Titration Curve: A graph of pH (of an acid or base) as a function of the volume of base (or acid) added to neutralize the substance Important Terms and Concepts to Know: - Equivalence Point : Point in a titration where the added solute reacts completely with the solute present in the solution. - K a =Acid-Dissociation Constant - K b =Base-Dissociation Constant -pH=-log[H + ] -pK a =-log[K a ] -K b =1.0x10 -14 / K a -ALWAYS WRITE OUT CHEMICAL EQUATIONS WHEN COMPLETING A TITRATION PROBLEM!
1. ) Solution with weak acid+strong base 2.) HX+OH - X - +H 2 O3.) Calculate [HX] and [X - ] after rxn 4.) Use K a,, [HX], and [X - ] to calculate [H + ] 5.) Arrive at pH
Calculation of a solution’s pH when 45.0 mL of.100 M NaOH is added to 50.0 mL of.100 M HC 2 H 3 O 2 (K a =1.8x10 -5 ): .0500L x (.100 mol HC 2 H 3 O 2 /1L soln) =5.00x10 -3 mol HC 2 H 3 O 2 .0450L x (.100 mol NaOH /1L soln) =4.50x10 -3 mol NaOH Total volume of solution: 45.0mL +50.0mL=.0950L [HC 2 H 3 O 2 ] =.50x10 -3 mol /.0950L =.0053 M [C 2 H 3 O 2 - ] = 4.50x10 -3 mol /.0950L=.0474 M K a = [H + ] [C 2 H 3 O 2 - ] / [HC 2 H 3 O 2 ] =1.8x10 -5 [H + ]= K a x[ [HC 2 H 3 O 2 ] / [C 2 H 3 O 2 - ]] =2.0x10 -6 M pH=-log[2.0x10 -6 ]=5.70
Solubility of Compound (g/L)Molar Solubility of Compound (mol/L) Molar Concentration of Ions K sp (Solubility Product Constant)
H 2 SeO 3 + H 2 O == H 3 O + +HSeO 3 - HSeO 3 - +H 2 O == SeO 3 2- + H 3 O + SeO 3 2- +H 2 O == HSeO 3 - + OH -
Solution involving the weak acid=higher initial pH than solution of a strong acid of equal concentration pH change at “rapid-rise” portion of curve near equivalence point is smaller for weak acid than it is for strong acid pH at equivalence point is over 7.00 for weak acid-strong base titration
Thymol Blue 1.2-2.8 Methyl yellow 2.9-4.0 Methyl orange 3.1-4.4 Methyl red 4.4-6.2 Phenol red 6.4-8.0 Phenolphthalein 8.0-10.0 Alizarin yellow 10.0-12.0
pH Meter: Electronic instrument used to measure the acidity or basicity of a liquid (consists of a glass electrode & electronic meter) Potentiometric Titration: “A volumetric method in in which the potential between two electrodes is measured as a function of the added reagent volume. Types of potentiometric titrations include acid-base, redox, precipitation, and complexometric” (Chemistry Dictionary & Glossary).
Involves determining concentration of a solution by reacting it with a certain # of moles of excess reagent. Excess reagent is later titrated with another reagent. Source: diracdelta.co.uk
The Common Ion Effect: The extent of ionization of a weak electrolyte is decreased by adding a strong electrolyte to the solution that has an ion in common with the weak electrolyte. 0.10 M HC 2 H 3 O 2 solution has a pH of 2.9, while a solution containing 0.10 M HC 2 H 3 O 2 and 0.01 M NaHC 2 H 3 O 2 has a pH of 4.7. Why has the addition of NaHC 2 H 3 O 2 to a HC 2 H 3 O 2 solution caused a decrease in H+ concentration (increase in pH)? When C 2 H 3 O 2 - (from the strong electrolyte NaC 2 H 3 O 2 ) is added to an acetic acid solution at equilibrium, the equilibrium condition is displaced. According to LeChatelier’s principle, the system reestablishes equilibrium by removing some of the added acetate ion. HC 2 H 3 O 2 (aq) H + (aq) + C 2 H 3 O 2 - (aq) Equilibrium C 2 H 3 O 2 - is added shifts left to remove some C 2 H 3 O 2 - This simultaneously removes H + ions from solution with the formation of HC 2 H 3 O 2 and thereby increases the pH.
Solutions containing a weak acid and its conjugate base or a weak base and its conjugate acid Used to control the pH of a solution Because of the common ion effect the mixture of a weak acid and its conjugate base inhibits the ionization of both A buffer also reduces changes in pH due to the addition of a strong acid or strong base. Strong acid added to a buffer reacts with the conjugate (weak) base to form more of the conjugate (weak) acid. Strong base added to a buffer reacts with the conjugate (weak) acid to form more of the conjugate (weak) base.
HC 2 H 3 O 2 + OH - == C 2 H 3 O 2 - + H 2 O Before reaction: 0.50 mol 0.010 mol 0.50 mol After reaction: 0.49 mol 0 0.51 mol HC 2 H 3 O 2 (aq) == H + (aq) + C 2 H 3 O 2 - (aq) Initial: 0.49 0 0.51 Change: -x x x Equilibrium: 0.49 - x x 0.51 + x K a = 1.8 x 10 -5 = [H + ][C 2 H 3 O 2 - ] / [HC 2 H 3 O 2 ] = (x) (0.51 + x) / (0.49 - x) [H + ] = x = 1.7 x 10 -5 M pH = 4.76 The change in pH from adding 0.010 mol OH - to the buffered solution is only: 4.76 - 4.74 = +0.02 What is the change in pH when 0.010 mol solid NaOH is added to 1.0L of a buffered solution containing 0.50M acetic acid (HC 2 H 3 O 2, K a = 1.8 x 10 -5 ) and 0.50M sodium acetate (NaC 2 H 3 O 2 ) with a pH of 4.74?
The amount of acid or base the buffer can neutralize before the pH begins to experience a noticeable change Depends on the amount of acid and base from which the buffer is made The greater the amounts of the conjugate acid-base pair, the more resistant the ratio of their concentrations (and therefore pH) is to change