Acid-Base Titrations. Terms. Terms. Terms

Transcription

1 Isoionic Point: The ph of pure, neutral polyprotic acid (the natural zwitterion) HAwater: H A +, A -, H +, and OH - Most HA and [H A + ][A - ] Isoelectric Point: ph at which the average charge of the polyprotic acid is zero (amino acids) Most HA and [H A + ]=[A - ] Acid-Base Titrations Titration: Increments of reagent solution-the titrant-are added to analyte until their reaction is complete. Titration type/endpoint determination: Volumetric: volume of standard is measured Gravimetric: mass of standard is measured, e.g., Mohr method (Cl-Ag-CrO 4 ). Coulometric: time at constant current is measured, i.e., amount of charge. Spectrophotometric: Beer s Law, the absorbance Terms Titration: Performed slowly by adding standard solution from a buret until the reaction is judged to be complete. Standard Solution (Standard Titrant): Reagent of known concentration that is used to carry out the analysis. Indicators: Added to a tritration to give an observable physical change at or near the equivalence point. Usually a color or turbidity change. Terms Equivalence Point: The point when the amount of added titrant is chemically equivalent to the amount of analyte in the sample. End Point: The point at which an indicator physically changes representing the estimated equivalence point (color or turbidity change). Titration error = Vep - Veq Terms Blank titration: Carry out the same procedure without analyte. To estimate the titration error. Back-titration: Add a known excess of one standard titrant to the analyte, and then titrate a second standard titrant to determine the excess reagent. May be required when a reaction is slow or the standard solution lacks stability. 1

2 Standard Solutions (Titrants) Standard Solutions (Titrants) Desirable Properties of Standard Solutions 1. Sufficiently stable (only need to determine concentration once).. React rapidly with the analyte (less time between additions). 3. React more or less completely with the analyte (satisfactory end points). 4. Undergo a selective reaction with analyte described by a simple chemical equation Methods for Establishing the Concentration of Standard Solutions 1. Direct Method: A carefully weighed quantity of primary standard is diluted to a known volume. First choice if possible.. Standardization: A titrant to be standardized is used to titrate a weighed quantity of a primary, secondary standard or a known volume of another standard solution. Standard Solutions (Titrants) Primary Standard: Highly purified compound that serves as a reference material. High purity, stability (air, heat, vacuum), Standard solution (direct method) To standardize a standard solution (standardization) Secondary Standard: A reference material (a compound) whose purity has been established by a chemical analysis. Often used in lieu of a primary standard because there are not many primary standards. Summary titration titrant analyte indicator equivalence point vs. end point titration error blank titration Standard: Primary standard, second standard, standard solution Acid-Base Titrations Titration of Strong Base with Strong Acid How does the ph change as titrant is added? Titration Curve: A graph showing how the concentration of one of the reactants varies as titrant is added. (how the ph changes as titrant is added in acid and base titration)

3 Titration of Strong Base with Strong Acid Titration of Strong Acid with Strong Base titration of 0.00 ml of 1. Before the equivalence point: 0.00 M HCl with 0.00,.00, 19.98, 0.00, 0.0 and ml of excess OH M NaOH.. At the equivalence point: ph = , initial point 3. After the equivalence point:.00 ml, 50% titration excess H ml, half-drop before equivalence point 0.00 ml, 0% titration (equivalence point) 0.0 ml, half-drop after equivalence point ml, 00% titration (far-over) titration of 0.00 ml of 0.00 M HCl with 0.00,.00, 19.98, 0.00, 0.0 and ml of 0.00 M NaOH. titration of 0.00 ml of 0.00 M HCl with 0.00,.00, 19.98, 0.00, 0.0 and ml of 0.00 M NaOH. At 0.00 ml of NaOH added, initial point [H 3 O + ] = C HCl = 0.00 M ph = 1.00 at.00 ml of NaOH added, V a * M a > V b * M b (V a M a ) - (V b M b ) [H 3 O + ] = ((0.00mL)*(0.00M)) - ((.00mL)*(0.00M)) = ( )mL = 3.33x - M ph =1.48 titration of 0.00 ml of 0.00 M HCl with 0.00,.00, 19.98, 0.00, 0.0 and ml of 0.00 M NaOH. at ml of NaOH added, V a * M a > V b * M b (V a M a ) - (V b M b ) [H 3 O + ] = ((0.00mL)*(0.00M)) - ((19.98mL)*(0.00M)) = ( )mL =5.0x -5 M ph =4.30 titration of 0.00 ml of 0.00 M HCl with 0.00,.00, 19.98, 0.00, 0.0 and ml of 0.00 M NaOH. at 0.00 ml of NaOH added V a * M a = V b * M b, equivalence point at equivalence point of a strong acid - strong base titration ph =

4 titration of 0.00 ml of 0.00 M HCl with 0.00,.00, 19.98, 0.00, 0.0 and ml of 0.00 M NaOH. titration of 0.00 ml of 0.00 M HCl with 0.00,.00, 19.98, 0.00, 0.0 and ml of 0.00 M NaOH. at 0.0 ml of NaOH added, V b * M b > V a * M a, post equivalence point (V b * M b ) - (V a * M a ) [OH - ] = ((0.0mL)*(0.00M)) - ((0.00mL)*(0.00M)) = ( )mL =5.0x -5 M poh =4.30 ph=9.70 at ml of NaOH added, V b * M b > V a * M a, post equivalence point (V b * M b ) - (V a * M a ) [OH - ] = ((40.00mL)*(0.00M)) - ((0.00mL)*(0.00M)) = ( )mL =3.33x - M, poh =1.48 ph=1.5 ph Strong Acid - Strong Base Titration Titration of Strong Acid with Strong Base The Effect of Concentration The change in ph at the equivalence point is less pronounced at decreased concentrations. The ph at the equivalence point is still 7. 1 (M) Titration of Weak Acid with Strong Base Volume of Base Added (ml) Titration of Weak Acid with Strong Base Titration of Weak Acid with Strong Base The ph change at the equivalence point decreases (the end point become less distinct) as the acid becomes weaker or more dilute The Effect of Concentration: the ph at the equivalence point is lower for lower initial weak acid concentrations and initial ph is higher. 4

5 Titration of Weak Acid with Strong Base 0.00, initial point ml, before equivalence point ml, 0% titration (equivalence point) ml, after equivalence point at 0.00 ml of NaOH added, initial point [H + ] K a C a = (1.75 x -5 M) x (0.00 M) = 1.3 x -3 M ph =.878 at ml of NaOH added, V a C a > V b C b (V a C a ) - (V b C b ) [HAc] ex. = [(35.00 ml)x(0.00 M)) - ((15.00 ml)x(0.00m)] = ( ) ml = 4.000x - M at ml of NaOH added, V a C a > V b C b [HAc] ex = 4.000x - M (V b C b ) [Ac - ] = (15.00 ml)(0.00 M) = = 3.000x - M ( ) ml at ml of NaOH added, V a C a > V b C b [HAc] ex = x - M [Ac - ] = x - M [HAc] ex [H + ] = K a [ Ac - ] x - M = 1.75 x -5 M =.33 x -5 M x - M at ml of NaOH added, V a C a > V b C b [HAc] ex = x - M [Ac - ] = x - M [H + ] =.33 x -5 M ph = 4.63 ph [ A ] log = log = 4.63 [ HA] = pka 5

6 at ml of NaOH added, V a C a = V b C b, equivalence point at ml of NaOH added, V a C a = V b C b, equivalence point K w V a C a [OH - ] K = b C b K a at ml of NaOH added, V a * C a = V b * C b, equivalence point 1.00 x -14 M * (35.00mL)(0.00M) [OH - ] = x -5 M * ( )mL = 5.34x -6 M at ml of NaOH added, V a C a = V b C b, equivalence point [OH - ] = 5.34 x -6 M poh = 5.7 ph = 8.78 V b C b > V a C a, post equivalence point V b C b > V a C a, post equivalence point (V b C b ) - (V a C a ) [OH - ] =

7 V b C b > V a C a (V b C b ) - (V a C a ) [OH - ] = [(50.00mL)*(0.00M)) - ((35.00mL)*(0.00M)] = ( ) ml = x - M V b C b > V a C a [OH - ] = x - M poh = ph = = 1.5 The following plot shows two titration curves, each representing the titration of 50.0 ml of 0.0 M acid with 0.0 M NaOH: Acid - Strong Base Titration Curve (a) Which of the two curves represents the titration of a strong acid, and which a weak acid? (b) What is the approximate ph at the equivalence point for each of the acids? (c) What is the approximate pk a of the weak acid? Vb (ml) Titration of Weak Weak Base - Strong Base Acid with Titration Strong Acid ph Volume of Acid Added (ml) Much the same shape, except the titration would start at high ph and decrease as acid is added Diprotic Systems: Titration of Weak Acid with Strong Base 0.00, Initial point ml, Before the 1 st equivalence point ml, At the 1 st equivalence point ml, After the 1 st equivalence point ml, At the nd equivalence point ml, After the nd equivalence point 7

8 at 0.00 ml of NaOH added, initial point + [ H ][ HA ] x Ka1 = = = 7.11 [ H A] 0.00 x x =.33 + = [ H ] 3 at ml of NaOH added, Va Ca > Vb Cb, the 1 st buffer region (Va Ca) - (Vb Cb) [H A] ex = (Va + Vb) ph = 1.63 at ml of NaOH added, Va Ca > Vb Cb, the 1 st buffer region [(35.00 ml) (0.00 M) - (15.00 ml) (0.00 M)] [H A] ex = ( ) ml = 4.000x - M at ml of NaOH added, Va Ca > Vb Cb, the 1 st buffer region [H A] ex = 4.000x - M (Vb Cb) (15.00 ml)(0.00 M) [HA - ] = = = 3.000x - M (Va + Vb) ( )mL at ml of NaOH added, Va Ca > Vb Cb, the 1 st buffer region [H A] ex = x - M; [HA - ] = x - M [H A] ex x - M [H + ] = Ka1 x = 7.11x -3 M x [HA - ] x - M = 9.48 x -3 M ph =.03 [ HA ] M ph = pka1 + log = log =.03 [ H A] M at ml of NaOH added, Va Ca = Vb Cb, the 1 st equivalence point [H + ] Ka1 Ka = (7.11 x -3 M)(6.34 x -8 M) =.1 x -5 M ph =

9 (Va Ca) > Vb Cb, the nd buffer region (Va Ca) - (Vb Cb) [HA - ] ex = (Va + Vb) (Va Ca) > Vb Cb, the nd buffer region [(35.00mL)*(0.00M)) - ((50.00mL)*(0.00M)] [HA - ] ex = ( ) ml =.353x - M (Va Ca) > Vb Cb, the nd buffer region [HA - ] ex =.353x - M (Vb Cb) - (Va Ca) [A - ] = (Va + Vb) (Va Ca) > Vb Cb, the nd buffer region [HA - ] ex =.353x - M (50.00 ml)(0.00 M) - (35.00 ml)(0.00 M) [A - ] = ( ) ml = 1.764x - M (Va Ca) > Vb Cb, the nd buffer region [HA - ] ex =.353x - M [A - ] = 1.764x - M Ka [HA - ] ex 6.34 x -8 M x.353x - M [H + ] = = [A - ] 1.764x - M = 8.46x -8 M ph = [ A ] M ph = pk + log = log = [ HA ].353 M at ml of NaOH added, (Va Ca) = Vb Cb, the nd equivalence point Kw Va Ca [OH - ] = Ka (Va + Vb) 9

10 at ml of NaOH added, (Va Ca) = Vb Cb, the nd equivalence point 1.00 x -14 M x (35.00 ml)x(0.00 M) [OH - ] = = 7.5x -5 M 6.34 x -8 M * ( ) ml at ml of NaOH added, Vb Cb > (Va Ca), after nd equivalence point Vb Cb - (Va Ca) [OH - ] = (Va + Vb) poh = ph = at ml of NaOH added, Vb Cb > (Va Ca), after nd equivalence point (90.00 ml)(0.00 M) - x (35.00 ml)(0.00 M) [OH - ] = ( ) ml ph Diprotic Systems: Diprotic Weak Acid - Strong Base Titration Titration of Weak Acid Base with Strong Base = x - M poh = ph = Volume of Base Added (ml) Diprotic Systems: Titration of Weak Base with Strong Acid Standard Solutions for Acid/Base Titrations Standard acid or base reagents are always strong acids or bases HCl, HClO 4, H SO 4, NaOH, and KOH They yield sharper end points. Weak reagents react incompletely. Nitric acid and hot sulfuric and perchloric yield undesirable ox/red side reactions. Primary standards for acid standardization Na CO 3 Tris (THAM) tris(hydroxymethyl)aminomethane, (HOCH ) 3 CNH Sodium tetraborate decahydrate, Na B 4 O 7.H O

11 Endpoint Determination 1. Indicators. Derivatives of titration curve 3. Gran Plot Acid/Base Indicators Weak acids or bases themselves The conjugate forms differs in color HIn In - + H + acid Color1 Using Indicators to Determine the End Point base Color In + H O InH + + OH - base acid Color1 Color ph = pka HIn ± 1 poh = pkb In ± 1 Indicator Behavior Indicators HIn In - + H + acid Color1 base Color [In - ][H + ] K a = (1) [HIn] Indicators Indicator Behavior In + H O InH + + OH - base acid Color1 Color [InH + ][OH - ] K b = () [In] Indicator Behavior Indicators rearranging Equation (1) gives [In - ] K a = [HIn] [H + ] Indicators Indicator Behavior acid color shows when [In-] 1 [H + ][In - ] <= = ---*[H + ] = K a [HIn] [HIn] base color shows when [In-] [H + ][In - ] >= = *[H + ] = K a [HIn] 1 [HIn] 11

12 Indicator Behavior acid color shows when Indicators and base color shows when ph + 1 = pk a ph - 1 = pk a Color change range is ph = pk a + 1 (Transition range) Choosing the Proper Indicator Color change range should be in area where titration curve is most vertical. Indicator error: the difference between the observed end point (color change) and the true equivalence point. Common Indicators Using Derivatives of Titration Curve to Determine the End Point See also p.15 Table 11-4 Using Derivatives of Titration Curve to Determine the End Point Titration curve for a polyprotic acid - strong base titration. Using Derivatives of Titration Curve to Determine the End Point First derivative of a weak acid- strong base titration Second derivative of a weak acid - strong base titration 1

13 Using the Gran Plot to Determine the End Point Using the Gran Plot to Determine the End Point The titration of a weak acid V b ph γ = γ HA A K ( V V ) a e b Using the Gran Plot to Determine the End Point The titration of a weak acid V b ph γ = Ka γ HA A ( V e V ) b The titration of a weak base V a + ph 1 γ B = K γ a + BH ( V e V ) a 13