Equilibrium · Lesson 14 of 14

Chapter Summary and Practice

“Connect the complete chapter through one equilibrium framework and a broad set of mixed problems.”

Learning Objectives

• Connect physical, chemical, acid–base and solubility equilibria using equal opposing rates and equilibrium ratios. • Select K, Q, Gibbs-energy, pH or solubility methods for unfamiliar situations. • Keep thermodynamic direction, equilibrium extent and reaction speed distinct. • Check standard states, stoichiometry, charge, dilution and approximations in numerical work. • Solve mixed problems spanning all major ideas in the chapter.

One idea links all the equilibria

A sealed liquid–vapour system, an ammonia synthesis mixture, a weak-acid solution and a saturated salt solution look different. Each nevertheless has competing processes whose rates balance at equilibrium. The observable state becomes steady while molecular exchange or reaction continues. The equilibrium relation changes form because the participating species and phases differ, not because equilibrium acquires a new meaning.

For physical equilibria, look for a fixed vapour pressure or saturation concentration at specified conditions. In chemical equilibria, the balanced equation determines an activity ratio. In water, proton transfer produces Ka, Kb and Kw expressions. In a saturated salt solution, dissolution gives Ksp. The same quotient comparison underlies all these applications. The final practice asks you to identify which version fits the situation.

Question to answer	Tool	Important check
Has equilibrium been reached?	Equal rates and Q = K	Constant values are not necessarily equal values
What is the final composition?	Stoichiometric change table and K	Accept only physically possible roots
Which way does the current mixture change?	Compare Q with K	Use the same reaction and conventions
Is the change thermodynamically favoured?	Actual reaction Gibbs energy	Do not replace actual with standard blindly
What happens after a disturbance?	Recalculate Q or consider changed K	Only temperature generally changes K here
What is the pH?	Kw with the appropriate acid/base equilibrium	Neutral pH is temperature dependent
Will a solid precipitate?	Compare Qsp with Ksp	Use free ions after mixing and dilution

Example — A physical equilibrium after expansion

Problem
A liquid and its vapour are at equilibrium at fixed temperature. The gas space expands. What has changed and what will happen?

1.The immediate vapour concentration and pressure fall because the same gas amount occupies more volume.
2.Condensation initially slows relative to evaporation. More liquid enters the vapour phase.
3.If liquid remains, the equilibrium vapour pressure returns to the same temperature-dependent value. This is the physical counterpart of changing a quotient while retaining the same equilibrium condition.

Choose a written reaction and conserve material

Balance atoms and charges before writing K. Reversing a reaction reciprocates K; multiplying coefficients by a factor raises K to that factor. For ideal gases, replace partial pressure by cRT to connect the two conventional numerical forms. In heterogeneous systems omit pure-phase activities but keep the phases in the physical interpretation. Components of a liquid mixture are not all automatically unit-activity pure liquids.

Chemical-equilibrium toolkitLaTeX

K = \frac{\prod a _{products}^{ν}}{\prod a _{reactants}^{ν}}, K_{reverse} = K^{- 1}, K_{scaled} = K^{n}, K_{p} = K_{c} (R T)^{Δ n_{g}}

The last relation is the conventional numerical ideal-gas form in matched units. The standard-state form includes RTc°/p° rather than dimensional RT alone.

Example — A simple equilibrium calculation

Problem
Initially 1.00 M A is present with no B for $A ⇌ B$ , and Kc = 4.00. Find both equilibrium concentrations.

1.Let x M A convert into B. The concentrations become $1.00 - x$ and x.
2.Set $x / (1.00 - x) = 4.00$ . Rearranging gives $5 x = 4.00$ .
3.Thus B is 0.800 M and A is 0.200 M. Their ratio is 4.00 and their sum preserves the initial 1.00 M total.

Separate direction, extent and speed

K tells us the equilibrium ratio at the specified temperature. Q tells us how the current mixture compares with it. Actual reaction Gibbs energy expresses the same direction thermodynamically. A catalyst changes how quickly the destination is approached, without changing K. Heating can change both K and speed, so do not apply one conclusion to both questions.

Direction and thermodynamicsLaTeX

Δ_{r} G = R T ln (Q / K), Δ_{r} G^{\circ} = - R T ln K

Q < K gives negative actual reaction Gibbs energy. Standard Gibbs energy and actual Gibbs energy are equal only when Q = 1.

Example — A standard-favourable reaction can run backwards

Problem
At 298 K a reaction has K = 100, but the current mixture has Q = 1000. Determine the direction and signs of its Gibbs-energy quantities.

1.The standard change is $- R T ln 100$ , which is negative because K exceeds one.
2.The actual change is $R T ln (1000/100) = R T ln 10 \approx + 5.71 kJ mo l^{- 1}$ .
3.The net reaction is reverse even though the standard reaction Gibbs energy is negative. Current composition supplies the missing information.

For concentration changes, use the expression for Q to see why adding reactant or removing product often promotes forward change. Compression favours fewer gaseous moles in the ideal-gas model; an inert gas at fixed volume leaves reacting partial pressures unchanged. Heating favours an endothermic direction. The FeSCN2+ and NO2 colour observations make these shifts visible. Ammonia and sulfur trioxide production show why equilibrium yield and acceptable reaction rate must be considered together.

Follow protons and choose the right pH model

Arrhenius descriptions focus on hydronium and hydroxide in water. Brønsted–Lowry descriptions identify proton donors, acceptors and conjugate partners. Lewis descriptions extend the discussion to electron-pair acceptors and donors, such as BF3 and ammonia. Electrolyte strength, acid strength, concentration and the number of transferable protons describe different properties.

For simple hydrides, bond weakening down a group and increasing polarity across a period explain different acidity trends. Polyprotic acids have sequential constants, usually decreasing as negative charge builds. Common ions suppress additional ionization. Salt ions can themselves be weak acids or bases. These ideas explain why dissolving a salt need not produce a neutral solution.

Acid–base toolkitLaTeX

K_{w} = [H^{+}] [O H^{-}], pH + pOH = p K_{w}, K_{a} K_{b} = K_{w}, K_{a} = \frac{c α ^{2}}{1 - α}

These are the conventional dilute concentration forms. The final formula is for a simple initially acid-only monoprotic solution, not an arbitrary buffer or common-ion mixture.

Example — Checking a weak-acid approximation

Problem
For a 0.100 M monoprotic acid with $K_{a} = 1 0^{- 5}$ at 298 K, estimate pH and check the assumption.

1.The square-root estimate gives $[H^{+}] \approx 1 0^{- 5} \times 0.100 = 0.00100 M$ .
2.The fraction ionized is $0.00100/0.100 = 0.0100$ , or 1.00%, supporting the denominator approximation.
3.The approximate pH is 3.00. If the same calculation had given a large fraction or an ion concentration near that of pure water, a more complete model would be required.

A pH reading must be interpreted at its temperature. Pure water is neutral when hydronium and hydroxide balance, even if pH is not seven. Indicators change colour because their two conjugate forms have different colours. pH paper estimates that change, while a calibrated meter measures an electrical response. Scientific interpretation combines the observation with the equilibrium model rather than relying on the colour alone.

Buffers and solubility connect several equilibria at once

For a buffer, perform strong-acid or strong-base neutralization first and then use the surviving conjugate pair. Hydrolysis explains the starting pH of many salt solutions. Solubility calculations need free-ion concentrations, which may be altered by protonation, hydrolysis or complex formation. The common-ion effect is the same equilibrium principle in both ionization and dissolution problems.

Buffer and solubility toolkitLaTeX

pH \approx p K_{a} + lo g \frac{[ A ^{-} ]}{[ HA ]}, pH \approx p K_{a} (B H^{+}) + lo g \frac{[ B ]}{[ B H ^{+} ]}, K_{s p} = (m S)^{m} (n S)^{n}

The S formula applies only to the simple salt model without initial common ions or important side reactions. For precipitation compare the current ion product with Ksp.

Example — Using post-reaction buffer amounts

Problem
An acid buffer has 30 mmol HA and 20 mmol A− with pKa = 4.80. After adding 5 mmol NaOH, estimate its pH if volume changes are unimportant to the ratio.

1.Hydroxide consumes HA and forms A−, leaving 25 mmol of each partner.
2.Their concentration ratio is one because both occupy the same final volume.
3.The new pH is approximately pKa, or 4.80. A calculation using the original ratio would ignore neutralization.

Example — Precipitation after dilution

Problem
Equal volumes of a $2.0 \times 1 0^{- 5}$ M solution of M2+ and a $2.0 \times 1 0^{- 5}$ M solution of X2− are mixed. For MX, Ksp = $4.0 \times 1 0^{- 11}$ . Predict precipitation.

1.Immediately after mixing, both ion concentrations are $1.0 \times 1 0^{- 5} M$ .
2.The ion product is $Q_{s p} = 1 0^{- 5} \times 1 0^{- 5} = 1 0^{- 10}$ .
3.Because this exceeds $4.0 \times 1 0^{- 11}$ , precipitation is favoured under the simple concentration model. The final concentrations adjust towards the saturation relation if solid and solution coexist.

Use logarithms and make a final reasonableness check

The logarithm and antilogarithm reference tables appended to the PDF are calculation aids. For a number written as $m \times 1 0^{n}$ , its base-ten logarithm is $lo g_{10} m + n$ . An antilogarithm reverses the operation. For example, pH 3.76 gives hydrogen-ion concentration $1 0^{- 3.76} = 1.74 \times 1 0^{- 4} M$ in the dilute approximation. Use the full power of ten, not just a table mantissa. Natural logarithms in Gibbs-energy equations differ from base-ten logarithms by the factor 2.303.

Before accepting an answer, check atom and charge balance, the total mixed volume, pressure and energy units, the direction predicted by Q, and the physical range of every concentration and ionized fraction. Reinsert calculated equilibrium values into the defining expression. Identify which neglected quantities are actually small. An internally consistent model is more useful than a memorised formula applied outside its assumptions.

Common mistake

Keep these distinctions visible: equal rates versus equal concentrations; K versus Q; actual versus standard Gibbs energy; strength versus concentration; free-ion concentration versus total dissolved amount; and pH versus pOH. Each pair answers a different question.

Quick check

Which feature is shared by physical and chemical dynamic equilibria?

Quick check

If a reaction equation is reversed and then doubled, its new constant is

Quick check

A reaction with Q > K has net change

Quick check

A catalyst can help a synthesis by

Quick check

At any temperature, neutral water is defined by

Quick check

A weak acid with smaller pKa is generally

Quick check

Which must be done before applying a buffer ratio after strong-acid addition?

Quick check

For a salt MX2 in the simple model, ion concentrations are

Quick check

Which comparison can favour dissolution of a salt with a basic anion?

Quick check

For a conjugate pair at 298 K, pKa + pKb is approximately

Mixed Chapter Practice

A liquid and its vapour are at equilibrium in a sealed vessel at fixed temperature. The gas volume suddenly increases. Explain the immediate pressure and rate changes and the final pressure if liquid remains.
For 2SO2(g) + O2(g) forming 2SO3(g), calculate Kc from equilibrium concentrations 0.60 M SO2, 0.82 M O2 and 1.90 M SO3.
Iodine vapour at total pressure 1.00 bar contains 40% atomic iodine by gas volume at equilibrium for I2(g) forming 2I(g). Find dimensionless Kp using p° = 1 bar.
Construct Kc expressions for 2NOCl(g) forming 2NO(g) + Cl2(g); decomposition of solid Cu(NO3)2 to solid CuO, NO2(g) and O2(g); dilute aqueous ester hydrolysis; Fe3+(aq) + 3OH−(aq) forming Fe(OH)3(s); and I2(s) + 5F2(g) forming 2IF5(g). State the phase assumptions.
Convert supplied numerical Kp values to Kc for 2NOCl(g) forming 2NO(g) + Cl2(g), Kp = $1.8 \times 1 0^{- 2}$ at 500 K, and CaCO3(s) forming CaO(s) + CO2(g), Kp = 167 at 1073 K. Use bar–litre units consistently.
For NO(g) + O3(g) forming NO2(g) + O2(g), Kc = $6.3 \times 1 0^{14}$ at 1000 K. Find the constant for the reverse reaction.
Explain why pure solid and pure liquid activities are omitted from an equilibrium expression, and why this is not a rule for omitting every component labelled liquid in a mixture.
For 2N2(g) + O2(g) forming 2N2O(g), Kc = $2.0 \times 1 0^{- 37}$ . Starting with 0.482 mol N2 and 0.933 mol O2 in 10.0 L, estimate the equilibrium composition, justifying any small-change approximation.
Initially 0.0870 mol NO and 0.0437 mol Br2 react by 2NO + Br2 forming 2NOBr. At equilibrium 0.0518 mol NOBr is present. Find the remaining reactant amounts.
For 2SO2(g) + O2(g) forming 2SO3(g), the numerical Kp is $2.0 \times 1 0^{10}$ with pressures in bar at 450 K. Find Kc with concentrations in moles per litre.
Pure HI begins at pressure 0.200 atm and its equilibrium partial pressure is 0.0400 atm for 2HI(g) forming H2(g) + I2(g). Find Kp using a consistent pressure convention.
At 500 K, a 20.0 L vessel contains 1.57 mol N2, 1.92 mol H2 and 8.13 mol NH3. For ammonia synthesis Kc = $1.7 \times 1 0^{2}$ . Test for equilibrium and predict any net direction.
Write a balanced gas reaction whose Kc expression is [NH3]^4[O2]^5 divided by [NO]^4[H2O]^6.
One mole each of H2O(g) and CO(g) is placed in 10.0 L at 725 K. For H2O + CO forming H2 + CO2, 40% of the original water reacts. Calculate Kc.
For H2(g) + I2(g) forming 2HI(g), Kc = 54.8 at 700 K. A preparation began with HI only and has equilibrium [HI] = 0.500 M. Find [H2] and [I2].
Pure ICl begins at 0.780 M. For 2ICl(g) forming I2(g) + Cl2(g), Kc = 0.140. Find all equilibrium concentrations.
For C2H6(g) forming C2H4(g) + H2(g) at 899 K, the numerical Kp is 0.0400 in atm units. Pure C2H6 begins at 4.00 atm at fixed volume. Find its equilibrium pressure and concentration using R = 0.082057 litre atm per mole kelvin.
For acetic acid + ethanol forming ethyl acetate + water in one liquid mixture, retain water as a varying component. Write the ideal concentration quotient. At 293 K, starting amounts 1.00 mol acid and 0.180 mol ethanol give 0.171 mol ester at equilibrium. Find K and test whether a mixture started with 1.00 mol acid and 0.500 mol ethanol has reached equilibrium when 0.214 mol ester has formed. Use the stated ideal-mixture approximation.
For PCl5(g) forming PCl3(g) + Cl2(g) at 473 K, Kc = $8.3 \times 1 0^{- 3}$ . A pure-PCl5 preparation has equilibrium [PCl5] = 0.0500 M. Find the two product concentrations.
For FeO(s) + CO(g) forming Fe(s) + CO2(g), dimensionless Kp = 0.265 at 1050 K. Initial partial pressures are 1.40 atm CO and 0.800 atm CO2. Find their equilibrium pressures while both solids remain.
At 500 K, Kc for ammonia synthesis is 0.0610. A mixture has [N2] = 3.0 M, [H2] = 2.0 M and [NH3] = 0.50 M. Use Q to predict its direction.
For 2BrCl(g) forming Br2(g) + Cl2(g), Kc = 32.0 at 500 K. Starting from $3.3 \times 1 0^{- 3}$ M BrCl only, find the equilibrium BrCl concentration.
At 1127 K and 1.00 atm, a CO/CO2 mixture in equilibrium with solid carbon contains 90.55% CO by mass. For C(s) + CO2(g) forming 2CO(g), calculate Kc. Convert mass fractions into mole fractions before calculating partial pressures.
At 298 K, standard formation Gibbs energies are 52.0 kJ per mole for NO2, 87.0 kJ per mole for NO and zero for O2. Find the standard Gibbs energy and K for NO(g) + one-half O2(g) forming NO2(g).
Predict the effect of expansion at fixed temperature on product amounts for PCl5(g) forming PCl3(g) + Cl2(g), CaO(s) + CO2(g) forming CaCO3(s), and 3Fe(s) + 4H2O(g) forming Fe3O4(s) + 4H2(g).
Predict compression effects for COCl2(g) decomposition to CO + Cl2; CH4(g) + 2S2(g) forming CS2(g) + 2H2S(g); C(s) + CO2(g) forming 2CO(g); 2H2(g) + CO(g) forming CH3OH(g); CaCO3(s) decomposition; and 4NH3(g) + 5O2(g) forming 4NO(g) + 6H2O(g).
At 1024 K, K = $1.6 \times 1 0^{5}$ for H2(g) + Br2(g) forming 2HBr(g). Pure HBr begins at 10.0 bar at fixed volume. Find all equilibrium partial pressures.
For endothermic CH4(g) + H2O(g) forming CO(g) + 3H2(g), write Kp and discuss separately how compression, heating and a catalyst affect composition, K and speed.
For 2H2(g) + CO(g) forming CH3OH(g), predict the effects of adding H2, adding methanol, removing CO and removing methanol at fixed temperature and volume.
For endothermic PCl5(g) decomposition to PCl3(g) + Cl2(g), Kc = $8.3 \times 1 0^{- 3}$ at 473 K and the standard reaction enthalpy is +124.0 kJ per mole. Write Kc, find the reverse constant and discuss the effects on Kc of added PCl5, compression and heating.
For CO(g) + H2O(g) forming CO2(g) + H2(g), Kp = 10.1 at 400 degrees Celsius. Starting partial pressures of CO and steam are each 4.00 bar with no products. Find the equilibrium hydrogen pressure.
Which equilibrium is likely to have appreciable reactant and product amounts: Cl2(g) forming 2Cl(g), Kc = $5 \times 1 0^{- 39}$ ; Cl2 + 2NO forming 2NOCl, Kc = $3.7 \times 1 0^{8}$ ; or Cl2 + 2NO2 forming 2NO2Cl, Kc = 1.8? Explain the use and limits of this comparison.
At 298 K, Kc = $2.0 \times 1 0^{- 50}$ for 3O2(g) forming 2O3(g). If equilibrium [O2] = $1.6 \times 1 0^{- 2}$ M, calculate [O3].
For CO(g) + 3H2(g) forming CH4(g) + H2O(g), Kc = 3.90 at 1300 K. A 1.00 L equilibrium vessel has 0.30 mol CO, 0.10 mol H2 and 0.020 mol H2O. Find methane concentration.
Define a conjugate pair and give the appropriate conjugate acid or base of HNO2, CN−, HClO4, F−, OH−, $C O_{3}^{2 -}$ and S2−.
Discuss Lewis-acid behaviour for H2O, BF3, H+ and NH4+. Identify the accepting site in any proton-transfer interpretation rather than assigning an expanded octet to nitrogen.
Find the conjugate bases of HF, H2SO4 and HCO3−.
Find the conjugate acids of NH2−, NH3 and HCOO−.
For H2O, HCO3−, HSO4− and NH3, give both a conjugate acid and a conjugate base with correct charges.
Classify OH−, F−, H+ and BCl3 as Lewis acids or bases and explain the relevant lone-pair behaviour.
Calculate the dilute-model pH of a soft drink with [H+] = $3.8 \times 1 0^{- 3}$ M.
Convert vinegar pH 3.76 into hydrogen-ion concentration using the dilute approximation.
Using the supplied exercise data Ka(HF) = $6.8 \times 1 0^{- 4}$ , Ka(HCOOH) = $1.8 \times 1 0^{- 4}$ and Ka(HCN) = $4.8 \times 1 0^{- 9}$ at 298 K, calculate each conjugate-base Kb. Treat these as the explicitly given datasets rather than substituting a different reference value.
For phenol, Ka = $1.0 \times 1 0^{- 10}$ . Find phenolate concentration in 0.050 M phenol, then find its degree of ionization when 0.010 M sodium phenolate is also present.
Use Ka1(H2S) = $9.1 \times 1 0^{- 8}$ and Ka2 = $1.2 \times 1 0^{- 13}$ to estimate HS− and S2− in 0.10 M H2S, first alone and then with 0.10 M HCl. State the approximations.
For 0.050 M acetic acid with Ka = $1.74 \times 1 0^{- 5}$ , calculate degree of ionization, acetate concentration and pH.
A 0.010 M monoprotic organic acid has pH 4.15. Find its anion concentration, Ka and pKa.
Assuming complete dissociation and dilute behaviour at 298 K, calculate pH for 0.0030 M HCl, 0.0050 M NaOH, 0.0020 M HBr and 0.0020 M KOH.
Using complete-dissociation concentration models, find pH at 298 K for 2.0 g TlOH in 2.00 L; 0.300 g Ca(OH)2 in 500 mL; 0.300 g NaOH in 200 mL; and 1.00 mL of 13.6 M HCl diluted to 1.00 L. State the molar masses you use.
A 0.100 M bromoacetic-acid solution has degree of ionization 0.132. Calculate pH and pKa without replacing one minus alpha by one.
A 0.0050 M codeine solution has pH 9.95 at 298 K. Determine Kb and pKb.
For 0.0010 M aniline with Kb = $4.27 \times 1 0^{- 10}$ at 298 K, determine pH, degree of ionization and the conjugate-acid Ka, checking water’s contribution.
Acetic acid has pKa = 4.74. Calculate its ionized fraction in a 0.050 M solution, then estimate the fraction with 0.010 M HCl and with 0.100 M HCl added as part of the final solution composition.
Dimethylamine has Kb = $5.4 \times 1 0^{- 4}$ . Find its degree of ionization in 0.020 M solution and its percentage ionization when that solution also contains 0.100 M NaOH.
Find approximate hydrogen-ion concentrations for biological-fluid pH values 6.83, 1.20, 7.38 and 6.40, corresponding to the supplied muscle-fluid, stomach-fluid, blood and saliva examples.
Find approximate hydrogen-ion concentrations corresponding to the sample pH values milk 6.8, coffee 5.0, tomato juice 4.2, lemon juice 2.2 and egg white 7.8.
At 298 K, 0.561 g KOH is dissolved to make 200 mL of solution. Find potassium, hydroxide and hydrogen-ion concentrations and pH in the complete-dissociation concentration model.
The supplied Sr(OH)2 solubility is 19.23 g per litre at 298 K. Find strontium and hydroxide concentrations and a concentration-model pH. State that activity corrections may matter at the resulting concentration.
For 0.050 M propanoic acid with Ka = $1.32 \times 1 0^{- 5}$ , find degree of ionization and pH. Then find the ionized fraction when the final solution also contains 0.010 M HCl.
A 0.100 M HCNO solution has pH 2.34. Find its ionization constant and degree of ionization using the specified data.
For 0.040 M sodium nitrite, calculate pH and degree of hydrolysis at 298 K using Ka(HNO2) = $4.5 \times 1 0^{- 4}$ .
A 0.020 M pyridinium chloride solution has pH 3.44 at 298 K. Determine the conjugate-base Kb of pyridine.
Predict whether NaCl, KBr, NaCN, NH4NO3, NaNO2 and KF solutions are acidic, neutral or basic, and identify the relevant ions.
For chloroacetic acid with Ka = $1.35 \times 1 0^{- 3}$ , find the pH of 0.100 M acid and of its 0.100 M sodium salt solution at 298 K. Check approximation validity separately.
At 310 K, Kw = $2.7 \times 1 0^{- 14}$ . Calculate the pH of neutral water and explain why it differs from seven.
At 298 K, analyse these mixtures: 10.0 mL of 0.200 M Ca(OH)2 with 25.0 mL of 0.100 M HCl; 10.0 mL of 0.0100 M H2SO4 with 10.0 mL of 0.0100 M Ca(OH)2; and 10.0 mL of 0.100 M H2SO4 with 10.0 mL of 0.100 M KOH. First show the fully dissociated stoichiometric model; then discuss the sulfuric-acid cases using Ka2 = 0.012 and the resulting sulfate or bisulfate equilibrium.
Using the reference table, calculate formal simple-model solubilities and ion concentrations for Ag2CrO4, BaCrO4, Fe(OH)3, PbCl2 and Hg2I2 at 298 K. Test water autoionization explicitly for the very insoluble hydroxide, and discuss the limits from hydrolysis and other side reactions before treating a formal result as a physical prediction.
With Ksp(Ag2CrO4) = $1.1 \times 1 0^{- 12}$ and Ksp(AgBr) = $5.0 \times 1 0^{- 13}$ , calculate the ratio of their molar solubilities in the simple model.
Equal volumes of 0.0020 M NaIO3 and 0.0020 M Cu(ClO3)2 are mixed. Does Cu(IO3)2 precipitate if Ksp = $7.4 \times 1 0^{- 8}$ ? Account for dilution.
Use Ka(benzoic acid) = $6.46 \times 1 0^{- 5}$ and Ksp(silver benzoate) = $2.5 \times 1 0^{- 13}$ . Estimate how many times more soluble the salt is in a maintained pH 3.19 buffer than the simple pure-water reference.
Equimolar FeSO4 and Na2S solutions are mixed in equal volumes. Find the largest initial concentration each can have without exceeding Ksp(FeS) = $6.3 \times 1 0^{- 18}$ , using the stipulated free-ion model and neglecting sulfide hydrolysis.
Estimate the minimum water volume needed to dissolve 1.00 g CaSO4 at 298 K using Ksp = $9.1 \times 1 0^{- 6}$ , the appropriate molar mass and the simple solubility model.
A stipulated acidic H2S solution has free sulfide concentration $1.0 \times 1 0^{- 19}$ M. Mix 10.0 mL of it with 5.00 mL of 0.0400 M FeSO4, MnCl2, ZnCl2 or CdCl2, considered separately. Use Ksp values FeS $6.3 \times 1 0^{- 18}$ , MnS $2.5 \times 1 0^{- 13}$ , ZnS $1.6 \times 1 0^{- 24}$ and CdS $8.0 \times 1 0^{- 27}$ to predict precipitation under the simple diluted-free-ion model. Explain why a fully re-equilibrated H2S model must also update protonation and pH.

Key Takeaways

All chapter equilibria combine continuing molecular processes with a steady macroscopic condition. Choose the equilibrium expression from the balanced chemistry and the phases actually present. Use Q for current direction, K for the equilibrium ratio, and kinetic information for speed. Track proton transfer, charge balance and common ions in acid–base and salt solutions. For reliable calculations, state assumptions, check approximations and compare the final result with the defining equilibrium relation.

Previous · Lesson 13

Solubility Products and Precipitation

End of chapter

One idea links all the equilibria

Choose a written reaction and conserve material

Separate direction, extent and speed

Follow protons and choose the right pH model

Buffers and solubility connect several equilibria at once

Use logarithms and make a final reasonableness check

Quiz

Practice Problems

Key Takeaways