Solubility, Dissociation or Ionization Constant, Partition Coefficient in Preformulation
CH 104: DETERMINATION OF A SOLUBILITY PRODUCT CONSTANT
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• A solution is A solution is saturated saturated if it has a solid at equilibrium if it has a solid at equilibrium with its solute. The concentration of solute in a with its solute. The concentration of solute in a saturated solution saturated solution is the largest that is normally is the largest that is normally possible. possible. • For a saturated solution of ionic solid in water: For a saturated solution of ionic solid in water: C C x A A y(s) y(s) = xC = xC n+ n+ (aq) (aq) + yA + yA m– m– (aq) (aq) • Solubility Product Constant = K Solubility Product Constant = K sp sp = [C = [C n+ n+ ] ] x [A [A m– m– ] ] y • Where Where • C C x A A y(s) y(s) is a slightly soluble ionic solid. is a slightly soluble ionic solid. • [C [C n+ n+ ] and [A ] and [A m– m– ] are the equilibrium concentrations of ] are the equilibrium concentrations of ions in moles per liter. ions in moles per liter. • x and y are the stoichiometric coefficients from the x and y are the stoichiometric coefficients from the balance reaction. balance reaction. • The The solubility product constant solubility product constant (K (K sp sp ) is the equilibrium ) is the equilibrium constant for the dissolution of a slightly soluble ionic constant for the dissolution of a slightly soluble ionic compound at a specified temperature. compound at a specified temperature. • By convention the “concentration” of the solid, C By convention the “concentration” of the solid, C A A , is , is CH 104: DETERMINATION OF A CH 104: DETERMINATION OF A SOLUBILITY PRODUCT CONSTANT SOLUBILITY PRODUCT CONSTANT
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CH 104: DETERMINATION OF A SOLUBILITY PRODUCT CONSTANT. A solution is saturated if it has a solid at equilibrium with its solute. The concentration of solute in a saturated solution is the largest that is normally possible. For a saturated solution of ionic solid in water: - PowerPoint PPT Presentation
Transcript of CH 104: DETERMINATION OF A SOLUBILITY PRODUCT CONSTANT
• A solution is A solution is saturatedsaturated if it has a solid at equilibrium with its solute. if it has a solid at equilibrium with its solute. The concentration of solute in a The concentration of solute in a saturated solutionsaturated solution is the largest that is is the largest that is normally possible.normally possible.
• For a saturated solution of ionic solid in water:For a saturated solution of ionic solid in water:CCxxAAy(s)y(s) = xC = xCn+n+
(aq)(aq) + yA + yAm–m–(aq)(aq)
• Solubility Product Constant = KSolubility Product Constant = Kspsp = [C = [Cn+n+]]xx[A[Am–m–]]yy
• WhereWhere• CCxxAAy(s)y(s) is a slightly soluble ionic solid. is a slightly soluble ionic solid.• [C[Cn+n+] and [A] and [Am–m–] are the equilibrium concentrations of ions in moles ] are the equilibrium concentrations of ions in moles
per liter.per liter.• x and y are the stoichiometric coefficients from the balance x and y are the stoichiometric coefficients from the balance
reaction.reaction.
• The The solubility product constantsolubility product constant (K (Kspsp) is the equilibrium constant for the ) is the equilibrium constant for the
dissolution of a slightly soluble ionic compound at a specified dissolution of a slightly soluble ionic compound at a specified temperature.temperature.
• By convention the “concentration” of the solid, CBy convention the “concentration” of the solid, CxxAAyy, is NOT used to , is NOT used to
calculate Kcalculate Kspsp. (That is, the activity of a pure solid is 1.). (That is, the activity of a pure solid is 1.)
CH 104: DETERMINATION OF A CH 104: DETERMINATION OF A SOLUBILITY PRODUCT CONSTANTSOLUBILITY PRODUCT CONSTANT
• What is the reaction for a saturated solution of BiWhat is the reaction for a saturated solution of Bi22SS3(s)3(s) in in
water?water?BiBi22SS3(s)3(s) = 2Bi = 2Bi3+3+
(aq)(aq) + 3S + 3S22––(aq)(aq)
• What is the KWhat is the Kspsp for this reaction? for this reaction?
KKspsp = [Bi = [Bi3+3+]]22[S[S2–2–]]33
• Notice the “concentration” of BiNotice the “concentration” of Bi22SS3(s)3(s) is NOT used to is NOT used to
calculate Kcalculate Kspsp..
• What is the reaction for a saturated solution of AgWhat is the reaction for a saturated solution of Ag22CrOCrO4(s)4(s) in in
water?water?AgAg22CrOCrO4(s)4(s) = 2Ag = 2Ag++
(aq)(aq) + CrO + CrO4422––
(aq)(aq)
• What is the KWhat is the Kspsp for this reaction? for this reaction?
KKspsp = [Ag = [Ag++]]22[CrO[CrO4422––]]
SOLUBILITY PRODUCT CONSTANTSOLUBILITY PRODUCT CONSTANT
• Calcium fluoride (CaFCalcium fluoride (CaF22) is slightly soluble ) is slightly soluble
in water. In a saturated solution the in water. In a saturated solution the CaFCaF2(s)2(s) is dissolving at the same rate that is dissolving at the same rate that
CaCa2+2+(aq)(aq) and F and F––
(aq)(aq) crystallize. That is, the crystallize. That is, the
solid and solute are at equilibrium.solid and solute are at equilibrium.
CaFCaF2(s)2(s) = Ca = Ca2+2+(aq)(aq) + 2F + 2F––
(aq)(aq)
SOLUBILITY PRODUCT CONSTANTSOLUBILITY PRODUCT CONSTANT
• A saturated solution is made by adding excess CaFA saturated solution is made by adding excess CaF2(s)2(s) to to
distilled water.distilled water. What is the solubility of this CaFWhat is the solubility of this CaF2(s)2(s) at 25° C? at 25° C?
• Step #1:Step #1: Write the balanced reaction and K Write the balanced reaction and Kspsp equation. equation.
CaFCaF2(s)2(s) = Ca = Ca2+2+(aq)(aq) + 2F + 2F––
(aq)(aq)
KKspsp = [Ca = [Ca2+2+][F][F––]]22 = 2.7x10 = 2.7x10–11–11 at 25° C at 25° C
• Step #2:Step #2: The initial concentrations of Ca The initial concentrations of Ca2+2+(aq)(aq) and F and F––
(aq)(aq) are 0. are 0.
The equilibrium concentrations of CaThe equilibrium concentrations of Ca2+2+(aq)(aq) and F and F––
(aq)(aq) are given are given
algebraic variables based on the stoichiometric coefficients algebraic variables based on the stoichiometric coefficients from the balance reaction. Write these equilibrium from the balance reaction. Write these equilibrium concentrations of Caconcentrations of Ca2+2+
(aq)(aq) and F and F––(aq)(aq)..
[Ca[Ca2+2+] = x] = x[F[F––] = 2x] = 2x
CALCULATING SOLUBILITY FROM KCALCULATING SOLUBILITY FROM Kspsp
• Step #3:Step #3: Use the K Use the Kspsp equation to solve for [Ca equation to solve for [Ca2+2+] and [F] and [F––].].
KKspsp = 2.7x10 = 2.7x10–11–11 = [Ca = [Ca2+2+][F][F––]]22 = (x)(2x) = (x)(2x)22 = 4x = 4x33
xx33 = 2.7x10 = 2.7x10–11–11 / 4 = 6.7 / 4 = 6.755x10x10–12–12
[F[F––] = 2x = 3.8x10] = 2x = 3.8x10–4–4 M M
• Step #4:Step #4: Solve for the solubility of CaF Solve for the solubility of CaF2(s)2(s)..
One mole of CaOne mole of Ca2+2+(aq)(aq) is produced for every mole of CaF is produced for every mole of CaF2(s)2(s)
that dissolves; therefore, that dissolves; therefore, the solubility of CaFthe solubility of CaF2(s)2(s) = [Ca = [Ca2+2+] = 1.9x10] = 1.9x10–4–4 M. M.
CALCULATING SOLUBILITY FROM KCALCULATING SOLUBILITY FROM Kspsp
• In the previous example the pure solid (CaFIn the previous example the pure solid (CaF2(s)2(s)) was the only ) was the only
source of its dissolved ions (Casource of its dissolved ions (Ca2+2+(aq)(aq) and F and F––
(aq)(aq)).).
• However, if However, if the common ionthe common ion F F––(aq)(aq) is added it will react with is added it will react with
CaCa2+2+(aq)(aq) to decrease the solubility of CaF to decrease the solubility of CaF2(s)2(s). The new . The new
concentration of Caconcentration of Ca2+2+(aq)(aq) is less than in the original is less than in the original
equilibrium. And the new concentration of Fequilibrium. And the new concentration of F––(aq)(aq) is greater is greater
than in the original equilibrium. This is called than in the original equilibrium. This is called Le Châtelier’s Le Châtelier’s principleprinciple..
CaFCaF2(s)2(s) = Ca = Ca2+2+(aq)(aq) + 2F + 2F––
(aq)(aq)
KKspsp = [Ca = [Ca2+2+][F][F––]]22 = 2.7x10 = 2.7x10–11–11 at 25° C at 25° C
• Similarly, if Similarly, if the common ionthe common ion Ca Ca2+2+(aq)(aq) is added it will react is added it will react
with Fwith F––(aq)(aq) to the solubility of to the solubility of CaFCaF2(s)2(s). The new . The new
concentration of Fconcentration of F––(aq)(aq) is than in the original is than in the original
equilibrium. And the new concentration of Caequilibrium. And the new concentration of Ca2+2+(aq)(aq) is is
than in the original equilibrium.than in the original equilibrium.
THE COMMON ION EFFECTTHE COMMON ION EFFECT
decreasedecreaselessless
greatergreater
• The common ion FThe common ion F––(aq)(aq) is added to a saturated solution of is added to a saturated solution of
CaFCaF2(s)2(s) in distilled water. in distilled water. What is the concentration of CaWhat is the concentration of Ca2+2+
(aq)(aq) in equilibrium with 1.0 M F in equilibrium with 1.0 M F––(aq)(aq) and CaF and CaF2(s)2(s) at 25° C? at 25° C?
KKspsp = 2.7x10 = 2.7x10–11–11 = [Ca = [Ca2+2+][F][F––]]22 = [Ca = [Ca2+2+](1](122))
[Ca[Ca2+2+] = 2.7x10] = 2.7x10–11–11 M M
• Compared to the previous example, did the concentration of Compared to the previous example, did the concentration of CaCa2+2+
(aq)(aq) increase or decrease? increase or decrease?• It decreased from 1.9x10It decreased from 1.9x10–4–4 M to 2.7x10 M to 2.7x10–11–11 M. M.
• Did the concentration of FDid the concentration of F––(aq)(aq) increase or decrease? increase or decrease?
• It increased from 3.8x10It increased from 3.8x10–4–4 M to 1.0 M. M to 1.0 M.
• Does this agree with the common ion effect?Does this agree with the common ion effect?• Yes. The concentration of CaYes. The concentration of Ca2+2+
(aq)(aq) decreased. The decreased. The
concentration of Fconcentration of F––(aq)(aq) increased. And the solubility of increased. And the solubility of
CaFCaF2(s)2(s) decreased. decreased.
THE COMMON ION EFFECTTHE COMMON ION EFFECT
• Common ions decrease the solubility of ionic solids.Common ions decrease the solubility of ionic solids.
• In contrast, the presence of “uncommon” ions tends to In contrast, the presence of “uncommon” ions tends to increase solubility of ionic solids. This is called the “salt increase solubility of ionic solids. This is called the “salt effect”, the “uncommon ion effect”, or the “diverse ion effect”, the “uncommon ion effect”, or the “diverse ion effect”.effect”.
• Soluble uncommon ions increase the interionic attractions Soluble uncommon ions increase the interionic attractions of a solution. As a result, these uncommon ions decrease of a solution. As a result, these uncommon ions decrease the effective concentrations (or activities) of other solutes the effective concentrations (or activities) of other solutes and increase the solubility of ionic solids.and increase the solubility of ionic solids.
THE SALT EFFECTTHE SALT EFFECT
• The salt effect is not as striking as the common ion effect.The salt effect is not as striking as the common ion effect.• The presence of the common ion CrOThe presence of the common ion CrO44
22––(aq)(aq), from K, from K22CrOCrO44, decreases the , decreases the
solubility of Agsolubility of Ag22CrOCrO44 by a factor of 35. by a factor of 35.• In contrast, the presence of the uncommon ions KIn contrast, the presence of the uncommon ions K++
(aq)(aq) and NO and NO33––
(aq)(aq), from , from
KNOKNO33, increase the solubility of Ag, increase the solubility of Ag22CrOCrO44 by a factor of only 0.25. by a factor of only 0.25.
THE SALT EFFECTTHE SALT EFFECT
• In today’s experiment you will measure the solubility of potassium In today’s experiment you will measure the solubility of potassium hydrogen tartrate (KOOC(CHOH)hydrogen tartrate (KOOC(CHOH)22COOH).COOH).
KOOC(CHOH)KOOC(CHOH)22COOH(s) = KCOOH(s) = K++(aq)(aq) + + ––OOC(CHOH)OOC(CHOH)22COOHCOOH(aq)(aq)
• What is the KWhat is the Kspsp for this reaction? for this reaction?
KKspsp = [K = [K++][][––OOC(CHOH)OOC(CHOH)22COOH]COOH]• You will make a saturated solution of KOOC(CHOH)You will make a saturated solution of KOOC(CHOH)22COOH in 0.10 M COOH in 0.10 M
NaCl and in 0.10 M KNONaCl and in 0.10 M KNO33..
• Is the NaCl a source of a common ion or uncommon ions?Is the NaCl a source of a common ion or uncommon ions?• NaNa++ and Cl and Cl–– are uncommon ions. are uncommon ions.• This NaCl should increase or decrease the solubility of This NaCl should increase or decrease the solubility of
KOOC(CHOH)KOOC(CHOH)22COOH?COOH?• Increase.Increase.• Is the KNOIs the KNO33 a source of a common ion or uncommon ions? a source of a common ion or uncommon ions?• KK++ is a common ion. NO is a common ion. NO33
–– is an uncommon ion. is an uncommon ion.• This KNOThis KNO33 should increase or decrease the solubility of should increase or decrease the solubility of
KOOC(CHOH)KOOC(CHOH)22COOH?COOH?• Decrease. The common ion effect is usually greater than the Decrease. The common ion effect is usually greater than the
uncommon ion effect.uncommon ion effect.
CALCULATING KCALCULATING Kspsp FROM SOLUBILITY FROM SOLUBILITY
• In today’s experiment you will measure the concentration of In today’s experiment you will measure the concentration of ––OOC(CHOH)OOC(CHOH)22COOHCOOH(aq)(aq) by titration with standardized by titration with standardized
sodium hydroxide (NaOH) to a phenolphthalein endpoint.sodium hydroxide (NaOH) to a phenolphthalein endpoint.
• Potassium hydrogen tartrate is a monoprotic acid; that is, Potassium hydrogen tartrate is a monoprotic acid; that is, only 1 hydrogen will be neutralized by titration with NaOH.only 1 hydrogen will be neutralized by titration with NaOH.
––OOC(CHOH)OOC(CHOH)22COOHCOOH(aq)(aq) + OH + OH––(aq)(aq) → →
––OOC(CHOH)OOC(CHOH)22COOCOO––(aq)(aq) + H + H22OO(l)(l)
CALCULATING KCALCULATING Kspsp FROM SOLUBILITY FROM SOLUBILITY
• Stop adding base when the indicator just begins to turn a Stop adding base when the indicator just begins to turn a faint but stable pink. This is the endpoint.faint but stable pink. This is the endpoint.
TITRATION USING PHENOLPHTHALEIN AS AN INDICATORTITRATION USING PHENOLPHTHALEIN AS AN INDICATOR
• Give at least 1 safety concern for the following procedure.Give at least 1 safety concern for the following procedure.
• Using KOOC(CHOH)Using KOOC(CHOH)22COOH, NaCl, KNOCOOH, NaCl, KNO33, NaOH, and , NaOH, and
phenolphthalein.phenolphthalein.
• These are irritants. Wear your goggles at all times. These are irritants. Wear your goggles at all times. Immediately clean all spills. If you do get either of these in Immediately clean all spills. If you do get either of these in your eye, immediately flush with water.your eye, immediately flush with water.
• Your laboratory manual has an extensive list of safety Your laboratory manual has an extensive list of safety procedures. Read and understand this section.procedures. Read and understand this section.
• Ask your instructor if you ever have any questions about Ask your instructor if you ever have any questions about safety.safety.
SAFETYSAFETY
SOURCESSOURCES
• McMurry, J., R.C. Fay. 2004. Chemistry, 4th ed. Upper McMurry, J., R.C. Fay. 2004. Chemistry, 4th ed. Upper Saddle River, NJ: Prentice Hall.Saddle River, NJ: Prentice Hall.
• Petrucci, R.H. 1985. General Chemistry Principles and Petrucci, R.H. 1985. General Chemistry Principles and Modern Applications, 4th ed. New York, NY: Macmillan Modern Applications, 4th ed. New York, NY: Macmillan Publishing Company.Publishing Company.
• Traverso M. 2006. Titration using Phenolphthalein as an Traverso M. 2006. Titration using Phenolphthalein as an Indicator. Available: Indicator. Available: www.chemistry.wustl.edu/.../AcidBase/phph.htm [accessed 14 September 2006]. [accessed 14 September 2006].
