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AP Unit 7 Solutions

Worked solutions to all 14 Unit 7 problems. These answers and reasons were added for the AP edition. Check the setup, units, assumptions and explanation alongside the final result.

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Problem 1

A sealed flask at constant temperature contains A(g) ⇌ B(g). A sensor records [A] = 0.80, 0.62, 0.53, 0.50, 0.50 M at 0, 10, 20, 40, 60 s; [B] = 0.20, 0.38, 0.47, 0.50, 0.50 M. Sketch both concentration curves with labeled axes. Explain why the flat part is consistent with dynamic equilibrium rather than a stopped reaction. Give a second reversible physical process that can reach dynamic equilibrium in a closed container.

  1. Put time (s) on the horizontal axis and concentration (M) on the vertical axis. A decreases and B increases toward horizontal plateaus by about 40 s; do not join the data with an invented oscillation.
  2. Equal forward and reverse conversion rates give zero net concentration change. The equal 0.50 M concentrations are a feature of this example, not a requirement for equilibrium.
  3. In a sealed container with liquid water and water vapor at fixed temperature, evaporation and condensation continue at equal rates at vapor–liquid equilibrium. Constant sensor readings alone cannot prove microscopic motion, so this interpretation also relies on the reversible-process model.

Final answer

Curves plateau at about 40 s; the two processes continue at equal rates. Evaporation/condensation is a valid physical example.

What earns credit
  • Graph both concentration series with time and concentration units and correct plateau behavior.
  • Explain equal nonzero forward/reverse rates while distinguishing rates from concentrations.
  • Provide a valid reversible physical process and acknowledge that constant readings alone do not prove microscopic equilibrium.

Watch out: Horizontal concentration curves mean zero forward and reverse rates.

Return to problem 1 → · Review Topic 7.1

Problem 2

For A(g) ⇌ B(g), the forward and reverse rates at one instant are 0.030 and 0.012 mol L−1 s−1. Determine the signs and magnitudes of the instantaneous changes in [A] and [B]. A student says the reaction is irreversible because the net change is forward. Evaluate the claim.

  1. Subtract competing rates: net forward rate = 0.030 − 0.012 = 0.018 mol L−1 s−1.
  2. For 1:1 stoichiometry, d[B]/dt = +0.018 M s−1 and d[A]/dt = −0.018 M s−1.
  3. A nonzero reverse rate is explicitly given. Net forward conversion means the forward process currently wins; it does not eliminate reverse events. Equilibrium is reached only when the rates become equal.

Final answer

d[A]/dt = −0.018 M s−1; d[B]/dt = +0.018 M s−1. The process remains reversible.

What earns credit
  • Subtract reverse rate from forward rate to obtain 0.018 M s−1 net progress.
  • Assign opposite signs to the 1:1 species changes.
  • Reject irreversibility by identifying the explicitly nonzero reverse rate.

Watch out: A net direction excludes simultaneous reverse events.

Return to problem 2 → · Review Topic 7.2

Problem 3

  1. Write Qc for CaCO3(s) ⇌ CaO(s) + CO2(g), and Qp for N2(g) + 3H2(g) ⇌ 2NH3(g). Explain each omitted term and exponent. When can either quotient be called its corresponding equilibrium constant?
  2. For H2O(l) ⇌ H2O(g), write Qp for the ideal water vapor above a pure-liquid phase. Explain why one water phase appears but the other does not.
  3. For HA(aq) + H2O(l) ⇌ H3O+(aq) + A−(aq) in a dilute aqueous solution, write Qc using the usual solvent-activity approximation. Explain why the dissolved HA is not omitted with the solvent.

Assume the stated pure phases are present and the gas and solution approximations are appropriate. Use the AP concentration/partial-pressure forms; thermodynamic activities are referenced to their standard states.

  1. For the carbonate reaction, Qc = [CO2]. Both pure solids have constant activity while their phases are present, so their terms are absorbed into the constant.
  2. For ammonia formation, Qp = (PNH3)2 / (PN2(PH2)3). Balanced coefficients become exponents; use individual partial pressures, not total pressure.
  3. Qc = Kc or Qp = Kp only for equilibrium values at the stated temperature. A gas concentration n/V is not omitted from Qc because its state label is (g).
  4. For water liquid–vapor equilibrium, the AP pressure form is Qp = PH2O. Pure H2O(l) has constant activity; the vapor has a variable partial pressure and remains. In a dimensionless thermodynamic expression the gas pressure is divided by its reference pressure.
  5. For the dilute weak-acid solution, Qc = [H3O+][A−]/[HA]. Water is the solvent and its nearly constant activity is incorporated into the equilibrium constant. HA is a dissolved solute with a variable concentration, so it remains. The omission is about phase/activity, not simply whether a formula is written as a liquid or contains hydrogen.

Final answer

Carbonate: Qc = [CO2]. Ammonia: Qp = (PNH3)2/(PN2(PH2)3). Water liquid–vapor: Qp = PH2O. Weak acid: Qc = [H3O+][A−]/[HA]. Pure condensed phases and a nearly constant solvent are omitted; variable gases/solutes remain.

What earns credit
  • Write carbonate Qc including gaseous CO2 and omitting both pure solids for constant-activity reasons.
  • Write ammonia Qp with individual partial pressures and coefficient exponents; distinguish current Q from equilibrium K.
  • Omit pure H2O(l) but retain water-vapor partial pressure and explain the phase/activity distinction.
  • Omit the nearly constant dilute solvent but retain dissolved HA, hydronium and A−.

Watch out: Qc includes only aqueous species and excludes every gas.

Return to problem 3 → · Review Topic 7.3

Problem 4

For 2NO2(g) ⇌ N2O4(g), equilibrium samples at one temperature give [NO2] = 0.0400 M and [N2O4] = 0.120 M. Calculate Kc. A second run reports 0.0600 M and 0.270 M, respectively. Does it support the same Kc? Explain why using the initial NO2 concentration would invalidate this determination.

  1. Kc = [N2O4]/[NO2]2 = 0.120/(0.0400)2 = 75.0.
  2. The second ratio is 0.270/(0.0600)2 = 75.0, consistent with the same equilibrium constant at that temperature.
  3. Initial and equilibrium measurements describe different states. All species must be measured at equilibrium in the same mixture; substituting an initial value produces neither the actual equilibrium quotient nor a valid K.

Final answer

Both runs give Kc = 75.0.

What earns credit
  • Use the squared NO2 denominator with simultaneous equilibrium measurements.
  • Calculate both runs as Kc = 75.0 and explain fixed-temperature agreement.
  • Reject mixing initial reactant and equilibrium product values.

Watch out: Use equilibrium product concentrations with initial reactant concentrations.

Return to problem 4 → · Review Topic 7.4

Problem 5

Two 1:1 isomerizations A ⇌ B begin with only A. At the same temperature Kc is 2.0 × 105 for one and 4.0 × 10−6 for the other. Compare their equilibrium compositions. Can these K values tell you which reaches equilibrium first? Justify both answers.

  1. For either 1:1 equation, Kc = [B]/[A]. The first equilibrium is overwhelmingly B-rich; the second is overwhelmingly A-rich.
  2. A large K does not require literally zero A. A small K does not require literally zero B.
  3. K describes composition at equilibrium. Rate constants and activation barriers determine the time to get there, so no rate ordering follows from these values.

Final answer

The first is product-rich, the second reactant-rich; relative speed cannot be determined from K.

What earns credit
  • Interpret the 1:1 K values as equilibrium concentration ratios and identify each favored side.
  • Avoid equating finite K with an exactly absent species.
  • Separate composition from the activation-barrier/rate question.

Watch out: A larger equilibrium constant means a faster reaction.

Return to problem 5 → · Review Topic 7.5

Problem 6

At one temperature, A ⇌ B has K1 = 4.0 and B ⇌ C has K2 = 9.0. Determine K for (a) C ⇌ A and (b) 2A ⇌ 2C. For a current mixture with [A] = 0.20 M, [B] = 0.30 M, [C] = 0.60 M, show that the analogous Q expressions multiply for A ⇌ C. Explain why multiplying an equation by two does not merely double K.

  1. Add the first two equations and cancel B: A ⇌ C, K = K1K2 = 36. Reverse the sum for C ⇌ A: K = 1/36 = 0.028, two significant figures.
  2. Double A ⇌ C: 2A ⇌ 2C, so K = 362 = 1.3 × 103, two significant figures.
  3. Q1 = [B]/[A] = 1.5; Q2 = [C]/[B] = 2.0. Their product is 3.0 = [C]/[A], because the B factors cancel.
  4. Doubling coefficients doubles each exponent. The whole quotient is squared; multiplying by two would not describe the new equation.

Final answer

(a) K = 0.028; (b) K = 1.3 × 103; Qoverall = 3.0.

What earns credit
  • Multiply component K values after cancelling B, invert for C ⇌ A, and square for doubled coefficients.
  • Demonstrate the analogous current Q multiplication and intermediate cancellation.
  • Explain coefficient scaling through quotient exponents rather than doubling K.

Watch out: Add K values when adding reactions or multiply K by a stoichiometric multiplier.

Return to problem 6 → · Review Topic 7.6

Problem 7

A(g) ⇌ B(g) has Kc = 4.00 at a fixed temperature. A sealed constant-volume flask initially has [A] = 0.300 M and [B] = 0.200 M. Determine the net direction and both equilibrium concentrations. Check the ratio and atom balance.

  1. Q = 0.200/0.300 = 0.667 < 4.00, so net conversion is A to B.
  2. Let x M convert forward. [A]eq = 0.300 − x and [B]eq = 0.200 + x; one mole of A produces one mole of B.
  3. 4.00 = (0.200 + x)/(0.300 − x), so 1.200 − 4x = 0.200 + x; x = 0.200 M.
  4. [A]eq = 0.100 M, [B]eq = 0.400 M. Both are nonnegative, the sum stays 0.500 M, and 0.400/0.100 = 4.00.

Final answer

[A]eq = 0.100 M; [B]eq = 0.400 M; net forward.

What earns credit
  • Compute initial Q and choose net forward progress.
  • Use opposite 1:1 changes, solve the equilibrium ratio and obtain both concentrations.
  • Check nonnegative values, total concentration and Q = K.

Watch out: Use the initial concentrations directly as the equilibrium concentrations.

Return to problem 7 → · Review Topic 7.7

Problem 8

  1. For N2O4(g) ⇌ 2NO2(g), Kc = 0.100 at the stated temperature. Initially [N2O4] = 0.100 M and [NO2] = 0. Calculate both equilibrium concentrations. A student assumes x is negligible compared with 0.100 M. Check that approximation.
  2. A second sealed, constant-volume flask at the same temperature starts with [N2O4] = 0.0500 M and [NO2] = 0.200 M. Calculate its initial Qc and net reaction direction. Build a stoichiometric change table, solve for both equilibrium concentrations, and explain which root is physically possible. Verify conserved nitrogen units and final Qc = Kc.

Treat both gas mixtures with the same ideal concentration model. The flasks are independent samples.

  1. For flask (a), initially Q = 0 < K, so dissociation is net forward. Let x M N2O4 dissociate: final concentrations are 0.100 − x and 2x.
  2. 0.100 = (2x)2/(0.100 − x), giving 4x2 + 0.100x − 0.0100 = 0. The physical root is x = 0.03904 M; the negative root would give negative NO2. Final concentrations are 0.06096 M N2O4 and 0.07808 M NO2, reported as 0.0610 and 0.0781 M.
  3. Neglecting depletion would predict x = √(0.100 × 0.100/4) = 0.0500 M, 50% of initial N2O4. That is not small, so the complete expression is needed.
  4. For flask (b), initial Qc = (0.200)2/0.0500 = 0.800 > 0.100. The net reverse reaction consumes NO2 and forms N2O4. Choose x as the molarity of N2O4 formed.
Flask (b): molar concentration bookkeeping
StageN2O4NO2
Initial0.05000.200
Change+x−2x
Equilibrium0.0500 + x0.200 − 2x
  1. 0.100 = (0.200 − 2x)2/(0.0500 + x), so 4x2 − 0.900x + 0.0350 = 0. Its roots are 0.0500 and 0.175 M. Only 0 ≤ x ≤ 0.100 M is possible; the larger root would give [NO2] = −0.150 M.
  2. With x = 0.0500 M, final [N2O4] = 0.100 M and [NO2] = 0.100 M. Check Qc = (0.100)2/0.100 = 0.100.
  3. Conserved nitrogen-atom concentration is 2[N2O4] + [NO2]: initially 2(0.0500) + 0.200 = 0.300 M, and finally 2(0.100) + 0.100 = 0.300 M. Oxygen-atom concentration is twice this value and is also conserved.

Final answer

(a) [N2O4]eq = 0.0610 M and [NO2]eq = 0.0781 M; the small-change approximation fails. (b) Qc initially = 0.800, so net reverse; the physical x = 0.0500 M gives both equilibrium concentrations 0.100 M.

What earns credit
  • For flask (a), use −x and +2x, solve the complete quadratic and reject the negligible-depletion approximation.
  • For flask (b), calculate Qc = 0.800 > Kc and use +x for N2O4 and −2x for NO2.
  • Solve the reverse-change quadratic and reject x = 0.175 M because it gives negative NO2.
  • Verify both nonnegative final concentrations, conserved nitrogen units and Qc = 0.100.

Watch out: A small K automatically makes x negligible for any initial concentration.

Return to problem 8 → · Review Topic 7.7

Problem 9

In equal-volume boxes for the isomerization A ⇌ B, box I contains 8 A and 2 B particles, box II contains 4 A and 6 B, and box III contains 2 A and 8 B. Each symbol represents the same number of molecules. Kc = 4.0. Identify the equilibrium box, predict the direction for the others, and describe what a second snapshot of the equilibrium box would show. State a limitation of these small-number pictures.

Draw a later equilibrium snapshot with a legend and the same volume/particle scale. Include a forward and a reverse event arrow, and explain what stays constant on average.

  1. For equal-volume 1:1 isomers, Q = number B/number A. The boxes give 0.25, 1.5, and 4.0.
  2. Only box III has Q = K. I and II proceed net forward because Q < K.
  3. A later equilibrium snapshot has the same average proportion, but particular molecules may have changed identity; a single tiny snapshot may fluctuate.
  4. Small drawings discretize huge populations and cannot by themselves give rate constants or prove dynamic events. Atom balance must still be respected; isomers contain the same atoms.
  5. A valid later average snapshot contains 2 A and 8 B in the same volume, with one event arrow A → B and another B → A. Individual particles can change identity; the average population remains in the 4:1 equilibrium proportion.

Final answer

III is the equilibrium representation; I and II move net toward B. Equilibrium preserves average counts while identities continue to change.

What earns credit
  • Compute each scaled 1:1 particle ratio and identify the equilibrium box and the other net directions.
  • Create a valid later same-scale particle representation with event arrows and conserved atoms.
  • Explain dynamic identity changes and a small-number/statistical limitation.

Watch out: A diagram at equilibrium must have equal symbol counts or fixed molecule identities.

Return to problem 9 → · Review Topic 7.8

Problem 10

For 2NO2(g) ⇌ N2O4(g), the forward reaction is exothermic; NO2 is brown and N2O4 is colorless. Predict the equilibrium response to (a) heating at fixed volume and (b) decreasing volume at fixed temperature. In (b), distinguish the immediate change in NO2 concentration from its later change during re-equilibration. Explain why the final brown intensity relative to the original cannot be concluded just from the word shift.

  1. Heating favors the endothermic reverse direction, producing more NO2; the brown component increases at the fixed volume.
  2. Compression first raises both gas concentrations. The later net response forms N2O4, the side with fewer gas particles.
  3. During this later response [NO2] falls relative to its immediate post-compression value, but its final concentration is higher than before compression in the closed ideal-gas model at fixed T. Let c = [NO2]eq and d = [N2O4]eq = Kc c². Conserved nitrogen units per volume are C = c + 2d = c + 2Kc c². Compression raises C; the right side increases monotonically with c, so the new equilibrium c must be larger. Brown intensity also depends on optical path and geometry.
  4. A shift describes chemical conversion after the stress, not the full observed effect of the stress plus response.

Final answer

(a) Net reverse, more brown NO2. (b) Immediate concentration increase, then a partial decrease during net forward response. Final equilibrium [NO2] exceeds its original value in the stated ideal model; brown intensity comparisons also require optical path control.

What earns credit
  • Relate heating to the endothermic reverse direction and brown NO2.
  • Separate immediate compression from subsequent net forward conversion.
  • Use the conserved nitrogen-unit relation to establish increased final NO2 versus the original, while identifying optical-path conditions for measured intensity.

Watch out: A shift consuming NO2 guarantees its final concentration is lower than before compression.

Return to problem 10 → · Review Topic 7.9

Problem 11

  1. A(g) ⇌ 2B(g) is at equilibrium with Kp = 0.800. The volume is doubled instantly at constant temperature. Determine the new Qp in terms of its old equilibrium value and predict the net direction. Explain separately the changes to Q and K if reactant A is added at fixed volume, or if temperature changes.
  2. For a different equilibrium, A(aq) + B(aq) ⇌ AB(aq), Qc initially equals Kc. In separate trials at constant temperature, (i) half the dissolved AB is selectively removed without changing solution volume or other species amounts, and (ii) enough pure water is added to double the volume. For each, obtain the immediate Qc as a multiple of Kc and predict the later net direction. Assume dilute solutions and no competing reactions.
  3. A saturated solution has MX(s) ⇌ M+(aq) + X−(aq). In separate trials, pure MX solid is added or half the existing solid is removed. A macroscopic amount of the same solid phase remains in both trials, and dissolved concentrations, temperature and solution volume are unchanged by the manipulation. Predict whether either manipulation changes Qsp or the saturated dissolved composition. Why does the condition that solid remains matter?
  1. For the gas, doubling volume halves both reacting partial pressures. Qnew = (PB/2)2/(PA/2) = Qold/2 = 0.400. Fixed temperature keeps Kp = 0.800, so net forward reaction forms more B.
  2. Adding gas A at fixed volume raises the denominator and lowers Q while K remains constant at fixed temperature. Changing temperature changes K; the physical concentrations or pressures can also change, so compare the appropriate new Q and new K.
  3. For dissolved A + B ⇌ AB, Qc = [AB]/([A][B]). Selectively removing half the product gives Qnew = Kc/2. Qnew < Kc, so the later net reaction forms AB. This response need not replace all of the removed amount.
  4. Doubling solution volume halves all three concentrations immediately: Qnew = ([AB]/2)/(([A]/2)([B]/2)) = 2Kc. Qnew > Kc, so the later net reaction dissociates AB into A and B. A blanket rule that dilution always favors products is incorrect.
  5. For MX, Qsp = [M+][X−]. Pure solid does not appear as a variable term. Adding solid or removing part of it leaves Qsp equal to Ksp and does not change the saturated dissolved composition under the stated conditions. It changes the solid amount, not its activity.
  6. If all solid were removed, there would be no solid reservoir available to dissolve following a later stress. Saturation with a present solid phase must not be assumed without that phase; the given unchanged-composition conclusion concerns the manipulations specified, with solid remaining.

Final answer

Gas expansion: Qp = 0.400, net forward. Selective AB removal: Qc = Kc/2, net forward. Aqueous dilution: Qc = 2Kc, net reverse. Changing pure-solid amount gives no Qsp/composition response while that phase remains and solution conditions stay fixed.

What earns credit
  • Apply the volume factor to every reacting gas pressure and distinguish fixed-temperature Q changes from temperature-dependent K.
  • For selective product removal, obtain Kc/2 and predict AB formation.
  • For aqueous dilution, halve all concentrations, obtain 2Kc and predict AB dissociation.
  • Explain unchanged pure-solid activity and dissolved quotient for both solid manipulations; retain the solid-present and unchanged-solution conditions.

Watch out: Every stress changes the equilibrium constant.

Return to problem 11 → · Review Topic 7.10

Problem 12

At a stated temperature, MX2(s) ⇌ M2+(aq) + 2X−(aq) has Ksp = 3.20 × 10−11. Find its molar solubility in pure water, ignoring side reactions. Then a separate salt NY(s) ⇌ N+ + Y− has molar solubility 3.00 × 10−5 M; calculate its Ksp and compare the two salts’ molar solubilities. Explain why sorting Ksp values alone is insufficient across different formulas.

For a third salt described by the same course model, explain what Ksp > 1 implies for the qualitative soluble/insoluble classification, and identify a limitation of concentration-based predictions when the resulting solution is concentrated.

  1. For MX2, [M2+] = s and [X−] = 2s. Ksp = s(2s)2 = 4s3, so s = (3.20 × 10−11/4)1/3 = 2.00 × 10−4 M.
  2. For NY, Ksp = s2 = (3.00 × 10−5)2 = 9.00 × 10−10.
  3. MX2 is more soluble in mol L−1 despite having the smaller Ksp. Different stoichiometries produce different powers and numerical factors.
  4. A very large Ksp, such as a value greater than 1 in the course’s standard-state model, is consistent with substantial dissolution/soluble behavior. Solubility rules give qualitative expectations; the balanced dissolution expression gives the quantitative interpretation.
  5. Ksp > 1 is consistent with soluble behavior in the AP standard-state model. A concentrated solution can have significant ion interactions, so activities may differ from molarity and the ideal concentration calculation can become inaccurate.

Final answer

s(MX2) = 2.00 × 10−4 M; Ksp(NY) = 9.00 × 10−10; MX2 has the larger molar solubility.

Ksp > 1 corresponds to soluble behavior in this course model; concentrated-solution activity effects can limit concentration-only predictions.

What earns credit
  • Write Ksp = 4s³ for MX2 and s² for NY using ion stoichiometry.
  • Calculate both requested quantities and correctly rank molar solubility despite the Ksp ordering.
  • Relate very large Ksp to soluble behavior while explaining why formula-dependent exponents matter.
  • Answer the explicitly requested large-Ksp classification and concentrated-solution model limitation.

Watch out: All ion concentrations equal the molar solubility, regardless of formula.

Return to problem 12 → · Review Topic 7.11

Problem 13

Equal 50.0 mL volumes of 2.00 × 10−4 M M+ and 4.00 × 10−4 M X− solutions are mixed. For MX(s) ⇌ M+ + X−, Ksp = 1.00 × 10−8. Assuming additive volumes, predict whether precipitation begins and justify using Qsp. If a student uses the concentrations before mixing, what error does that produce?

A separate saturated pure-water sample of MX at the same temperature contains undissolved solid and measured [M+] = [X−] = 1.00 × 10−4 M. Determine its molar solubility and infer Ksp from this measurement.

  1. Total volume is 100.0 mL, so concentrations immediately after mixing are 1.00 × 10−4 M M+ and 2.00 × 10−4 M X−.
  2. Qsp = (1.00 × 10−4)(2.00 × 10−4) = 2.00 × 10−8 > Ksp; net precipitation reduces the ion product toward Ksp.
  3. Using premixing values gives 8.00 × 10−8, four times too large. The prediction happens to stay the same here, but the method is wrong and can reverse a prediction near the threshold.
  4. The separate pure-water saturated sample has s = [M+] = [X−] = 1.00 × 10−4 M because the dissolution is 1:1 and there are no other ion sources. Ksp inferred from the saturated measurement is s² = 1.00 × 10−8, consistent with the stated constant.

Final answer

Precipitation begins because Qsp = 2.00 × 10−8 > 1.00 × 10−8.

The separate saturated sample has s = 1.00 × 10−4 M and measured Ksp = 1.00 × 10−8.

What earns credit
  • Dilute both stock-ion concentrations into the combined volume before forming Qsp.
  • Obtain Qsp = 2.00 × 10−8 and predict precipitation from Qsp > Ksp.
  • Quantify the fourfold premixing-concentration error and explain its potential to alter a near-threshold decision.
  • Use the separate saturated pure-water measurement to obtain molar solubility and infer Ksp with the 1:1 stoichiometry.

Watch out: Use stock-solution concentrations directly after mixing.

Return to problem 13 → · Review Topic 7.11

Problem 14

MX(s) ⇌ M+(aq) + X−(aq) has Ksp = 4.00 × 10−10 at 25 °C. Compare its molar solubility in pure water with that in 0.0200 M NaX. Show and check an approximation. Predict how replacing water with NaX solution would change the measured dissolved M+ concentration in a controlled solubility investigation, and whether Ksp itself changes.

  1. In pure water, s = √Ksp = 2.00 × 10−5 M.
  2. In NaX solution, [M+] = s and [X−] = 0.0200 + s. Ksp = s(0.0200 + s). If s ≪ 0.0200, s ≈ 4.00 × 10−10/0.0200 = 2.00 × 10−8 M.
  3. The neglected fraction is s/0.0200 = 1.00 × 10−6, so this approximation is excellent under the stated ideal dilute model.
  4. A common ion lowers the dissolved M+ concentration by roughly a factor of 1000 here. Ksp remains fixed at the same temperature; use the total equilibrium X concentration when checking it.

Final answer

Pure-water s = 2.00 × 10−5 M; common-ion s ≈ 2.00 × 10−8 M. Ksp does not change.

What earns credit
  • Find pure-water s from √Ksp and include both common-ion sources in s(c+s).
  • Calculate and check the negligible-s approximation quantitatively.
  • Predict the lower measured M+ concentration while keeping fixed-temperature Ksp unchanged.

Watch out: A common ion lowers Ksp rather than changing equilibrium concentrations.

Return to problem 14 → · Review Topic 7.12

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