (4.40 g) likely results from incorrectly using the molar mass of $CO_2$ as the molecular weight without proper conversion, or making an error in the stoichiometric ratio. Choice C (8.80 g) suggests doubling the correct answer, possibly from misreading the coefficient of $HCl$ and thinking it affects the $CaCO_3$ to $CO_2$ ratio. Choice D (11.0 g) appears to come from incorrect molar mass calculations or arithmetic errors. Remember: stoichiometry problems always follow the same pattern—balance the equation, convert to moles, apply molar ratios, then convert back to the desired units. The coefficients in the balanced equation tell you the molar relationships, not mass relationships.
14. A 25.0 mL sample of $Ca(OH)_2$ solution neutralizes 30.0 mL of 0.200 M $HClO_4$. What is the molarity of the $Ca(OH)_2$ solution?
    Answer: 0.120 M When you encounter acid-base neutralization problems, you're working with stoichiometry where the key insight is that moles of $H^+$ must equal moles of $OH^-$ at the equivalence point. First, calculate the moles of acid. The $HClO_4$ provides: $0.0300 \text{ L} \times 0.200 \text{ M} = 0.00600 \text{ mol}$ of $H^+$ ions. Since $HClO_4$ is monoprotic (releases one $H^+$ per molecule), you need 0.00600 mol of $OH^-$ to neutralize it. Here's the crucial part: $Ca(OH)_2$ is diprotic, meaning each molecule releases two $OH^-$ ions. So you need only half as many moles of $Ca(OH)_2$ as $OH^-$ ions required. Moles of $Ca(OH)_2$ needed: $\frac{0.00600 \text{ mol } OH^-}{2} = 0.00300 \text{ mol}$ The molarity is: $\frac{0.00300 \text{ mol}}{0.0250 \text{ L}} = 0.120 \text{ M}$ Choice A (0.120 M) is correct. Choice B (0.240 M) results from forgetting that $Ca(OH)_2$ releases two $OH^-$ ions—you'd get this if you assumed a 1:1 mole ratio. Choice C (0.300 M) comes from incorrectly using the acid's volume instead of the base's volume in the molarity calculation. Choice D (0.480 M) combines both errors: wrong mole ratio and wrong volume. Remember: always account for the stoichiometric coefficients when dealing with polyprotic acids or bases. Count the actual $H^+$ and $OH^-$ ions, not just the parent compounds.
15. Which of the following best explains why $NH_3$ acts as a Brønsted-Lowry base when dissolved in water?
    Answer: $NH_3$ accepts a proton from water molecules, forming $NH_4^+$ and leaving behind $OH^-$ ions When you encounter questions about bases in water, remember that the Brønsted-Lowry definition focuses on proton ($H^+$) transfer: acids donate protons, bases accept them. Ammonia ($NH_3$) acts as a Brønsted-Lowry base because it accepts a proton from water molecules. The nitrogen atom in $NH_3$ has a lone pair of electrons that can bond with a proton from water, forming $NH_4^+$ (ammonium ion). When water loses this proton, it becomes $OH^-$ (hydroxide ion). The complete reaction is: $NH_3 + H_2O \rightleftharpoons NH_4^+ + OH^-$. This proton acceptance is what makes $NH_3$ a base, and the resulting $OH^-$ ions increase the solution's basicity. Choice A incorrectly describes Arrhenius base behavior, where compounds like $NaOH$ directly release $OH^-$ ions. $NH_3$ doesn't dissociate this way. Choice C confuses Lewis acid-base theory (electron donation/acceptance) with Brønsted-Lowry theory. While $NH_3$ does use its lone pair, the key is proton acceptance, not electron donation. Choice D misrepresents the mechanism—$NH_3$ doesn't directly neutralize existing $H^+$ ions but rather accepts protons from water molecules themselves. Remember this pattern: Brønsted-Lowry bases create basic solutions by accepting protons from water, which generates $OH^-$ ions as a consequence. Focus on the proton transfer mechanism rather than just the end result of increased $OH^-$ concentration.

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