Presentation on theme: "Topic 1C – Part II Acid Base Equilibria and Ksp Applying Equilibrium Titrations Buffers Solubility Product."— Presentation transcript:
Topic 1C – Part II Acid Base Equilibria and Ksp Applying Equilibrium Titrations Buffers Solubility Product
Titrations Titration is an experimental technique used to perform a neutralization reaction (see unit 3). Used to determine the unknown concentration of an acid or base. As a base or acid is added to an acid or base in a titration, there is very little change in pH initially When the moles of titrant (from the buret) are in the exact stoichiometric proportion with the analyte (in the flask) (100% of the acid or base has been neutralized) This is the equivalence point has been reached. At this point there is a rapid change in pH
These changes can be summarized using titration curve plots.
Most acid-base titrations involve the addition of one colorless solution to another colorless solution We need a method of determining when the equivalence point has been reached. We use an indicator, a chemical that changes colors at various pH’s. Indictors are often weak acids, the ionized and the unionized form have different colors. will change color over a small, given range of pH, for example;
The equivalence point is located in the middle of the vertical portion of the titration curves. We need to choose an indicator that changes color at a pH value as close to the equivalence point as possible
This observable color change of the indicator is called the end point, Should correspond to the equivalence point as closely as possible. Using the titration plots and the table on the previous slide, suitable indicators can be chosen; Strong acid - Strong base most indicators Weak acid - Strong base phenolphthalein Strong acid - Weak base methyl orange
Titration Calculations 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 HNO 3 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 HC 2 H 3 O 2 (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
pH mL of Base added 7 Titration Curves Strong acid with strong Base Equivalence at pH 7
pH mL of Base added >7 l Weak acid with strong Base l Equivalence at pH >7
pH mL of Base added 7 l Strong base with strong acid l Equivalence at pH 7
pH mL of Base added <7 l Weak base with strong acid l Equivalence at pH <7
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 Ka Equation or Henderson-Hasselbach ** Weak acid at ½ equivalence pt [H+] = [Ka] or 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.
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 (HC 2 H 3 O 2, Ka = 1.8 x 10 -5 ) and 0.50 M sodium acetate (NaC 2 H 3 O 2 ). 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- + H 2 O 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])
pH = pKa + log([A - ]/[HA]) pH = pKa + log(base/acid) This is called the Henderson- Hasselbach equation
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 NH 3 (K b = 1.8 x 10 -5 ) and 0.40 M NH 4 Cl. 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?
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 CaF 2 (s) Ca 2+ (aq) + 2F - (aq) Ultimately this reaches equilibrium Solubility Product Constant (Ksp) is K sp = [Ca 2+ ] [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 K sp from Solubility I Copper (I) Bromide has a measured solubility of 2.0 x 10 -4 mol/L at 25 o C. Calculate its Ksp value
15.13 - Calculating K sp from Solubility II Calculate the Ksp value for bismuth sulfide (Bi 2 S 3 ), which has a solubility of 1.0 x 10 -15 mol/L at 25 o C.
15.14 - Calculating Solubility from K sp The Ksp value for copper (II) iodate, Cu(IO 3 ) 2, is 1.4 x 10 -7 at 25 o C. Calculate its solubility at 25 o C.
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 CaF 2 (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) Mg 2+ (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 + + PO 4 3- HPO 4 2- Ag 3 PO 4 (s) 3Ag + (aq) + PO 4 3- (aq)
Precipitation and Qualitative Analysis Ion Product, Q =[M + ] a [N - ] b If Q>Ksp a precipitate forms. If Q
"name": "Precipitation and Qualitative Analysis Ion Product, Q =[M + ] a [N - ] b If Q>Ksp a precipitate forms.",
"description": " If Q
Salts as acids and bases Salts are ionic compounds. Salts of the cation of strong bases and the anion of strong acids are neutral. for example: NaCl (salt) – NaOH or HCl KNO 3 (salt) – KOH or HNO 3 There is no equilibrium for strong acids and bases. Na +, K + and Cl -, NO 3 - have no affinity for OH - ions or H + ions.
Salts that Produce a Basic Soln If the anion of a salt is the conjugate base of a weak acid – this produces a basic solution. In an aqueous solution of NaF (Salt) F- (anion) / HF (weak acid) The major species are Na +, F -, and H 2 O F - + H 2 O HF + OH - K b = [HF][OH - ] [F - ]
K a tells us K b K a x K b = K W = 1.0 x 10 -14 The anion of a weak acid is a strong base. Calculate the pH of a solution of 0.30 M NaF solution. (Ka of HF is 7.2 x 10 -4 ) (Ex. 14.18) The F - ion competes with OH - for the H +
Acidic salts A salt with the conj. acid of a weak base and the anion of a strong acid will be acidic. NH 4 Cl (salt) NH 4 + (conj. acid of weak base – NH 3 ) Cl - (anion of a strong acid - HCl) Calculate the pH of a solution of 0.10 M NH 4 Cl (K b of NH 3 1.8 x 10 -5 ) (Ex. 14.19) Other acidic salts are those of highly charged metal ions.
Anion of weak acid, cation of weak base If the salts in solution produce both cations and anions with acidic and basic properties. K a > K b acidic K a < K b basic K a = K b Neutral Example: NH 4 C 2 H 3 O 2 or NH 4 CN
The Lewis Acid-Base Model ModelDefinition of AcidDefinition of Base ArrheniusH+ producerOH- producer Bronsted-LowryH+ donorH+ acceptor LewisElectron-pair acceptorElectron-pair donor
Lewis Acid Has an empty orbital that can accept an electron pair Lewis Base Has a lone pair of electrons that can be shared