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Concepts in Acid-Base Titration
A titration is a technique where a solution of known concentration is used to determine the concentration of an unknown solution.
Titrant (titrator) is a standard solution of known concentration.
Analyte (titrand) is a solution of unknown concentration.
An acid-base indicator is a weak organic acid (HIn) whose color differs from that of its conjugate base (In-).
The equivalence point (stoichiometric point) of a chemical reaction is the point at which chemically equivalent quantities of acid and base have been mixed.
The end point (related to, but not the same as the equivalence point) refers to the point at which the indicator changes colour in a colourimetric titration.
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The color change of the acid-base indicator bromthymol blue.
pH < 6.0 pH = 6.0-7.5 pH > 7.5 22-Mar-16 22
Curve for a strong acid–strong base titration.
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Curve for a weak acid–strong base titration.
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Curve for a weak base–strong acid titration.
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Example 6:
Calculate the pH during the titration of 40.00 mL of 0.1000 M HCl is titrated with 0.1000 M NaOH, after adding the following volumes:
a) 0 mL b) 20 mL. c) 40 mL. d) 50 mL.
--- a) pH = -log[H+] = -log(0,1) = 1.
b) A reaction happens:
pH = -log[H+] = -log(0,033) = 1.477 H+(aq) + OH-(aq) → H2O (aq)
Initial (mol) 0.004 0.002
Reacted (mol) 0.002 0.002
Equilibrium (mol) 0.002 0
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c) A reaction happens:
pH = -log[H+] = -log(10-7) = 7 d) A reaction happens:
pOH= -log[OH-]= -log(0.011)= 1.95pH = 14-pOH = 12.05 H+(aq) + OH-(aq) → H2O (aq)
Initial (mol) 0.004 0.004
Reacted (mol) 0.004 0.004
Equilibrium (mol) 0 0
H+(aq) + OH-(aq) → H2O (aq)
Initial (mol) 0.004 0.005
Reacted (mol) 0.004 0.004
Equilibrium (mol) 0 0.001
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Exercise 2
Calculate the pH during the titration of 40.00 mL of 0.1000 M propanoic acid (HPr; Ka= 1.3x10-5) after adding the following volumes of 0.1000 M NaOH:
a) 0 mL b) 20 mL. c) 40 mL. d) 50 mL.
--- Suggestion:
If the solution after the reaction is just a weak acid:
If the solution after the reaction is just a weak base:
pH = 14 - pOH
If the solution after the reaction is a buffer solution:
19.3
Equilibria of Slightly Soluble Ionic Compounds
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Any “insoluble” ionic compound is actually slightly soluble in aqueous solution.
We assume that the very small amount of such a compound that dissolves will dissociate completely.
For a slightly soluble ionic compound in water, equilibrium exists between solid solute and aqueous ions.
Example 7:
PbF2is a Slightly Soluble Ionic Compound PbF2(s) ⇌Pb2+(aq)+ 2F-(aq) Ksp= [Pb2+]eq[F-]2eq and
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Qspand Ksp
For any slightly soluble compound MpXq, which consists of ions Mn+and Xz-, so:
MpXq (s) ⇌ pMn+(aq)+ qXz-(aq)
Qspis called the ion-product expression for a slightly soluble ionic compound.
When the solution is saturated, the system is at equilibrium, and Qsp= Ksp, the solubility product constant.
• The Kspvalue of a salt indicates how far the dissolution proceeds at equilibrium (saturation).
Qsp= [Mn+]p[Xz-]q
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The elements effect on solubility
The effects of a common ion on solubility follow the Le Chatelier's principle.
The effect of pH on solubility:
• The addition of H3O+will increase the solubility of a salt that contains the anion of a weak acid.
Example 7: CaCO3(s) ⇌ Ca2+(aq) + CO32-(aq) If H+solution is added to a saturated solution of CaCO3:
CO32-(aq) + H3O+(aq) → HCO3-(aq) + H2O(l) HCO3-(aq) + H3O+(aq) → CO2(g) + 2H2O(l)
The net effect of adding H+to CaCO3solution is the removal of CO32- ions, which causes an equilibrium shift to the right
more CaCO3will dissolve.
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Example 8: PbCrO4(s) ⇌ Pb2+(aq)+ CrO42-(aq).
If Na2CrO4solution is added to a saturated solution of PbCrO4, it provides the common ion CrO42-, causing the equilibrium to shift to the left solubility decreases and solid PbCrO4 precipitates.
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Relationship between the solubility (S) and the solubility product constant (Ksp)
For any slightly soluble compound MpXq, which consists of ions Mn+and Xz-, so:
MpXq (s) ⇌ pMn+(aq) + qXz-(aq)
Solubility (M) S pS qS
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Example 9:
When lead(II) fluoride (PbF2) is shaken with pure water at 25°C, the solubility is found to be 0.064 g/100 mL. Calculate the Ksp of PbF2. MPbF2= 245.2 (g/mol).
---
Because the unit of solubility (S) in calculations is mol/L (M), so we need convert (g/mL) to (M).
PbF2 (s) ⇌ Pb2+(aq) + 2F-(aq)
Solubility (M) S S 2S
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Exercises Involving Ksp
1. Determining Kspfrom Solubility (S) 2. Determining Solubility (S) from Ksp 3. Using KspValues to Compare Solubilities
4. Calculating the Effect of a Common Ion on Solubility 5. Predicting the Effect on Solubility of Changing pH 6. Predicting Whether a Precipitate Will Form
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Exercise 3
a) Lead (II) sulfate (PbSO4) solubility in water at 25°C is 4.25x10-2g/L solution. What is theKspof PbSO4? MPbSO4= 303.3 g/mol
b) Calculate the molar solubility of Ca(OH)2in water if theKspis 6.5x10-6.
c) What is the solubility of Ca(OH)2in 0.10MCa(NO3)2? Kspof Ca(OH)2is 6.5x10-6.
--- Suggestion:
With part a) we need convert the solubility (g/L) to molar solubility (M).
Write dissociation equations for their dissolution. With part c) we need construct an ICE table for Ca(OH)2.
Write the relationship between the ion-product expression (Ksp) and solubility (S)calculate.
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Exercise 4
Explain affects the solubility of each ionic compound when addition of H3O+from a strong acid to solution of:
a) lead (II) bromide. b) copper (II) hydroxide.
c) iron (II) sulfide. d) Silver iodide --- Suggestion:
With part c) we know that S2-is the anion of HS-, a weak acid, and is a strong base. It will react completely with water to form HS-and OH-. Both these ions will react with added H3O+:
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Predicting the Formation of a Precipitate
For asaturated solutionof a slightly soluble ionic salt:
Qsp= Ksp.
When two solutions containing the ions of slightly soluble salts are mixed:
Qsp> Ksp • a precipitate will form until the solution is saturated
Qsp< Ksp • no precipitate will form because the solution is unsaturated
Qsp= Ksp • the solution is saturated and no change will occur
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Example 10:
Does a precipitate form when 0.100 L of 0.30 M Ca(NO3)2is mixed with 0.200 L of 0.060 M NaF? Ksp= 3.2x10-11.
---
The ions present are Ca2+, NO3-, Na+, and F-. All Na+and NO3-salts are soluble, so the only possible precipitate is CaF2(Ksp= 3.2x10-11).
Ca(NO3)2and NaF are soluble, and dissociate completely in solution.
Ca2+(aq) + 2F-(aq)⇌CaF2(s) Qsp= [Ca2+][F-]2 We need to calculate [Ca2+] and [F-] in the final solution.
Qsp= [Ca2+]init[F-]2init= (0.10)(0.040)2= 1.6x10-4 >Ksp= 3.2x10-11.
Since Qsp> Ksp, CaF2will precipitate until Qsp= 3.2x10-11.
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Selective Precipitation
Selective precipitation is used to separate a solution containing a mixture of ions.
A precipitating ion is added to the solution until the Qspof the more soluble compound is almost equal to its Ksp.
The less soluble compound will precipitate in as large a quantity as possible, leaving behind the ion of the more soluble compound.
Example 11
A solution consists of 0.20 MMgCl2and 0.10 MCuCl2. Calculate the [OH-] that would separate the metal ions as their hydroxides.
Ksp of Mg(OH)2= is 6.3x10-10; Kspof Cu(OH)2 is 2.2x10-20.
---
Mg(OH)2(s) ⇌Mg2+(aq) + 2OH-(aq) Ksp= [Mg2+][OH-]2= 6.3x10-10 Cu(OH)2(s) ⇌Cu2+(aq) + 2OH-(aq) Ksp= [Cu2+][OH-]2= 2.2x10-20 The maximum [OH-] that willnot precipitate Mg2+ion.
With this [OH-], the [Cu2+] remaining in solution:
Since the initial [Cu2+] is 0.10M, virtually all the Cu2+ion is precipitated.
So, the [OH-] need to control < 5.6x10-5M, then Cu(OH)2is precipitated.
[OH-] = = = 5.6x10-5M
Ksp
[OH-]2
[Cu2+] = = 2.2x10-20 (5.6x10-5)2
= 7.0x10-12M
19.4