High school · ScienceAbout 220 minutes
ScienceHigh schoolAbout 220 minutes

Describe dynamic equilibrium and its shifts, and explain acids, bases, pH and neutralization

Several sessions: a safe investigation, modeled examples, independent calculation practice and a written explanation. Atomic masses are used to two decimal places; every investigation uses household or teacher-supplied dilute materials, goggles, and no open flame without a supervising adult.

Ohio learning standards this draft targets

science-2018:C.IM.1science-2018:C.IM.3

Read the standard text

science-2018:C.IM.1 Chemical reactions

science-2018:C.IM.3 Stoichiometry

Provisional targets. Clause-by-clause educational review is still pending.

Section 1 · LearnLearn the idea
A single reversible reaction in a closed container, starting with 1.00 mol/L of reactant and no product. The reactant concentration falls to 0.40 mol/L and the product rises to 0.60 mol/L by time 6, and both stay constant afterwards. The curves cross at 0.50 mol/L at time 3, which is not the equilibrium point; equilibrium is where the curves flatten, because that is when the forward and reverse rates are equal. The flat lines do not mean reaction has stopped: both directions continue at the same rate. Values are illustrative.A single reversible reaction in a closed container, starting with 1.00 mol/L of reactant and no product. The reactant concentration falls to 0.40 mol/L and the product rises to 0.60 mol/L by time 6, and both stay constant afterwards. The curves cross at 0.50 mol/L at time 3, which is not the equilibrium point; equilibrium is where the curves flatten, because that is when the forward and reverse rates are equal. The flat lines do not mean reaction has stopped: both directions continue at the same rate. Values are illustrative.Concentrations approaching dynamicequilibrium02468101200.20.40.60.81reactantproductequilibrium reachedequal at 0.50time (arbitrary units)concentration (mol/L)
A single reversible reaction in a closed container, starting with 1.00 mol/L of reactant and no product. The reactant concentration falls to 0.40 mol/L and the product rises to 0.60 mol/L by time 6, and both stay constant afterwards. The curves cross at 0.50 mol/L at time 3, which is not the equilibrium point; equilibrium is where the curves flatten, because that is when the forward and reverse rates are equal. The flat lines do not mean reaction has stopped: both directions continue at the same rate. Values are illustrative.

Learning goal: Describe dynamic equilibrium as two opposing processes occurring at the same rate, predict qualitatively how an equilibrium shifts when temperature, pressure or concentration is changed, identify acids by the formation of hydronium ion in water and bases by the formation of hydroxide ion, calculate pH from hydronium ion concentration and the reverse, and write and interpret neutralization reactions including the idea of a titration.

Reactions that appear to stop. All reactions are reversible to some degree. Many do not run until the reactants are used up; instead the amounts of reactants and products stop changing while some of each remain. The reaction has not stopped. As products build up, the reverse reaction, products turning back into reactants, speeds up, while the forward reaction slows as reactants are used. When the two rates become equal, every product molecule that forms is matched by one that turns back, and the concentrations hold steady. This is dynamic equilibrium: two opposing processes occurring at the same time at the same rate, with no apparent change. Dynamic is the important word. A student who pictures equilibrium as everything sitting still, or as equal amounts of reactants and products, has both ideas wrong: the molecules are reacting continuously in both directions, and the amounts are usually unequal. The diagram shows one reaction approaching equilibrium: the reactant falls from 1.00 to 0.40 and the product rises from 0 to 0.60 over the first six time units and both are flat afterwards, not because reaction has stopped but because the two rates are equal.

Disturbing an equilibrium. If a system at equilibrium is disturbed by a change in conditions, it responds by shifting toward a new equilibrium in the direction that reduces the effect of the change; this is Le Chatelier's principle. Concentration: adding a reactant makes the forward reaction faster for a while, so the system shifts toward products until a new balance is reached; removing a product does the same. Removing a product as it forms therefore forces the system to keep making it, which is how an otherwise unfavorable reaction can be driven to completion. Pressure, for gases only: increasing the pressure by reducing the volume shifts the equilibrium toward the side with fewer moles of gas, because fewer particles exert less pressure; if both sides have the same number of gas moles, pressure has no effect on the position. Temperature: raising the temperature shifts the equilibrium in the endothermic direction, the direction that absorbs the added energy; lowering it favors the exothermic direction. Temperature is the only one of the three that changes the balance point itself (in later courses, the equilibrium constant); concentration and pressure changes shift the mixture to a new composition that satisfies the same balance point. A catalyst speeds both directions equally and does not shift the position; it only shortens the time to reach equilibrium.

A worked model: making ammonia. The synthesis N₂(g) + 3H₂(g) ⇌ 2NH₃(g) is exothermic in the forward direction and reaches equilibrium with much nitrogen and hydrogen unreacted. To maximize ammonia: increase the pressure, since four moles of gas become two, so the forward direction reduces pressure; remove ammonia continuously by cooling it to a liquid, so the system keeps shifting forward; and use a low temperature, which favors the exothermic direction. But a low temperature makes the reaction slow, so in practice a moderate temperature and a catalyst are used, trading some yield for speed. Every prediction follows from asking which direction reduces the disturbance. The mathematics of equilibrium constants is reserved for later study.

Acids and hydronium. Physical Science introduced the properties of acids and bases and the pH scale; this course explains them by structure. An acid is a substance that produces hydronium ions, H₃O⁺, when dissolved in water. This happens when hydrogen is covalently bonded to an electronegative atom, so that the bond is very polar and the hydrogen can leave as a bare proton, H⁺, which immediately bonds to a water molecule's lone pair: HCl + H₂O → H₃O⁺ + Cl⁻. Writing H⁺(aq) is a common shorthand for H₃O⁺. Hydrochloric and nitric acids, and sulfuric acid for its first hydrogen, dissociate essentially completely in water and are called strong; acetic acid in vinegar dissociates only slightly and is weak. A base is a substance that produces hydroxide ions, OH⁻, in water: sodium hydroxide dissociates directly, NaOH → Na⁺ + OH⁻, and ammonia does it by taking a proton from water, NH₃ + H₂O ⇌ NH₄⁺ + OH⁻. The Brønsted-Lowry and Lewis definitions are for later courses.

pH. Even pure water contains a tiny concentration of hydronium, 1.0 × 10⁻⁷ mol/L at 25 °C, balanced by the same concentration of hydroxide; the product of the two concentrations is always 1.0 × 10⁻¹⁴ at that temperature. Because concentrations range over many powers of ten, acidity is reported as pH = −log[H₃O⁺], where the brackets mean concentration in mol/L. A hydronium concentration of 1.0 × 10⁻³ mol/L gives pH 3.00; 1.0 × 10⁻⁷ gives 7.00, neutral; 1.0 × 10⁻¹¹ gives 11.00, basic. Each unit of pH is a factor of ten in concentration, so pH 2 is a hundred times more acidic than pH 4. Going the other way, [H₃O⁺] = 10⁻ᵖᴴ: a solution of pH 5.5 has [H₃O⁺] = 10⁻⁵·⁵ = 3.2 × 10⁻⁶ mol/L. For a strong acid such as 0.010 mol/L hydrochloric acid, which dissociates completely, [H₃O⁺] = 0.010 and pH = 2.00. Indicators are dyes whose color changes over a particular pH range; red cabbage juice runs from red in acid through purple to green and yellow in base, and a pH meter reads the value directly.

Neutralization and titration. An acid reacts with a hydroxide base to form a salt and water: HCl + NaOH → NaCl + H₂O, or H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O, where the hydronium and hydroxide ions combine to make water and the remaining ions form the salt. Antacids work this way: baking soda neutralizes stomach acid, NaHCO₃ + HCl → NaCl + H₂O + CO₂, and magnesium hydroxide does so with Mg(OH)₂ + 2HCl → MgCl₂ + 2H₂O, one formula unit neutralizing two acid particles. Neutralization is studied quantitatively by titration: a solution of known concentration, the standard, is added from a burette to a measured volume of the unknown until the indicator changes color (the endpoint), with the indicator chosen so that the endpoint falls at the equivalence point, where the moles of acid and base match the equation. The volume used, the known concentration and the mole ratio then give the unknown concentration. Suppose 25.00 mL of hydrochloric acid needs 18.40 mL of 0.100 mol/L sodium hydroxide. Moles of NaOH: 0.100 × 0.01840 = 0.00184 mol. The ratio is 1 : 1, so moles of HCl = 0.00184 mol in 0.02500 L, a concentration of 0.0736 mol/L. This is why a standardized solution matters: without a known concentration on one side, the volume tells you nothing.

Acid rain and limestone. Lakes on limestone beds suffer less from acid rain than lakes on granite because calcium carbonate reacts with and neutralizes acid, CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂ in simplified form, while granite's silicate minerals neutralize acid only very slowly. The same reaction erodes marble statues. Roofing and building materials can be tested for acid resistance with dilute vinegar and a balance.

Where the rest of the standards is taught. From C.IM.3 this lesson teaches titration; mole ratios, yields, limiting reagents, molarity and dilution are in the stoichiometry lesson. Reaction types and rates from C.IM.1 are in the two earlier reaction lessons.

Investigation: a household pH survey and a neutralization. Materials: red cabbage indicator (chop cabbage, steep in hot water from a kettle handled by an adult, cool and strain), white vinegar, lemon juice, baking soda solution, soapy water, distilled water, clear cups, a dropper, goggles. Do not use bleach, drain cleaner, oven cleaner or any concentrated acid; they are hazardous and must not be mixed. Steps: add a few drops of indicator to each solution and record the color; arrange the cups from most acidic to most basic; then add baking soda solution dropwise to a cup of vinegar plus indicator, swirling, and count drops until the color reaches the neutral purple. Record colors and drop counts and repeat the neutralization twice. A result that supports the claim: vinegar and lemon juice give red-pink, baking soda and soap give green-blue, distilled water stays purple, and a definite number of drops turns the vinegar neutral, more drops for more vinegar. What the model cannot show: the indicator gives a color range, not a pH value, and drops are not calibrated volumes, so this is a titration in method but not in precision; it cannot show the hydronium ions themselves or the dynamic equilibrium of the weak acid.

Section 2 · ModelWork through the examples

Worked example 1 — recognize equilibrium from data. A sealed flask contains a colored gas mixture. The color deepens for the first minute and then stays constant for an hour. A student concludes the reaction has stopped. The better conclusion is that the observation is consistent with dynamic equilibrium: the constant color means the concentrations are steady, which is what equal forward and reverse rates would produce. Steady color alone does not prove it, because a reaction that had run to completion or simply stalled would look the same. Evidence for equilibrium comes from disturbing it: warming the flask changes the color to a new steady value and cooling it again brings the first color back, as a reversible equilibrium should; experiments with labelled atoms give the direct evidence that both directions continue.

Worked example 2 — predict shifts. For 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), exothermic forward: (a) adding O₂ increases the collision rate of the forward reaction, so the system shifts toward SO₃; (b) compressing the mixture shifts toward the side with fewer gas moles, three on the left against two on the right, so toward SO₃; (c) raising the temperature favors the endothermic direction, which is the reverse, so the amount of SO₃ falls; (d) adding a catalyst shifts nothing but brings the system to equilibrium sooner.

Worked example 3 — pH both ways. (a) A soft drink has [H₃O⁺] = 2.5 × 10⁻⁴ mol/L: pH = −log(2.5 × 10⁻⁴) = 3.60. (b) Blood has pH 7.40: [H₃O⁺] = 10⁻⁷·⁴⁰ = 4.0 × 10⁻⁸ mol/L. (c) 0.0010 mol/L nitric acid, a strong acid, gives [H₃O⁺] = 0.0010 = 1.0 × 10⁻³, pH 3.00. (d) A solution with pH 11.00 has [H₃O⁺] = 1.0 × 10⁻¹¹ and, since the product of the two concentrations is 1.0 × 10⁻¹⁴, [OH⁻] = 1.0 × 10⁻³ mol/L.

Worked example 4 — a titration with a 2 : 1 ratio. 20.00 mL of sulfuric acid requires 30.00 mL of 0.150 mol/L sodium hydroxide. Moles NaOH = 0.150 × 0.03000 = 0.00450 mol. From H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O, one mole of acid needs two of base, so moles of acid = 0.00450 / 2 = 0.00225 mol. Concentration = 0.00225 / 0.02000 = 0.113 mol/L. Skipping the ratio step would double the answer; the equation is part of the calculation.

Section 3 · DoPractice, demonstrate, and explain

Representation task: an equilibrium sketch and a pH ladder. Sketch a graph of concentration against time for a reaction reaching equilibrium, label the point where the rates become equal, and beneath it draw arrows showing the shift for each of four disturbances to the ammonia synthesis. Then draw a pH ladder from 0 to 14, mark the hydronium concentration at each whole number in scientific notation, and place vinegar (about 3), pure water (7), blood (7.4) and a baking soda solution (about 8.5) on it.

Independent performance: pH survey and neutralization report. Carry out the cabbage indicator investigation with at least five safe household solutions and the drop-count neutralization in triplicate. Success criteria: a color table and an ordering from most acidic to most basic with reasons; an equation for the neutralization of acetic acid by sodium hydrogen carbonate; an explanation of why an indicator can order solutions but not give a pH; three pH calculations set by your teacher, in both directions, with the logarithm work shown; a titration calculation from data your teacher supplies that uses the mole ratio from the balanced equation; and a paragraph explaining what dynamic equilibrium means and one way the acetic acid in vinegar illustrates it.

Section 4 · Check8-question lesson practice test

An adult may read the prompts aloud. Answer independently with words, objects, drawings or a demonstration before opening the answer guide. Award 1 point for a correct result or claim and 1 point for a matching explanation or demonstration; where a question asks only for a result (a corrected sentence, a spelling, a count), a fully correct result earns both points, and the answer guide shows what a complete response contains. Record prompting separately. The total (two points per question) is local practice feedback, not an Ohio proficiency level.

Problem 1

Describe the key features of dynamic equilibrium and explain why constant concentrations do not mean the reaction has stopped.

Show worked solution

At dynamic equilibrium two opposing processes, the forward and reverse reactions, occur at the same time and at the same rate. Because every product particle formed is matched by one converted back, the concentrations of reactants and products stay constant, but molecules are continuously reacting in both directions. The amounts of reactants and products are constant, not necessarily equal. Constant amounts alone are consistent with equilibrium but do not prove it; the response to a disturbance, shifting to a new steady state in the direction Le Chatelier's principle predicts, is further evidence, and experiments with labelled atoms show directly that both reactions continue.

Problem 2

The reaction N₂O₄(g) ⇌ 2NO₂(g) is endothermic in the forward direction. Predict the direction of shift when (a) the temperature is raised, (b) the volume is halved, (c) NO₂ is removed.

Show worked solution

(a) Raising the temperature favors the direction that absorbs energy, the endothermic forward direction, so more NO₂ forms. (b) Halving the volume raises the pressure; the system shifts toward the side with fewer gas moles, one mole of N₂O₄ against two of NO₂, so toward N₂O₄. (c) Removing a product lowers the reverse rate, so the system shifts forward to replace NO₂. In each case the shift partly counteracts the disturbance.

Problem 3

Explain what makes a molecule an acid in water, using hydrogen chloride as the example, and write the equation.

Show worked solution

In HCl the hydrogen is covalently bonded to chlorine, a much more electronegative atom, so the bond is very polar and the hydrogen can separate as a proton. In water that proton bonds to a lone pair on a water molecule, forming the hydronium ion: HCl + H₂O → H₃O⁺ + Cl⁻. The hydronium ion is what makes the solution acidic. In ordinary water chemistry, methane is not an acid because its C–H bonds are nearly nonpolar and the hydrogen does not leave (bond polarity is not the whole story for acid strength, which also depends on bond strength and the stability of the ion left behind).

Problem 4

Calculate the pH of a solution with [H₃O⁺] = 4.0 × 10⁻⁵ mol/L, and find [H₃O⁺] for a solution of pH 9.20.

Show worked solution

pH = −log(4.0 × 10⁻⁵) = −(log 4.0 + log 10⁻⁵) = −(0.60 − 5) = 4.40. For pH 9.20, [H₃O⁺] = 10⁻⁹·²⁰ = 6.3 × 10⁻¹⁰ mol/L. The first solution is acidic, below 7; the second is basic, above 7, and its hydroxide concentration is 1.0 × 10⁻¹⁴ / 6.3 × 10⁻¹⁰ = 1.6 × 10⁻⁵ mol/L.

Problem 5

A solution of pH 2 and a solution of pH 5 are compared. How many times more concentrated in hydronium ion is the first, and why?

Show worked solution

pH is a logarithmic scale: each unit of pH is a factor of ten in hydronium concentration. pH 2 corresponds to 1 × 10⁻² mol/L and pH 5 to 1 × 10⁻⁵ mol/L, so the pH 2 solution has 10⁻² / 10⁻⁵ = 10³, one thousand times, the hydronium concentration. Three pH units difference is three factors of ten, not a factor of three.

Problem 6

Write the neutralization equation for magnesium hydroxide with hydrochloric acid and explain why one formula unit of the antacid neutralizes two units of acid.

Show worked solution

Mg(OH)₂ + 2HCl → MgCl₂ + 2H₂O. Each formula unit of magnesium hydroxide releases two hydroxide ions, and each hydroxide combines with one H⁺ from HCl to make one water molecule (H⁺ + OH⁻ → H₂O; written with hydronium it is H₃O⁺ + OH⁻ → 2H₂O), so two acid particles are consumed per formula unit of base. The magnesium and chloride ions that remain form the salt magnesium chloride. Comparing antacids per gram therefore requires both the mole ratio and the molar mass.

Problem 7

In a titration, 25.00 mL of an unknown hydrochloric acid solution needs 21.50 mL of 0.200 mol/L sodium hydroxide to reach the indicator's color change. Find the acid concentration and explain why the base must be a standardized solution.

Show worked solution

Moles of NaOH = 0.200 mol/L × 0.02150 L = 0.00430 mol. HCl and NaOH react 1 : 1, so 0.00430 mol of HCl was in 25.00 mL, giving 0.00430 / 0.02500 = 0.172 mol/L. The calculation converts the base volume to moles using its known concentration; if that concentration were unknown, the volume added could not be turned into moles, and the acid concentration could not be found. A standardized solution is the fixed reference the unknown is measured against.

Problem 8

Explain why lakes on limestone beds are less harmed by acid rain than lakes on granite beds.

Show worked solution

Limestone is largely calcium carbonate, which reacts with acid in a neutralization: CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂ in simplified form, consuming hydronium ions and keeping the lake's pH near neutral. Granite contains no carbonate; its silicate minerals react with acid only very slowly and neutralize little of it, so hydronium ions from acid rain accumulate and the pH falls, harming aquatic life. The limestone acts as a natural antacid; the same reaction is used deliberately when crushed limestone is added to acidified lakes.