Topic 10 – Part II • Applying Equilibrium • Buffers • Titrations • Solubility Product
The Common Ion Effect • When the salt with the anion of a weak acid is added to that acid, • It reverses the dissociation of the acid. • Lowers the percent dissociation of the acid. • The same principle applies to salts with the cation of a weak base. • The calculations are the same as last chapter.
HF (aq) H+ (aq) + F- (aq) • If NaF is added to the solution • [F-] increases and the equilibrium shifts left • This makes the solution less acidic • Soln of NaF + HF less acidic than soln of HF
Buffered solutions • A solution that resists a change in pH. • Either a weak acid and its salt or a weak base and its salt. • We can make a buffer of any pH by varying the concentrations of these solutions. • Same calculations as before.
15.2 – The pH of a Buffered Solution • A buffered solution contains 0.50 M acetic acid (HC2H3O2, Ka = 1.8 x 10-5) and 0.50 M sodium acetate (NaC2H3O2). • Calculate the pH of the solution.
Adding a strong acid or base • Do the stoichiometry first. • A strong base will grab protons from the weak acid reducing [HA]0 • A strong acid will add its proton to the anion of the salt reducing [A-]0 • Then do the equilibrium problem.
15.3 – pH Changes in Buffered Solutions • Calculate the change in pH that occurs when 0.010 mol solid NaOH is added to 1.0 L of the buffered solution in sample exercise 15.2.
Buffers – How do they work? • A buffered solution contains large quantities of a weak acid (HA) and its conjugate base (A-) from the salt. • When OH- ions are added they pull off the H+ from the HA • OH- + HA A- + H2O • OH- cannot accumulate. It reacts with HA to produce A-
General equation • Ka = [H+] [A-] [HA] • so [H+] = Ka [HA] [A-] • The [H+] depends on the ratio [HA]/[A-] • taking the negative log of both sides • pH = -log(Ka [HA]/[A-]) • pH = -log(Ka)-log([HA]/[A-]) • pH = pKa + log([A-]/[HA])
This is called the Henderson-Hasselbach equation • pH = pKa + log([A-]/[HA]) • pH = pKa + log(base/acid)
15.4 – The pH of a Buffered Solution II • Calculate the pH of a solution containing 0.75 M lactic acid (Ka = 1.4 x 10-4) and 0.25 M sodium lactate. 15.5 – The pH of a Buffered Solution III • A buffered solution contains 0.25 M NH3 (Kb = 1.8 x 10-5) and 0.40 M NH4Cl. Calculate the pH of this solution.
15.6 – Adding Strong Acid to a Buffered Solution I • What would the pH be if 0.10 mol of HCl is added to 1.0 L of the buffered solution from Ex. 15.5.
Buffering Capacity • The pH of a buffered solution is determined by the ratio [A-]/[HA]. • As long as this doesn’t change much the pH won’t change much. • The more concentrated these two are the more H+ and OH- the solution will be able to absorb. • Larger concentrations bigger buffer capacity.
15.7 – Adding Strong Acid to a Buffered Solution II • Calculate the change in pH that occurs when 0.010 mol of HCl(g) is added to 1.0L of each of the following: • 5.00 M HAc and 5.00 M NaAc • 0.050 M HAc and 0.050 M NaAc • Ka= 1.8x10-5
Buffer capacity • The best buffers have a ratio [A-]/[HA] = 1 • This is most resistant to change • True when [A-] = [HA] • Make pH = pKa (since log1=0)
15.8 – Preparing a Buffer • A chemist needs a solution buffered at pH 4.30 and can choose from the following acids (and their sodium salts): • a. chloroacetic acid (Ka = 1.35 x 10-3) • b. propanoic acid (Ka = 1.3 x 10-5) • c. benzoic acid (Ka = 6.4 x 10-5) • d. hypochlorous acid (Ka = 3.5 x 10-8) • Caclulate the ratio [HA] / [A-] required for each to yield a pH of 4.30. Which system will work best?
Titrations • Millimole (mmol) = 1/1000 mol • Molarity = mmol/mL = mol/L • Makes calculations easier because we will rarely add Liters of solution. • Adding a solution of known concentration until the substance being tested is consumed. • This is called the equivalence point. • Graph of pH vs. mL is a titration curve.
Strong acid with Strong Base • Do the stoichiometry. • There is no equilibrium . • They both dissociate completely. • The titration of 50.0 mL of 0.200 M HNO3 with 0.100 M NaOH • Calculate the pH after the following ml of NaOH are added: • A. no NaOH • B. 10ml of NaOH • C. 20ml of NaOH • D. 50ml of NaOH • E. 100ml of NaOH • F. 150ml of NaOH
Weak acid with Strong base • There is an equilibrium. • Do stoichiometry. • Then do equilibrium. • Titrate 50.0 mL of 0.10 M HC2H3O2 (Ka = 1.8 x 10-5) with 0.10 M NaOH. Calculate the pH at the following points. • A. No NaOH • B. 10ml of NaOH • C. 25ml of NaOH • D. 40ml of NaOH • E. 50ml of NaOH • F. 60ml of NaOH
Titration Curves • Strong acid with strong Base • Equivalence at pH 7 7 pH mL of Base added
Weak acid with strong Base • Equivalence at pH >7 >7 pH mL of Base added
Strong base with strong acid • Equivalence at pH 7 7 pH mL of Base added
Weak base with strong acid • Equivalence at pH <7 <7 pH mL of Base added
Summary: Titration of a Weak Acid with a Strong Base • At 0 ml base find pH with ICE and Ka • Weak acid before equivalence point • Stoichiometry first • Then Henderson-Hasselbach • Weak acid at ½ equivalence pt pH=pKa • Weak acid at equivalence point find pH with ICE and Kb • Weak base after equivalence - leftover strong base.
Indicators • Weak acids that change color when they become bases. • weak acid written HIn • Weak base • HIn H+ + In- clear red • Equilibrium is controlled by pH • End point - when the indicator changes color.
Indicators • Since it is an equilibrium the color change is gradual. • It is noticeable when the ratio of [In-]/[HIn] or [HIn]/[In-] is 1/10 • Since the Indicator is a weak acid, it has a Ka. • pH the indicator changes at is. • pH=pKa +log([In-]/[HIn]) = pKa +log(1/10) • pH=pKa - 1 on the way up (adding a base)
Indicators • pH=pKa + log([HIn]/[In-]) = pKa + log(10) • pH=pKa+1 on the way down (adding an acid) • For strong acid base titrations, the indicator color change is sharp and a wide range of indicators , might be suitable. • For titrations of weak acids choose the indicator with a pKa as close to equivalence point as possible.
Solubility Equilibria(Equilibria between solids and solutions) • All dissolving is an equilibrium. • If there is not much solid it will all dissolve. • As more solid is added the solution will become saturated. • Solid dissolved • The solid will precipitate as fast as it dissolves . • Equilibrium
Consider the following equilibrium • CaF2 (s) Ca2+ (aq) + 2F- (aq) • Ultimately this reaches equilibrium • Solubility Product Constant (Ksp) is • Ksp = [Ca2+] [F-]2 • Called the solubility product.
Solubility is not the same as solubility product • Solubility product is an equilibrium constant. • it doesn’t change except with temperature. • Solubility is an equilibrium position for how much can dissolve. • A common ion can change this.
15.12 - Calculating Ksp from Solubility I • Copper (I) Bromide has a measured solubility of 2.0 x 10-4 mol/L at 25oC. • Calculate its Ksp value
15.13 - Calculating Ksp from Solubility II • Calculate the Ksp value for bismuth sulfide (Bi2S3), which has a solubility of 1.0 x 10-15 mol/L at 25oC.
15.14 - Calculating Solubility from Ksp • The Ksp value for copper (II) iodate, Cu(IO3)2, is 1.4 x 10-7 at 25oC. Calculate its solubility at 25oC.
Relative solubilities • Ksp will only allow us to compare the solubility of solids the that fall apart into the same number of ions. • The bigger the Ksp of those the more soluble. • If they fall apart into different number of pieces you have to do the math.
Common Ion Effect • If we try to dissolve the solid in a solution with either the cation or anion already present less will dissolve. • 15.15 - Solubility and Common Ions • Calculate the solubility of solid CaF2 (Ksp = 4.0 x 10-11) in a 0.025 M NaF
pH and solubility • OH- can be a common ion. • More soluble in acid or less in base. • Mg(OH)2 (s) Mg2+ (aq) + 2OH- (aq) • For other anions if they come from a weak acid they are more soluble in a acidic solution than in water. • H+ + PO43- HPO42- • Ag3PO4(s) 3Ag+(aq) + PO43-(aq)
Precipitation and Qualitative Analysis • Ion Product, Q =[M+]a[N-]b • If Q>Ksp a precipitate forms. • If Q<Ksp No precipitate. • If Q = Ksp equilibrium. • 15.17 – Determining Precipitation Conditions • A solution is prepared by mixing 150.0 mL of 1.00 x 10-2 M Mg(NO3)2 and 250.0 mL of 1.00 x 10-1 M NaF. Calculate the concentrations of Mg2+ and F- at equilibrium with solid MgF2 (Ksp = 6.4 x 10-9).