[{"data":1,"prerenderedAt":277},["ShallowReactive",2],{"topic-acids-bases-and-ph":3},{"topic":4,"prev":262,"next":270},{"id":5,"slug":6,"lastAlignmentAudit":7,"title":8,"shortTitle":9,"lesson":5,"category":10,"description":11,"metaDescription":12,"objectives":13,"conceptSections":47,"workedExample":153,"oerSources":164,"relatedTopicSlugs":171,"relatedElements":175,"relatedReferences":181,"keyEquations":186,"howTo":207,"conceptProbe":217,"practiceTeaser":221,"seoKeywords":229,"selfStudyQuestions":235,"objectiveCount":261},22,"acids-bases-and-ph","2026-07-17","Acids, Bases and pH","Acids & Bases","equilibrium-acids","Study acid-base chemistry: pH calculations, strong/weak acids and bases, Ka/Kb, conjugate pairs, and neutralization.","Learn acids and bases: pH calculations, strong and weak acid/base equilibria, Ka and Kb values, conjugate acid-base pairs. Step-by-step worked examples.",[14,17,20,23,26,29,32,35,38,41,44],{"id":15,"text":16},"22.1","Classify a substance as an acid or base according to the Arrhenius, Brønsted-Lowry, and Lewis models, and identify the framing model",{"id":18,"text":19},"22.2","Identify conjugate acid-base pairs in a Brønsted-Lowry reaction",{"id":21,"text":22},"22.3","Calculate pH, pOH, [H⁺], and [OH⁻] and convert among them using Kw",{"id":24,"text":25},"22.4","Calculate the pH of strong acid and strong base solutions",{"id":27,"text":28},"22.5","Calculate the pH of a weak acid solution using Ka and an ICE table",{"id":30,"text":31},"22.6","Calculate the pH of a weak base solution using Kb and an ICE table",{"id":33,"text":34},"22.7","Relate Ka, Kb, and Kw for a conjugate acid-base pair",{"id":36,"text":37},"22.8","Predict whether an aqueous salt solution is acidic, basic, or neutral based on the identities of its ions",{"id":39,"text":40},"22.9","Calculate percent ionization of a weak acid or weak base",{"id":42,"text":43},"22.10","Identify polyprotic acids and calculate the pH of their solutions using stepwise Ka values",{"id":45,"text":46},"22.11","Predict whether a metal or nonmetal oxide is acidic, basic, or amphoteric in water based on its position on the periodic table",[48,57,70,78,87,97,106,115,125,129,137,145,149],{"heading":49,"content":50,"relatedObjectives":51,"deepDive":53},"Defining Acids and Bases","\u003Cp>There are three major frameworks for understanding acids and bases, each broader than the last:\u003C/p>\u003Cp>\u003Cstrong>Arrhenius definition:\u003C/strong> An acid produces H⁺ ions in water; a base produces OH⁻ ions in water. This is the simplest model but only works for aqueous solutions. HCl is an Arrhenius acid (produces H⁺), and NaOH is an Arrhenius base (produces OH⁻).\u003C/p>\u003Cp>\u003Cstrong>Brønsted-Lowry definition:\u003C/strong> An acid is a proton (H⁺) donor; a base is a proton acceptor. This is more general — it explains acid-base behavior in any solvent, not just water. For example, NH₃ is a Brønsted-Lowry base because it accepts a proton from water: NH₃ + H₂O → NH₄⁺ + OH⁻.\u003C/p>\u003Cp>\u003Cstrong>Lewis definition:\u003C/strong> An acid is an electron-pair acceptor; a base is an electron-pair donor. This is the broadest definition and includes reactions with no proton transfer at all, such as BF₃ + NH₃ → F₃B-NH₃.\u003C/p>\u003Cp>Water is \u003Cstrong>amphoteric\u003C/strong> — it can act as either an acid or a base depending on what it reacts with. With HF, water acts as a base (accepts H⁺). With NH₃, water acts as an acid (donates H⁺).\u003C/p>",[52],164,[54],{"label":55,"body":56},"Three definitions, three zoom levels","\u003Cp>The three frameworks are not rivals; they are nested lenses, each wider than the last. Ammonia shows why the widening was needed: NH₃ contains no OH⁻ to release, so the Arrhenius definition cannot call it a base, yet ammonia solutions are plainly basic. Brønsted-Lowry fixes this by watching the proton: NH₃ accepts one from water, base confirmed. Lewis widens once more to electron pairs, capturing reactions with no proton at all.\u003C/p>\u003Cp>In practice, chemists use the narrowest lens that fits the problem, and general chemistry lives mostly in Brønsted-Lowry territory: if you can track the proton, you can solve this topic.\u003C/p>",{"heading":58,"content":59,"relatedObjectives":60,"deepDive":63},"The pH and pOH Scales","\u003Cp>The concentration of H₃O⁺ (hydronium) ions in solution determines its acidity. Because these concentrations span many orders of magnitude, we use a logarithmic scale:\u003C/p>\u003Cp>\u003Cstrong>pH = −log[H₃O⁺]\u003C/strong>\u003C/p>\u003Cp>\u003Cstrong>pOH = −log[OH⁻]\u003C/strong>\u003C/p>\u003Cp>At 25 °C, these two values are always related by:\u003C/p>\u003Cp>\u003Cstrong>pH + pOH = 14.00\u003C/strong>\u003C/p>\u003Cp>This comes from the \u003Cstrong>ion-product constant of water\u003C/strong>: K\u003Csub>w\u003C/sub> = [H₃O⁺][OH⁻] = 1.0 × 10⁻¹⁴ at 25 °C.\u003C/p>\u003Cp>To convert between pH and ion concentrations:\u003C/p>\u003Cul>\u003Cli>[H₃O⁺] = 10\u003Csup>−pH\u003C/sup>\u003C/li>\u003Cli>[OH⁻] = 10\u003Csup>−pOH\u003C/sup>\u003C/li>\u003C/ul>\u003Cp>A neutral solution has pH = 7.00, acidic solutions have pH < 7, and basic solutions have pH > 7.\u003C/p>",[61,62],165,166,[64,67],{"label":65,"body":66},"Why this matters: a logarithmic scale runs your body","\u003Cp>Each pH unit is a factor of ten, which makes small-looking differences enormous. Rain at pH 5 carries one hundred times the H₃O⁺ of neutral water at pH 7. Your stomach, near pH 2, is over a hundred thousand times more acidic than your blood, which your body defends inside the razor-thin range of 7.35 to 7.45; drifting even a few tenths outside it is a medical emergency. Lemon juice (about 2), black coffee (about 5), baking-soda solution (about 8), and household ammonia (about 11) span nine orders of magnitude of H₃O⁺ on one kitchen counter: the logarithm is what makes such a range speakable.\u003C/p>",{"label":68,"body":69},"Common mistake: sig figs go sideways in a logarithm","\u003Cp>The pH of 0.025 M HCl is 1.60, and the number of decimal places is not an accident. In a logarithm, the digits \u003Cem>before\u003C/em> the decimal point only encode the power of ten (the exponent), so they carry no precision at all; the measurement’s quality lives entirely in the decimals. The rule: the pH gets as many \u003Cem>decimal places\u003C/em> as the concentration has \u003Cem>significant figures\u003C/em>. Two sig figs in 0.025 M means two decimals: 1.60.\u003C/p>\u003Cp>Writing pH 1.6021 for that solution fabricates precision, and writing pH 2 throws it away. The rule runs in reverse too: converting pH 4.35 back to [H₃O⁺] yields two sig figs, 4.5 × 10⁻⁵ M.\u003C/p>",{"heading":71,"content":72,"relatedObjectives":73,"deepDive":74},"Conjugate Acid-Base Pairs","\u003Cp>In every Brønsted-Lowry acid-base reaction, a proton is transferred from the acid to the base. This creates a \u003Cstrong>conjugate pair\u003C/strong>: the acid loses a proton to become its \u003Cstrong>conjugate base\u003C/strong>, and the base gains a proton to become its \u003Cstrong>conjugate acid\u003C/strong>.\u003C/p>\u003Cp class=\"chem-equation\">HF + H₂O ⇌ F⁻ + H₃O⁺\u003C/p>\u003Cp>In this reaction:\u003C/p>\u003Cul>\u003Cli>HF is the acid → F⁻ is its conjugate base\u003C/li>\u003Cli>H₂O is the base → H₃O⁺ is its conjugate acid\u003C/li>\u003C/ul>\u003Cp>\u003Cstrong>Key relationship:\u003C/strong> A strong acid has a very weak conjugate base, and a weak acid has a relatively stronger conjugate base. This inverse relationship is fundamental to understanding acid-base equilibrium.\u003C/p>\u003Cp>Every acid-base reaction has \u003Cem>two\u003C/em> conjugate pairs — always identify both when analyzing a reaction.\u003C/p>",[],[75],{"label":76,"body":77},"Common mistake: conjugate pairs differ by exactly one proton","\u003Cp>A conjugate pair is two species separated by a single H⁺: no more, no less. HF and F⁻ qualify; H₂SO₄ and SO₄²⁻ do not (they differ by two protons, with HSO₄⁻ standing between them as the true conjugate partner of each).\u003C/p>\u003Cp>Water belongs to two different pairs at once, which is the cleanest demonstration that “conjugate” describes a relationship, not an identity: in H₂O/H₃O⁺, water is the base of the pair; in H₂O/OH⁻, water is the acid. One substance, two pairings, one proton of separation in each: that is the whole grammar of the concept.\u003C/p>",{"heading":79,"content":80,"relatedObjectives":81,"deepDive":83},"pH of Strong Acids and Bases","\u003Cp>\u003Cstrong>Strong acids\u003C/strong> ionize completely in water — every molecule donates its proton. The common strong acids are HCl, HBr, HI, HNO₃, HClO₄, and H₂SO₄ (first proton).\u003C/p>\u003Cp>For a strong acid, [H₃O⁺] equals the initial acid concentration:\u003C/p>\u003Cp>\u003Cstrong>Example:\u003C/strong> 0.025 M HCl → [H₃O⁺] = 0.025 M → pH = −log(0.025) = 1.60\u003C/p>\u003Cp>\u003Cstrong>Strong bases\u003C/strong> also dissociate completely. The common ones are Group 1 and 2 hydroxides: NaOH, KOH, Ca(OH)₂, Ba(OH)₂.\u003C/p>\u003Cp>\u003Cstrong>Example:\u003C/strong> 0.010 M NaOH → [OH⁻] = 0.010 M → pOH = 2.00 → pH = 12.00\u003C/p>\u003Cp>For bases with two OH⁻ per formula unit (like Ca(OH)₂), remember to double: 0.005 M Ca(OH)₂ → [OH⁻] = 0.010 M.\u003C/p>\u003Cp>For a \u003Cstrong>mixture\u003C/strong> of two strong-acid (or two strong-base) solutions: convert each pH back to its ion concentration, add the moles of H₃O⁺ (or OH⁻), divide by the total volume, and re-take the log. Never average pH values directly — pH is logarithmic.\u003C/p>",[61,62,82],170,[84],{"label":85,"body":86},"Try it: two strong-acid pH values, one trap","\u003Cp>First, the straightforward one: what is the pH of 0.0050 M HNO₃? Then the trap: is the pH of 1.0 × 10⁻⁸ M HCl equal to 8?\u003C/p>\u003Cp>\u003Cstrong>Answers:\u003C/strong> HNO₃ is strong, so [H₃O⁺] = 0.0050 M and pH = −log(0.0050) = \u003Cstrong>2.30\u003C/strong>. The second answer is \u003Cstrong>no\u003C/strong>: an acid, however dilute, cannot make water basic. At 10⁻⁸ M, the acid contributes less H₃O⁺ than water’s own autoionization (10⁻⁷ M), which dominates; the true pH sits just below 7. The formula pH = −log(acid concentration) quietly assumes the acid is the main H₃O⁺ source, and this is the concentration where that assumption dies.\u003C/p>",{"heading":88,"content":89,"relatedObjectives":90,"deepDive":93},"Weak Acid Equilibrium (Ka)","\u003Cp>Weak acids only partially ionize in water. The extent of ionization is quantified by the \u003Cstrong>acid ionization constant\u003C/strong>, K\u003Csub>a\u003C/sub>:\u003C/p>\u003Cp class=\"chem-equation\">HA + H₂O ⇌ A⁻ + H₃O⁺\u003C/p>\u003Cp>\u003Cstrong>K\u003Csub>a\u003C/sub> = [H₃O⁺][A⁻] / [HA]\u003C/strong>\u003C/p>\u003Cp>A larger K\u003Csub>a\u003C/sub> means a stronger weak acid (more ionization). To find pH from K\u003Csub>a\u003C/sub> and initial concentration:\u003C/p>\u003Col>\u003Cli>Set up an ICE table (Initial, Change, Equilibrium)\u003C/li>\u003Cli>Let x = [H₃O⁺] at equilibrium\u003C/li>\u003Cli>Substitute into the K\u003Csub>a\u003C/sub> expression: K\u003Csub>a\u003C/sub> = x² / (C₀ − x)\u003C/li>\u003Cli>If K\u003Csub>a\u003C/sub> is small relative to C₀, simplify: x ≈ √(K\u003Csub>a\u003C/sub> × C₀)\u003C/li>\u003Cli>pH = −log(x)\u003C/li>\u003C/ol>\u003Cp>Always verify the simplification: if x is more than 5% of C₀, solve the full quadratic instead.\u003C/p>\u003Cp>The \u003Cstrong>inverse direction\u003C/strong> works too: given the pH of a solution of known initial concentration, x = 10\u003Csup>−pH\u003C/sup> gives the equilibrium [H₃O⁺], and K\u003Csub>a\u003C/sub> = x² / (C₀ − x) recovers the ionization constant. The same inversion with pOH recovers K\u003Csub>b\u003C/sub> for a weak base.\u003C/p>",[91,92],168,171,[94],{"label":95,"body":96},"Why this matters: strength and concentration are different axes","\u003Cp>Vinegar is concentrated but weak; the acid in a dilute HCl rinse is dilute but strong. The 5% acetic acid in vinegar is about 0.83 M, over thirty times more concentrated than a 0.025 M HCl solution, yet its pH (about 2.4) is \u003Cem>higher\u003C/em>, because only about half a percent of the acetic acid molecules bother to ionize. Strong versus weak describes the ionization fraction (the K\u003Csub>a\u003C/sub>); concentrated versus dilute describes how much you dissolved. The two axes are independent, and mixing them up is the most common conceptual error in the topic. K\u003Csub>a\u003C/sub> is the number that keeps them straight.\u003C/p>",{"heading":98,"content":99,"relatedObjectives":100,"deepDive":102},"Weak Base Equilibrium (Kb)","\u003Cp>A \u003Cstrong>weak base\u003C/strong> only partially accepts protons from water, establishing an equilibrium:\u003C/p>\u003Cp>B + H\u003Csub>2\u003C/sub>O ⇌ BH\u003Csup>+\u003C/sup> + OH\u003Csup>−\u003C/sup>\u003C/p>\u003Cp>The base-ionisation constant is \u003Cstrong>K\u003Csub>b\u003C/sub> = [BH\u003Csup>+\u003C/sup>][OH\u003Csup>−\u003C/sup>] / [B]\u003C/strong>. A small K\u003Csub>b\u003C/sub> means the base ionises very little and the solution is only mildly basic.\u003C/p>\u003Cp>Solving for pH follows the same \u003Cstrong>ICE-table\u003C/strong> approach as weak acids, but you solve for [OH\u003Csup>−\u003C/sup>] first:\u003C/p>\u003Col>\u003Cli>Set up the ICE table with initial base concentration C\u003Csub>0\u003C/sub>, change −x for B and +x for BH\u003Csup>+\u003C/sup> and OH\u003Csup>−\u003C/sup>.\u003C/li>\u003Cli>If the small-x approximation is valid (x < 5% of C\u003Csub>0\u003C/sub>): x = [OH\u003Csup>−\u003C/sup>] ≈ √(K\u003Csub>b\u003C/sub> · C\u003Csub>0\u003C/sub>).\u003C/li>\u003Cli>Calculate pOH = −log[OH\u003Csup>−\u003C/sup>], then \u003Cstrong>pH = 14.00 − pOH\u003C/strong>.\u003C/li>\u003C/ol>\u003Cp>\u003Cstrong>Example:\u003C/strong> 0.50 M NH\u003Csub>3\u003C/sub> with K\u003Csub>b\u003C/sub> = 1.8 × 10\u003Csup>−5\u003C/sup>: x = √(1.8 × 10\u003Csup>−5\u003C/sup> × 0.50) = 3.0 × 10\u003Csup>−3\u003C/sup> M. Check: 3.0 × 10\u003Csup>−3\u003C/sup>/0.50 = 0.6% < 5%, so the approximation holds. pOH = 2.52, pH = 11.48.\u003C/p>",[101,92],169,[103],{"label":104,"body":105},"Common mistake: stopping at pOH","\u003Cp>Weak-base problems have one extra step that exam adrenaline loves to delete. The ICE table on K\u003Csub>b\u003C/sub> hands you [OH⁻], which gives pOH, and the question almost always asked for pH: subtract from 14.00 (at 25 °C) before writing the answer. A pH of 3 reported for an ammonia solution is this slip in the wild: the solution is basic, so its pH must land \u003Cem>above\u003C/em> 7.\u003C/p>\u003Cp>Bake in the sanity check: bases finish above 7, acids below. Any answer on the wrong side of neutral means a conversion step went missing, and the check costs two seconds.\u003C/p>",{"heading":107,"content":108,"relatedObjectives":109,"deepDive":111},"The Ka–Kb–Kw Relationship","\u003Cp>For any conjugate acid-base pair, the product of their ionization constants equals the ion-product constant of water:\u003C/p>\u003Cp>\u003Cstrong>K\u003Csub>a\u003C/sub> × K\u003Csub>b\u003C/sub> = K\u003Csub>w\u003C/sub> = 1.0 × 10⁻¹⁴\u003C/strong> (at 25 °C)\u003C/p>\u003Cp>This means if you know K\u003Csub>a\u003C/sub> for an acid, you can calculate K\u003Csub>b\u003C/sub> for its conjugate base, and vice versa:\u003C/p>\u003Cul>\u003Cli>K\u003Csub>b\u003C/sub> = K\u003Csub>w\u003C/sub> / K\u003Csub>a\u003C/sub>\u003C/li>\u003Cli>K\u003Csub>a\u003C/sub> = K\u003Csub>w\u003C/sub> / K\u003Csub>b\u003C/sub>\u003C/li>\u003C/ul>\u003Cp>\u003Cstrong>Example:\u003C/strong> Acetic acid has K\u003Csub>a\u003C/sub> = 1.8 × 10⁻⁵. Its conjugate base (acetate, CH₃COO⁻) has K\u003Csub>b\u003C/sub> = (1.0 × 10⁻¹⁴) / (1.8 × 10⁻⁵) = 5.6 × 10⁻¹⁰.\u003C/p>\u003Cp>A useful consequence: the stronger an acid, the weaker its conjugate base — and this is quantitatively precise, not just qualitative.\u003C/p>",[110],167,[112],{"label":113,"body":114},"The seesaw built into Kw","\u003Cp>K\u003Csub>a\u003C/sub> × K\u003Csub>b\u003C/sub> = K\u003Csub>w\u003C/sub> is a seesaw with a fixed pivot: push one side up and the other must come down by exactly the same factor. A strong acid therefore has a spectacularly feeble conjugate base (Cl⁻ has effectively no basic behavior in water), while a very weak acid partners with a respectably strong conjugate base. The slogan “the stronger the acid, the weaker its conjugate base” is not a qualitative memory aid; it is exact arithmetic, with 1.0 × 10⁻¹⁴ as the constant product.\u003C/p>\u003Cp>Practically, the relation halves your table-lookup work: any K\u003Csub>a\u003C/sub> you know is a K\u003Csub>b\u003C/sub> you also know, one division away.\u003C/p>",{"heading":116,"content":117,"relatedObjectives":118,"deepDive":121},"Salt Hydrolysis: Acidic, Basic, or Neutral?","\u003Cp>When a salt dissolves in water, its ions may react with water (hydrolyze) to produce an acidic or basic solution. The outcome depends on the parent acid and base:\u003C/p>\u003Cul>\u003Cli>\u003Cstrong>Strong acid + strong base → neutral salt.\u003C/strong> Example: NaCl from HCl + NaOH. Neither Na⁺ nor Cl⁻ hydrolyzes.\u003C/li>\u003Cli>\u003Cstrong>Strong acid + weak base → acidic salt.\u003C/strong> Example: NH₄Cl from HCl + NH₃. The NH₄⁺ ion donates a proton to water.\u003C/li>\u003Cli>\u003Cstrong>Weak acid + strong base → basic salt.\u003C/strong> Example: CH₃COONa from CH₃COOH + NaOH. The CH₃COO⁻ ion accepts a proton from water.\u003C/li>\u003Cli>\u003Cstrong>Weak acid + weak base → depends on K\u003Csub>a\u003C/sub> vs K\u003Csub>b\u003C/sub>.\u003C/strong> Compare the ionization constants: if K\u003Csub>a\u003C/sub> > K\u003Csub>b\u003C/sub>, the solution is acidic; if K\u003Csub>b\u003C/sub> > K\u003Csub>a\u003C/sub>, it's basic.\u003C/li>\u003C/ul>\u003Cp>To predict the pH, use K\u003Csub>a\u003C/sub> or K\u003Csub>b\u003C/sub> of the hydrolyzing ion (found via the K\u003Csub>a\u003C/sub>·K\u003Csub>b\u003C/sub> = K\u003Csub>w\u003C/sub> relationship).\u003C/p>",[119,120],172,173,[122],{"label":123,"body":124},"Why this matters: hydrolysis in the garden","\u003Cp>Ammonium-based fertilizers gradually acidify soil: NH₄⁺ is the conjugate acid of a weak base, so it donates protons to water exactly as this section predicts, and after seasons of fertilizing, farmers spread crushed limestone to neutralize the accumulated acidity. Baking soda soothes an acidic stomach for the mirror-image reason: HCO₃⁻ is the conjugate base of a weak acid, so its solutions come out basic.\u003C/p>\u003Cp>Neither substance looks like a classic acid or base sitting dry in the bag or box. Salt hydrolysis is the reminder that the label “salt” says nothing about the pH of the solution it will make.\u003C/p>",{"heading":126,"content":127,"relatedObjectives":128},"Percent Ionization","\u003Cp>\u003Cstrong>Percent ionization\u003C/strong> measures how much of a weak acid or base actually ionizes in solution:\u003C/p>\u003Cp>\u003Cstrong>% ionization = ([H₃O⁺]\u003Csub>eq\u003C/sub> / [HA]\u003Csub>initial\u003C/sub>) × 100%\u003C/strong>\u003C/p>\u003Cp>Key observations:\u003C/p>\u003Cul>\u003Cli>Percent ionization \u003Cstrong>increases\u003C/strong> as the solution is \u003Cstrong>diluted\u003C/strong>. A 0.01 M weak acid is more ionized (percentage-wise) than a 1.0 M solution of the same acid.\u003C/li>\u003Cli>A larger K\u003Csub>a\u003C/sub> gives a higher percent ionization at any given concentration.\u003C/li>\u003Cli>For the 5% approximation used in ICE tables to be valid, the percent ionization must be ≤ 5%.\u003C/li>\u003C/ul>\u003Cp>\u003Cstrong>Example:\u003C/strong> If 0.10 M acetic acid gives [H₃O⁺] = 1.34 × 10⁻³ M, then % ionization = (1.34 × 10⁻³ / 0.10) × 100% = 1.34%.\u003C/p>",[91,101],{"heading":130,"content":131,"relatedObjectives":132,"deepDive":133},"Polyprotic Acids","\u003Cp>\u003Cstrong>Polyprotic acids\u003C/strong> have more than one ionizable proton. They ionize in successive steps, each with its own K\u003Csub>a\u003C/sub> value:\u003C/p>\u003Cp>\u003Cstrong>Diprotic acid\u003C/strong> (e.g., H₂SO₃):\u003C/p>\u003Cp class=\"chem-equation\">H₂SO₃ + H₂O ⇌ HSO₃⁻ + H₃O⁺    K\u003Csub>a1\u003C/sub>\u003C/p>\u003Cp class=\"chem-equation\">HSO₃⁻ + H₂O ⇌ SO₃²⁻ + H₃O⁺    K\u003Csub>a2\u003C/sub>\u003C/p>\u003Cp>\u003Cstrong>Triprotic acid\u003C/strong> (e.g., H₃PO₄) has three steps: K\u003Csub>a1\u003C/sub> > K\u003Csub>a2\u003C/sub> > K\u003Csub>a3\u003C/sub>.\u003C/p>\u003Cp>Each successive ionization is always weaker (smaller K\u003Csub>a\u003C/sub>) because it's harder to remove a proton from a species that's already negatively charged. In practice, this means:\u003C/p>\u003Cul>\u003Cli>The first ionization dominates the pH — you can usually ignore the second and third.\u003C/li>\u003Cli>Treat each step as a separate equilibrium, using the equilibrium concentrations from the previous step as the initial concentrations for the next.\u003C/li>\u003C/ul>",[110],[134],{"label":135,"body":136},"Common mistake: doubling the H3O+ for a diprotic acid","\u003Cp>Seeing two ionizable protons on H₂SO₃, students double the hydronium concentration. The successive K\u003Csub>a\u003C/sub> values forbid it: K\u003Csub>a2\u003C/sub> is typically ten thousand times smaller than K\u003Csub>a1\u003C/sub> or more, because pulling a second proton away from a now-negative ion is far harder than the first removal. The second ionization contributes a negligible sliver of H₃O⁺, so for pH purposes a diprotic weak acid is treated as monoprotic using K\u003Csub>a1\u003C/sub> alone.\u003C/p>\u003Cp>The exam-relevant exception is sulfuric acid, whose \u003Cem>first\u003C/em> proton is strong; even there, the second proton is a weak-acid equilibrium, not a free doubling.\u003C/p>",{"heading":138,"content":139,"relatedObjectives":140,"deepDive":141},"Acid-Base Properties of Oxides","\u003Cp>Oxides of elements show a clear pattern in their acid-base behavior that follows the periodic table:\u003C/p>\u003Cul>\u003Cli>\u003Cstrong>Metal oxides\u003C/strong> are typically \u003Cstrong>basic\u003C/strong>. They react with water to form hydroxides: Na₂O + H₂O → 2NaOH. They react with acids to form salt + water.\u003C/li>\u003Cli>\u003Cstrong>Nonmetal oxides\u003C/strong> are typically \u003Cstrong>acidic\u003C/strong>. They react with water to form oxyacids: SO₃ + H₂O → H₂SO₄. They react with bases to form salt + water.\u003C/li>\u003Cli>\u003Cstrong>Amphoteric oxides\u003C/strong> (like Al₂O₃ and ZnO) can react with both acids and bases.\u003C/li>\u003C/ul>\u003Cp>The trend across a period: oxides become more acidic from left to right as metallic character decreases. Down a group, oxides of the same type become more basic as metallic character increases.\u003C/p>\u003Cp>This is directly related to electronegativity: highly electronegative nonmetals form oxides where the O-H bond in the resulting oxyacid is highly polar and breaks easily, releasing H⁺.\u003C/p>",[],[142],{"label":143,"body":144},"Why this matters: acid rain and fizzy oceans are oxide chemistry","\u003Cp>The nonmetal-oxides-are-acidic rule is planetary news. Sulfur dioxide from smelters and coal plants, and nitrogen oxides from engines, meet cloud water and become sulfurous, sulfuric, and nitric acids: acid rain, which mobilized one of the first great international pollution treaties. Meanwhile the oceans absorb a large share of emitted CO₂, and carbonic acid formation has measurably lowered seawater pH, stressing shell-building organisms whose carbonate chemistry you have already met.\u003C/p>\u003Cp>Basic metal oxides close the loop: scrubbers in smokestacks and lime spread on acidified lakes are metal-oxide chemistry deployed as the antidote.\u003C/p>",{"heading":146,"content":147,"relatedObjectives":148},"Acid-Base Problem Selection Framework and Common Mistakes","\u003Cp>Acid-base chapters feel broad because different equations apply to different regimes. Use this selection framework:\u003C/p>\u003Col>\u003Cli>\u003Cstrong>Strong acid/base only?\u003C/strong> Use direct stoichiometry, then pH/pOH from concentration.\u003C/li>\u003Cli>\u003Cstrong>Weak acid/base in water?\u003C/strong> Use K\u003Csub>a\u003C/sub> or K\u003Csub>b\u003C/sub> with ICE-table logic (small-x check).\u003C/li>\u003Cli>\u003Cstrong>Conjugate-pair mixture?\u003C/strong> Use Henderson–Hasselbalch (buffer region).\u003C/li>\u003Cli>\u003Cstrong>Salt solution?\u003C/strong> Identify parent acid/base strength, then hydrolysis behavior.\u003C/li>\u003C/ol>\u003Cp>Quality checks: pH must be between 0 and 14 for standard aqueous problems, strong-acid solutions must yield pH below 7, and weak-acid pH should be higher than an equal-concentration strong acid. Include units and sig figs consistently when converting between [H\u003Csup>+\u003C/sup>], [OH\u003Csup>−\u003C/sup>], pH, and pOH.\u003C/p>",[61,62,91,101,82,92,119,120],{"heading":150,"content":151,"relatedObjectives":152},"The pKa Scale and Its Relation to Ka","\u003Cp>Just as the wide range of [H\u003Csub>3\u003C/sub>O\u003Csup>+\u003C/sup>] concentrations is compressed onto the logarithmic pH scale, the equally wide range of acid ionization constants is often reported as \u003Cstrong>pK\u003Csub>a\u003C/sub>\u003C/strong>:\u003C/p>\u003Cp>\u003Cstrong>pK\u003Csub>a\u003C/sub> = −log\u003Csub>10\u003C/sub>(K\u003Csub>a\u003C/sub>)\u003C/strong>\u003C/p>\u003Cp>The relationship inverts exactly, so a K\u003Csub>a\u003C/sub> can always be recovered from a pK\u003Csub>a\u003C/sub>:\u003C/p>\u003Cp>\u003Cstrong>K\u003Csub>a\u003C/sub> = 10\u003Csup>−pK\u003Csub>a\u003C/sub>\u003C/sup>\u003C/strong>\u003C/p>\u003Cp>Because the logarithm is negative, the scale runs \u003Cem>opposite\u003C/em> to K\u003Csub>a\u003C/sub>: a \u003Cstrong>smaller pK\u003Csub>a\u003C/sub> means a larger K\u003Csub>a\u003C/sub>, and therefore a stronger acid\u003C/strong>. For example, acetic acid (K\u003Csub>a\u003C/sub> = 1.8 × 10\u003Csup>−5\u003C/sup>) has pK\u003Csub>a\u003C/sub> = −log(1.8 × 10\u003Csup>−5\u003C/sup>) = 4.74, while the much stronger hydrofluoric acid (K\u003Csub>a\u003C/sub> = 6.8 × 10\u003Csup>−4\u003C/sup>) has the smaller pK\u003Csub>a\u003C/sub> = 3.17.\u003C/p>\u003Cp>For a \u003Cstrong>polyprotic acid\u003C/strong>, each ionization step has its own constant, so each step also has its own pK\u003Csub>a\u003C/sub>. Since K\u003Csub>a1\u003C/sub> > K\u003Csub>a2\u003C/sub> > K\u003Csub>a3\u003C/sub>, the pK\u003Csub>a\u003C/sub> values increase with each successive step: pK\u003Csub>a1\u003C/sub> < pK\u003Csub>a2\u003C/sub> < pK\u003Csub>a3\u003C/sub>. To convert in the other direction, raise 10 to the negative of each value. For phosphoric acid (H\u003Csub>3\u003C/sub>PO\u003Csub>4\u003C/sub>) with pK\u003Csub>a1\u003C/sub> = 2.12, pK\u003Csub>a2\u003C/sub> = 7.21, and pK\u003Csub>a3\u003C/sub> = 12.32:\u003C/p>\u003Cul>\u003Cli>K\u003Csub>a1\u003C/sub> = 10\u003Csup>−2.12\u003C/sup> = 7.6 × 10\u003Csup>−3\u003C/sup>\u003C/li>\u003Cli>K\u003Csub>a2\u003C/sub> = 10\u003Csup>−7.21\u003C/sup> = 6.2 × 10\u003Csup>−8\u003C/sup>\u003C/li>\u003Cli>K\u003Csub>a3\u003C/sub> = 10\u003Csup>−12.32\u003C/sup> = 4.8 × 10\u003Csup>−13\u003C/sup>\u003C/li>\u003C/ul>\u003Cp>The first step (smallest pK\u003Csub>a\u003C/sub>) is by far the strongest and dominates the pH, consistent with the rule that successive ionizations grow progressively weaker.\u003C/p>",[110],{"title":154,"problem":155,"steps":156,"answer":163},"Finding pH of a Weak Acid Solution","Calculate the pH of a 0.200 M solution of acetic acid (CH₃COOH).\nKa = 1.8 × 10⁻⁵",[157,158,159,160,161,162],"Write the ionization equation: CH₃COOH + H₂O ⇌ CH₃COO⁻ + H₃O⁺","Set up the Ka expression: Ka = [CH₃COO⁻][H₃O⁺] / [CH₃COOH] = x² / (0.200 − x)","Since Ka is small, assume x \u003C\u003C 0.200: Ka ≈ x² / 0.200","Solve for x: x = √(1.8 × 10⁻⁵ × 0.200) = √(3.6 × 10⁻⁶) = 1.90 × 10⁻³ M","Check assumption: 1.90 × 10⁻³ / 0.200 = 0.95% (\u003C 5%, valid)","pH = −log(1.90 × 10⁻³) = 2.72","pH = 2.72",[165,166,167,168,169,170],"OpenStax Chemistry 2e, Section 14.1: Brønsted-Lowry Acids and Bases (CC BY 4.0)","OpenStax Chemistry 2e, Section 14.2: pH and pOH (CC BY 4.0)","OpenStax Chemistry 2e, Section 14.3: Relative Strengths of Acids and Bases (CC BY 4.0)","OpenStax Chemistry 2e, Section 14.4: Hydrolysis of Salts (CC BY 4.0)","OpenStax Chemistry 2e, Section 14.5: Polyprotic Acids (CC BY 4.0)","OpenStax Chemistry 2e, Section 14.6: Buffers (CC BY 4.0)",[172,173,174],"equilibrium","buffers","chemical-reactions-in-solution",[176,177,178,179,180],"H","O","N","Cl","S",[182,183,184,185],"weak-acid-base-constants","strong-acids-bases","acid-base-indicators","titration-curves",[187,191,195,199,203],{"label":188,"equation":189,"note":190},"pH Definition","pH = -log[H3O+]","[H3O+] = 10^(-pH)",{"label":192,"equation":193,"note":194},"Ion-Product Constant of Water","Kw = [H3O+][OH-] = 1.0 x 10^-14","At 25 degrees C; pH + pOH = 14.00",{"label":196,"equation":197,"note":198},"Weak Acid Ka Expression","Ka = [H3O+][A-] / [HA]","Larger Ka means a stronger weak acid",{"label":200,"equation":201,"note":202},"Ka-Kb Relationship","Ka x Kb = Kw","For any conjugate acid-base pair at 25 degrees C",{"label":204,"equation":205,"note":206},"pKa Definition","pKa = -log(Ka)","Ka = 10^(-pKa); smaller pKa = stronger acid",{"title":208,"steps":209},"How to Calculate the pH of a Weak Acid Solution",[210,211,212,213,214,215,216],"Write the ionization equilibrium: HA + H2O ⇌ A- + H3O+.","Write the Ka expression: Ka = [H3O+][A-] / [HA].","Set up an ICE table with the initial acid concentration C0.","Let x = [H3O+] at equilibrium, so [A-] = x and [HA] = C0 - x.","If Ka is small relative to C0, approximate: x = sqrt(Ka x C0).","Verify the approximation: x must be less than 5% of C0. If not, solve the full quadratic.","Calculate pH = -log(x).",{"question":218,"answer":219,"type":220},"Hydrochloric acid (HCl) and acetic acid (CH3COOH) are both prepared at 0.10 M concentration. The HCl solution has pH = 1.00, while the acetic acid has pH = 2.87. Explain why the same concentration gives such different pH values.","HCl is a strong acid that ionizes completely, so all 0.10 mol of HCl produces 0.10 M H3O+, giving pH = 1.00. Acetic acid is a weak acid (Ka = 1.8 x 10^-5) that only partially ionizes. At equilibrium, only a small fraction of CH3COOH molecules donate a proton to water, producing much less H3O+ (about 1.3 x 10^-3 M), which gives a higher pH of 2.87.","conceptual",[222,225],{"id":223,"problem":224,"type":220},"pt-22-1","A 0.20 M solution of a mystery acid HA has a pH of 4.15. Is this a strong acid or a weak acid? Explain your reasoning.",{"id":226,"problem":227,"type":228},"pt-22-2","Calculate the pH of a 0.30 M solution of hydrofluoric acid (HF). Ka for HF = 6.8 x 10^-4.","calculation",[230,231,232,233,234],"acids and bases","pH calculation","Ka Kb","conjugate acid-base pairs","Henderson-Hasselbalch",[236,238,240,243,245,247,249,251,253,255,257,259],{"question":237},"What is an acid and a base according to the Arrhenius definition?",{"question":239},"What is an acid and a base according to the Brønsted–Lowry definition?",{"question":241,"hint":242},"What is a conjugate acid–base pair?","A conjugate pair differs by exactly one proton.",{"question":244},"What is the relationship between pH, pOH, [H⁺], and [OH⁻]?",{"question":246},"What is the difference between a strong acid and a weak acid?",{"question":248},"How do you calculate the pH of a strong acid or strong base solution?",{"question":250},"What is Ka and what does it measure?",{"question":252},"What is Kb and how is it related to Ka through Kw?",{"question":254},"What determines whether a salt solution is acidic, basic, or neutral?",{"question":256},"What is percent ionization and how does it change with concentration?",{"question":258},"What is a polyprotic acid and how does its ionization proceed?",{"question":260},"What is the Henderson–Hasselbalch equation and when is it used?",11,{"id":263,"slug":264,"lesson":263,"title":265,"shortTitle":266,"description":267,"category":10,"objectiveCount":268,"problemCount":269},21,"chemical-equilibrium","Chemical Equilibrium","Equilibrium","Understand chemical equilibrium: equilibrium constants, Le Chatelier's principle, ICE tables, and reaction quotients.",7,56,{"id":271,"slug":272,"lesson":271,"title":273,"shortTitle":274,"description":275,"category":10,"objectiveCount":268,"problemCount":276},23,"buffers-and-titration-curves","Buffers and Titration Curves","Buffers","Understand buffer solutions: Henderson-Hasselbalch equation, buffer capacity, and buffer preparation.",46,1785108608173]