Titrations In a titration a solution of accurately known concentration is added gradually added to another solution of unknown concentration until the chemical reaction between the two solutions is complete. Equivalence point the point at which the reaction is complete Indicator substance that changes color at (or near) the equivalence point Slowly add base to unknown acid UNTIL The indicator changes color (pink) 4.7 Titration Curves - - You should know how to draw them. That means - - you should understand what is happening; What is reacting with what? What equations do you use? In doing this you will learn lots of practical chemistry. 1. The initial Situation 2. The mid-point Situation 3. intermediate-points (just before & after the mid-point) 4. The end-point Situation The equivalence-point M 1 V 1 = M 2 V 2 5. Beyond the end point A student was asked to determine the molecular weight of an unknown acid. A 0.1508 gram sample was dissolved in 25 ml of water and titrated with 0.105 M NaOH solution. An indicator was used to determine the endpoint which occurred after 12.32 ml of the NaOH solution was added. What is the molecular weight of the acid? Moles of acid = moles of base Calculate & Draw a Titration Curve that represents the titration of HCl with NaOH [Assume 25 ml of 0.10 M HCl is titrated with 0.10 M NaOH] Identify the Major Species at various points in the Titration Curve 0.1508 gram 0.105 mole /L x 0.01232L = 117 g / mol A 95 g/mol B 102 /mol C 117 g/mol D 126 g/mol E 152 g/mol Strong Acid Strong Base Titration NaOH (aq) + HCl (aq) H 2 O (l) + NaCl (aq) 0.10 M NaOH added to 25 ml of 0.10 M HCl [H+] << [OH ] End Point Situation ph = 7.00 Calculate & Draw a Titration Curve that represents the titration of Acetic Acid (HA) with NaOH [Assume 25 ml of 0.10 M HA is titrated with 0.10 M NaOH] Identify the Major Species at various points in the Titration Curve [H + ] = [OH ] [H + ] >> [OH ] It is important to consider the effect of Volume change through out the calculation
HA CH 3 COOH (aq) + OH (aq) Weak acid CH 3 COO (aq) + H 2 O (l) Weak Acid Strong Base Titration CH 3 COO (aq) + H 2 O (l) At equivalence point (ph > 7): End Point Situation ph = 8.72 HA & A OH (aq) + CH 3 COOH (aq) A weak acid (conj. base) 0.10 M NaOH added to 25 ml of 0.10 M HOAc HCl (aq) + NH 3 (aq) H + (aq) + NH 3 (aq) NH 4 + (aq) + H 2 O (l) Weak Base Strong Acid Titration NH 4 Cl (aq) NH 4 Cl (aq) At equivalence point (ph < 7): B Weak base B & BH + NH 3 (aq) + H + (aq) 0.10 M HCl added to 25 ml of 0.10 M NH 3 At the End Point situation ph = 5.28 weak base (conj. acid) BH + [A ] [H + ] = K a = [x] [x] [0.2 x] = 1.34 x x = 10-3 [H+ ] = 1.63 x 10-2 ph = 1.8 Equal so ph = pk a [OH ] = K b = [x] [x] [A ] [0.1 x] = poh = 6.1 ph =7.9 7.41 x 102 x = [OH - ] = 8.61 x 10-7 a 1.4 b 1.8 c 2.9 d 3.6 e 7 f 7.9 g 8.1 a 3.6 b 6.1 c 6.9 d 7 e 7.2 f 7.9 g 8.1 Weak Acid Strong Base Titration CH 3 COOH (aq) + OH (aq) CH 3 COO (aq) + H 2 O (l) At equivalence point (ph > 7): CH 3 COO (aq) + H 2 O (l) OH (aq) + CH 3 COOH (aq) NaOH is dominant species The first 20mL is used up. 10 ml diluted to 50 ml so M of NaOH is.04m -log(0.04) = poh = 1.4 ph = 12.6 a. 9.7 b. 10.3 c. 11.4 d 12.6 e 13.1 f 13.4
Suppose you were carrying out a titration of a 20.0 ml solution of 0.200 M NH 3 using [OH ] = K b = [x] [x] [A ] [0.2 x] = poh = 2.7 ph =11.3 1.8 x 10-5 x = [OH - ] = 1.9 x 10-3 Suppose you were carrying out a titration of a 20.0 ml solution of 0.200 M NH 3 using Equal so poh = pk b [A ] [H + ] = K a = [x] [x] [0.1 x] = 5.6 x x = 100 [H+ ] = 7.48 x 10-6 ph = 5.1 a. 4.7 b 5.0 c. 5.1 d. 9.3 e. 11.3 f 11.6 a. 4.7 b 5.0 c. 5.1 d. 9.3 e. 11.3 f 11.6 Weak Base Strong Acid Titration Suppose you were carrying out a titration of a 20.0 ml solution of 0.200 M NH 3 using H + (aq) + NH 3 (aq) NH 4 Cl (aq) At equivalence point (ph < 7): NH + 4 (aq) + H 2 O (l) NH 3 (aq) + H + (aq) HCl is dominant species The first 20mL is used up. 10 ml diluted to 50 ml so M of HCl is.04m -log(0.04) = ph = 1.4 weak base (conj. acid) a. 0.7 b 0.9 c. 1.4 d. 2.7 e. 4.7 Acid/Base Indicators: substances that change color in a specific ph range and is used as a visual index of an approximate ph range. Phenolphthalein = ph Acid Form = HIn Colorless Conj. Base Form = In Pink
The useful ph ranges for several common indicators ph Curve for Titration of HCl with NaOH The titration curve is so steep between ph 4 and 10 that either indicator will give a satisfactory result. 8a 19 8a 20 ph Curve for Titration of Acetic Acid with NaOH The ph at the end point is 8.72. Methyl red will give a very diffuse color change around 40 ml rather than the correct 50 ml. Phenolphthalein will work well. 8a 21 Titration Curve for a Triprotic Acid (H 3 A) - - This Figure is very important - - as it indicates the Major Species @ at different ph values and how those ph values are calculated. #4 () #6 () #7 (3 rd end point) Phosphoric acid & Phosphate Groups are Important Industrially = About 10 million tons of phosphoric acid, H 3 PO 4, are produced in the United States each year. Mainly used as fertilizer. Biologically = ATP, ADP, DNA, RNA #5 (2 nd end point) #2 () #3 (1 st end point) #1 (Weak Acid)
DNA Titration Curve for a Triprotic Acid (H 3 A) - - This Figure is very important - - as it indicates the Major Species @ at different ph values and how those ph values are calculated. #4 () #6 () #5 (2 nd end point) #7 (3 rd end point) #2 () #3 (1 st end point) #1 (Weak Acid) Suppose you had a liter solution of 1.0 M phosphoric acid. How many moles of solid NaOH would need to add to the solution to make a ph 7.0 buffer? 1.387 moles H 3 PO 4 pk a1 = 2.12; pk a2 = 7.21; pk a3 = 12.32 The strange one Carbonic Acid K a pk a H 2 CO 3 HCO 3 + H + 4.3 x 10-7 6.4 HCO 3 CO -2 3 + H + 4.8 x 10 1 10.3 K A = 10-7.2 = [H+][A-] Or ph = pka + log [A-] [10-7 ][A-] Or 7.0 = 7.2 + log [A-] [A-] = 10-.2 =.63 x 1-x =.63 x =.387 mole + 1 mole to make H 2 PO 4 CO 2 (aq) H 2 CO 3 (aq) + H + 1.7 x 10-3 6.4 H 2 CO 3 (aq) HCO 3 (aq) + H + 2.5 x 10-4 3.6 What is the ph of a 0.1 M solution of NaHCO 3? K b pk b HCO 3 + H 2 O H 2 CO 3 + OH - 2.3 x 10-8 7.6 [x] [x] = 2.3 x 10-8 [.1 x] [OH-] = 4.8x10-5 poh = 4.3 ph = 9.7 Titration of H 2 CO 3 with OH For Inspiration information taken from a Biochemistry Text is given: Percentages of carbonic acid and its conjugate bases as a function of ph In an aqueous solution at ph 7.4 (the ph of blood), the concentrations of carbonic acid (H 2 CO 3 ) and bicarbonate (HCO 3 ) are substantial, but the concentration of carbonate (CO 2 3 ) is negligible. The carbonate system serves to buffer the blood ph to 7.4 through a balance of all three forms of carbonic acid. from Principles of Biochemistry - - Horton, et al