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Química Geral · Atoms, moles and stoichiometry

Estequiometria em solução e titulações

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O que você vai aprender

  • Calculate molarity, moles and solute mass for a solution, and plan how to prepare it accurately, including from hydrates and concentrated stock acids.
  • Use C₁V₁ = C₂V₂ for single and serial dilutions, and find concentrations after mixing solutions that do or do not react.
  • Calculate the concentration of each ion in a salt solution and the amounts left after a precipitation.
  • Use the balanced equation to turn titration volumes into concentrations, masses and purities for acid–base and precipitation titrations.
  • Analyse a back-titration and diagnose common errors in preparing solutions and titrating.

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Molarity

Molarity is moles of solute per litre of solution: c = n/V, in mol/L (M). The volume is that of the final solution, not of the water added. Example: 4.00 g of NaOH (40.00 g/mol) is 0.100 mol; made up to 250.0 mL of solution it is 0.100 mol ÷ 0.2500 L = 0.400 M. Rearranged, n = c × V gives moles from a volume, and mass = n × M gives the mass to weigh. Millimoles are convenient: mL × mol/L = mmol, so 25.0 mL of 0.400 M NaOH holds 10.0 mmol.

Preparing a solution

Weigh the solute, dissolve it in less water than the final volume in a volumetric flask, add water up to the mark and mix. For a hydrate, use the molar mass of the hydrate, because the water of crystallization is part of what you weigh: 100.0 mL of 0.0500 M CuSO₄ needs 0.00500 mol of CuSO₄·5H₂O (249.69 g/mol), which is 1.248 g. To use a concentrated stock acid given as a mass percent and a density, find the mass of acid needed, divide by the mass fraction to get the mass of solution, then divide by the density to get its volume. When the mass needed is too small to weigh accurately on the balance available, weigh a larger mass to make a concentrated stock and dilute part of it.

Dilution and mixing

Adding water changes the volume but not the moles of solute, so C₁V₁ = C₂V₂, with V₂ the final total volume. Example: 20.0 mL of 0.500 M diluted to 100.0 mL gives 0.500 × 20.0/100.0 = 0.100 M; the water added is 80.0 mL, not 100.0 mL. In a serial dilution the dilution factors multiply. When two solutions are mixed, add the moles of each species and divide by the total volume (dilute aqueous volumes are taken as additive). If the mixed solutes react, do the reaction stoichiometry in moles first, then divide what is left by the total volume: 10.0 mmol HCl plus 4.0 mmol NaOH in 50.0 mL leaves 6.0 mmol H₃O⁺, 0.12 M.

Ion concentrations and precipitation

A strong electrolyte dissociates completely, and each ion's concentration is the salt concentration times the number of that ion in the formula: 0.200 M K₂SO₄ is 0.400 M K⁺ and 0.200 M SO₄²⁻, 0.600 M ions in total. Concentrations of different ions are not added together unless you want the total ion concentration. When two salt solutions mix and a precipitate forms (Ag⁺ + Cl⁻ → AgCl, Ba²⁺ + SO₄²⁻ → BaSO₄), the ion present in the smaller stoichiometric amount is used up, the excess of the other remains, and the spectator ions are only diluted into the total volume.

Titrations

A titration adds a solution of known concentration (the titrant) from a burette until the equivalence point, where the titrant added exactly matches the analyte by the balanced equation; an indicator's colour change (the end point) marks it. Moles of titrant = c × V, then the mole ratio, then the unknown. HCl and NaOH react 1 : 1, but H₂SO₄ needs 2 NaOH and Ba(OH)₂ neutralizes 2 HCl. A NaOH solution is standardized against a primary standard, a pure, stable, non-hygroscopic solid such as potassium hydrogen phthalate (KHP, drawn, 204.22 g/mol, one acidic H): 0.4084 g of KHP is 2.000 mmol, so if it needs 20.00 mL of NaOH, the NaOH is 2.000 mmol ÷ 20.00 mL = 0.1000 M. In the Mohr precipitation titration, chloride is titrated with AgNO₃ (Ag⁺ + Cl⁻ → AgCl, 1 : 1) until red Ag₂CrO₄ appears. Repeat titrations and average only concordant titres.

Back-titration

When the analyte is an insoluble solid or a volatile base, or reacts too slowly for a sharp end point, add a measured excess of a standard reagent, let the reaction finish, then titrate the reagent that is left over. Reagent consumed = reagent added − reagent left. Example: a sample of Mg(OH)₂ is dissolved in 25.00 mL of 0.2000 M HCl (5.000 mmol); the excess needs 15.00 mL of 0.1000 M NaOH (1.500 mmol). The Mg(OH)₂ consumed 5.000 − 1.500 = 3.500 mmol HCl, and Mg(OH)₂ + 2HCl → MgCl₂ + 2H₂O gives 1.750 mmol Mg(OH)₂, or 1.750 × 58.32 = 102.1 mg.

What to carry forward

Work in moles (or millimoles): n = c × V for solutions and n = m/M for solids. Molarity uses the final volume of solution; dilution conserves moles (C₁V₁ = C₂V₂); mixing adds moles over the total volume, after any reaction has run. Multiply by the subscript for ion concentrations, and in every titration or back-titration let the balanced equation supply the mole ratio.

Problemas de prática (40)

Praticar é grátis com uma conta.

Preveja o produto 1

  • What is left after barium sulfate precipitates?Predict every ion concentration after mixing solutions that form a precipitate, including the excess ion and the spectators.

Classificar 1

  • Choosing a primary standardClassify substances as suitable or unsuitable primary standards for standardizing a base.

Comparar 4

Dilution and mixing

  • Rank diluted solutionsRank several diluted solutions by final concentration using C₁V₁ = C₂V₂.

Molarity and making solutions

  • Rank solutions by molarityRank solutions by molarity after converting masses and millimoles to moles per litre.

Ions in solution and precipitation

  • Rank by total ion concentrationRank equal-molarity solutions by total concentration of dissolved ions.
  • Which solution is 0.300 M in chloride?Compare chloride concentrations of salts with different formulas to find one matching a target value.

Calcular 22

Dilution and mixing

  • Excess acid after mixingFind the concentration of the excess reagent after mixing a strong acid with a strong base.
  • Concentration after dilutionCalculate the concentration of a solution after diluting it to a stated final volume.
  • Volume of stock acid for a dilutionUse C₁V₁ = C₂V₂ to find the volume of a stock solution needed for a dilution.
  • Mixing two NaCl solutionsFind the concentration after mixing two solutions of the same solute with unequal volumes.
  • Two-step serial dilutionCalculate the final concentration after a two-step serial dilution by multiplying the dilution factors.

Ions in solution and precipitation

  • Sulfate in aluminium sulfate solutionCalculate the concentration of the polyatomic anion in a solution of a salt with several anions per formula unit.
  • Chloride in a calcium chloride sampleCalculate the amount of one ion in a measured volume of a salt solution.
  • Chloride from two saltsFind the concentration of an ion supplied by two different salts after their solutions are mixed.
  • Mass of silver chloride precipitatedFind the limiting ion and the mass of precipitate when two salt solutions are mixed.

Titrations and back-titrations

  • Nitrogen by ammonia back-titrationCalculate the mass of nitrogen in a sample from a back-titration of absorbed ammonia.
  • Calcium carbonate in an antacid tabletCalculate the mass percent of an insoluble base from back-titration data.
  • Volume of barium hydroxide to the equivalence pointPredict the titrant volume at equivalence when the base supplies two hydroxide ions per formula unit.
  • Purity of sodium carbonateFind the mass percent of a dibasic salt in an impure sample from its titration with strong acid.
  • Concentration of HCl by titrationCalculate the concentration of a strong acid from a 1 : 1 titration with standard base.
  • Standardizing NaOH with KHPCalculate the exact concentration of a NaOH solution from a titration against a weighed primary standard.
  • Purity of salt by Mohr titrationCalculate the mass percent of a chloride salt in a sample from a precipitation titration with silver nitrate.
  • Titrating sulfuric acidUse a 1 : 2 mole ratio to find the concentration of a diprotic acid from titration data.

Molarity and making solutions

  • Diluting concentrated hydrochloric acidUse the mass percent and density of a concentrated stock acid to find the volume needed for a dilute solution.
  • Molarity of a sodium chloride solutionCalculate molarity from the mass of solute and the final volume of solution.
  • Mass of NaOH for a solutionCalculate the mass of solute needed to prepare a given volume of a solution of known molarity.
  • Topping up a stock solutionFind the extra solute needed when an existing solution is combined with solid and diluted to a new concentration.
  • Sodium ion from washing sodaCalculate an ion concentration in a solution prepared from a hydrated salt.

Encontre o erro 3

Titrations and back-titrations

  • A wet buretteDiagnose how rinsing a burette with water instead of titrant biases a titration result.

Molarity and making solutions

  • Is it really 0.250 M?Identify why molarity must be based on the final volume of solution rather than the volume of water used.

Dilution and mixing

  • Spectator ions after neutralizationDiagnose errors in the spectator-ion concentrations after a diprotic acid is exactly neutralized.

Interprete as evidências 1

  • Which titres to average?Select concordant titres from a set of titration results and use their mean to calculate a concentration.

Trabalhe de trás para frente 3

Titrations and back-titrations

  • Molar mass of an unknown diprotic acidWork back from titration data to the molar mass of an unknown acid with two acidic protons.

Ions in solution and precipitation

  • Salt mass for a target sodium concentrationWork back from a target ion concentration to the mass of salt to dissolve.

Dilution and mixing

  • How much water to add?Work back from a target concentration to the volume of water that must be added to a solution.

Planeje uma síntese 3

Dilution and mixing

  • Accurate tenfold dilutionChoose the glassware and procedure for an accurate dilution.

Molarity and making solutions

  • Making copper(II) sulfate solution from the hydrateChoose the correct mass and procedure to prepare a solution of known molarity from a hydrated salt.
  • Weighing too little solidPlan the preparation of a very dilute solution by making a weighable stock solution and diluting it.

Explique por quê 2

Dilution and mixing

  • Why C₁V₁ = C₂V₂ worksExplain the dilution equation as conservation of the moles of solute.

Titrations and back-titrations

  • Why not titrate chalk directly?Explain why an insoluble, slowly reacting analyte such as calcium carbonate is determined by back-titration.

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