MECH 310 Exam 2 with Solution, Fall 2011

1 Chemistry 203 Spectrophotometric Determination of the pKa of Bromothymol Blue Purpose • To review the concepts of pH calculations and measurements. • To learn about indicators and buffers. • To learn how to use a pH-meter. • To determine the pKa of Bromothymol Blue using the spectophotometer and the pH-meter. 2 Fig. 15.2 pH of a Strong Acid What is the pH of a 2 x 10-3 M HNO3 solution? HNO3 is a strong acid – 100% dissociation. HNO3 (aq) + H2O (l) H3O+ (aq) + NO3 - (aq) 0.002 M 0.0 M 0.0 M 0.002 M 0.002 M 0.0 M pH = -log [H+] = -log [H3O+] = -log(0.002) pH= 2.7 3 pH of a Strong Base Ba(OH)2 is a strong base – 100% dissociation. What is the pH of a 1.8 x 10-2 M Ba(OH)2 solution? Ba(OH)2 (s) Ba2+ (aq) + 2OH- (aq) 0.018 M 0.0 M 0.0 M 0.0 M 0.018 M 0.036 M pH = 14.00 – pOH = 14.00 + log(0.036) pH= 12.56 pH + pOH = 14.00 What is the pH of a 0.5 M HF solution (at 250C)? HF (aq) H+ (aq) + F- (aq) Ka = [H+][F-] [HF] = 7.1 x 10-4 HF (aq) H+ (aq) + F- (aq) Initial (M) Change (M) Equilibrium (M) 0.50 0.00 -x +x 0.50 - x 0.00 +x x x Ka = x2 0.50 - x = 7.1 x 10-4 Ka ≈ x2 0.50 = 7.1 x 10-4 0.50 – x ≈ 0.50 Ka << 1 x2 = 3.55 x 10-4 x = 0.019 M [H+] = [F-] = 0.019 M pH = -log [H+] = 1.72 [HF] = 0.50 – x = 0.48 M Ref. Chang, Sect. 15.5 pH of a Weak Acid 4 When can I use the approximation? 0.50 – x ≈ 0.50 Ka << 1 When x is less than 5% of the value from which it is subtracted. x = 0.019 0.019 M 0.50 M x 100% = 3.8% Less than 5% Approximation ok. What is the pH of a 0.05 M HF solution (at 250C)? Ka ≈ x2 0.05 = 7.1 x 10-4 x = 0.006 M 0.006 M 0.05 M x 100% = 12% More than 5% Approximation not ok. Must solve for x exactly using quadratic equation or method of successive approximation. Ref. Chang, Sect. 15.5 NH3 (aq) + H2O (l) NH4 + (aq) + OH- (aq) pH of Weak Bases Kb = [NH4 +][OH-] [NH3] Kb is the base ionization constant Kb weak base strength Ref. Chang, Sect. 15.6 Solve weak base problems like weak acids except solve for [OH-] instead of [H+]. 5 Ref. Chang, Sect. 15.7 Ionization Constants of Conjugate Acid-Base Pairs HA (aq) H+ (aq) + A- (aq) A- (aq) + H2O (l) OH- (aq) + HA (aq) Ka Kb H2O (l) H+ (aq) + OH- (aq) Kw KaKb = Kw Weak Acid and Its Conjugate Base Ka = Kw Kb Kb = Kw Ka Buffer Solutions • Definition: A buffer solution is a solution of: 1. A weak acid or a weak base and 2. The salt of the weak acid or weak base Both must be present! A buffer solution has the ability to resist changes in pH upon the addition of small amounts of either acid or base, and upon dilution. • Examples: HAc/NaAc NH3/NH4Cl • Importance: Most chemical reactions take place at constant pH Proper functioning of Enzymes Metabolic processes Calibration of pH-meters 6 Consider mixture of weak acid HA and its salt NaA HA (aq) H+ (aq) + A- (aq) NaA (s) Na+ (aq) + A- (aq) Ka = [H+][A-] [HA] [H+] = Ka [HA] [A-] -log [H+] = -log Ka - log [HA] [A-] -log [H+] = -log Ka + log [A-] [HA] pH = pKa + log [A-] [HA] Henderson-Hasselbalch equation Ref. Chang, Sect. 16.2 pH = pKa + log [conjugate base] [acid] pH of Buffers What is the pH of a solution containing 0.500 M CH3COOH and 0.500 M CH3COONa? CH3COOH(aq) H+ (aq) + CH3COO- (aq) Initial (M) Change (M) Equilibrium (M) 0.500 0.00 -x +x 0.500 - x 0.500 +x x 0.500 + x Neglect x 0.500 – x ≈ 0.500 0.500 + x ≈ 0.500 pH = pKa + log[CH3COO-] [CH3COOH] CH3COOH pKa = 4.76 pH = 4.76 + log[0.500] [0.500] Mixture of weak acid and conjugate base! pH = 4.76 An equal molar mixture of CH3COOH and CH3COONa 7 Buffer Action: Addition of Base Add strong base, the stoichiometric reaction: OH- (aq) + CH3COOH (aq) CH3COO- (aq) + H2O (l) Add 0.010mol of solid NaOH to 1.00L of the acetate buffer solution. Calculate the pH Before 0.010 0.500 0.500 After 0.0 0.490 0.510 Then, the equilibrium reaction is treated with new initial concentrations, [CH3COOH]0=0.490M [CH3COO-]0 =0.510M Plug into Henderson-Hasselbalch equation pH = pKa + log [conjugate base] [acid] pH=4.76+log(0.510)/(0.490) pH=4.78 Only 0.02 pH unit change (increase) Add this to 1.00L of pure water: pH rises from 7.00 to 12.00! 5 pH units. Buffer Action: Addition of Acid Add strong acid, the stoichiometric reaction: H+ (aq) + CH3COO- (aq) CH3COOH (aq) Add 0.010mol of gaseous HCl to 1.00L of the buffer solution Before 0.010 0.500 0.500 After 0.0 0.490 0.510 Then, the equilibrium reaction is treated with new initial concentrations, [CH3COOH]0=0.510M [CH3COO-]0 =0.490M Plug into Henderson-Hasselbalch equation pH = pKa + log [conjugate base] [acid] pH=4.76+log(0.490)/(0.510) pH=4.74 Only 0.02 pH unit change (decrease) Add this to 1.00L of pure water: pH drops from 7.00 to 2.00! 5 pH units. 8 HCl H+ + Cl- HCl + CH3COO- CH3COOH + Cl- Ref. Chang, Sect. 16.3 Strength of a Buffer • Buffer capacity represents the amount of protons or hydroxide ions the buffer can absorb without a significant change in pH. • The higher the initial concentrations of acid and salt, the stronger the buffer action (the higher the buffer capacity). • The pH of a buffer is determined by the ratio of [A-]/[HA], while the capacity of the buffer is determined by the magnitudes of [HA] and [A-]. • The optimal buffer solution is the equimolar mixture of acid and its salt. • When choosing a buffer with a desired pH, the pKa of the acid should be as close as possible to the desired pH; For example, if the needed pH is 4.00 then the best choice will be an equimolar mixture of the acid HA whose pKa is 4.00 and its salt. (Ka=1.0x10-4) pH=pKa + log(1) pH=pKa 9 = 9.20 Calculate the pH of the 0.30 M NH3/0.36 M NH4Cl buffer system. What is the pH after the addition of 20.0 mL of 0.050 M NaOH to 80.0 mL of the buffer solution? NH4 + (aq) H+ (aq) + NH3 (aq) pH = pKa + log [NH3] [NH4 +] pKa = 9.25 pH = 9.25 + log [0.30] [0.36] = 9.17 NH4 + (aq) + OH- (aq) H2O (l) + NH3 (aq) start (moles) end (moles) 0.029 0.001 0.024 0.028 0.0 0.025 pH = 9.25 + log [0.25] [0.28] [NH4 +] = 0.028 0.10 final volume = 80.0 mL + 20.0 mL = 100 mL [NH3] = 0.025 0.10 Ref. Chang, Sect. 16.3 Buffer of Weak Base and its Salt Indicators An acid-base indicator is a substance that displays different colors at different pH. Such indicators are weak acids or weak bases depending on whether they are present in solution as their acidic form (HIn) or as their basic form (In-). HIn + H2O � H3O+ + In� (acidic) (basic) Ka = Ka= Acid dissociation constant As the pH of a solution changes, the equilibrium shown above will be driven either towards reactants (HIn) or products (In-). At low pH values most of the indicator will be present in its acidic form, causing the solution color to correspond to that of HIn. At high pH values, most of the indicator will be in the In- form, causing the solution color to correspond to that of In-. [ HIn ] In H O ] ][ [ 3 − + 10 • Rearranging the above equation • ] [ ] [ In− HIn [H3O+] = Ka • Taking the -log of both sides - log [H3O+] = - log Ka - log pH = pKa + log [ ] [ HIn] In− when the concentrations of HIn and In- in solution are equivalent, pH will be equal to the pKa of the indicator. Spectrophotometrically, this corresponds to the pH where the absorbance of each form is half of its maximum absorbance. The indicator used in this experiment is Bromothymol Blue, BTB, also called 3,3’-dibromothymolsulfone-phthalein (pH range: 6.0-7.8). It is YELLOW in acidic solutions and BLUE in basic solutions. The intermediate color is the green. [ ] [ In− ] HIn Determine the wavelength (�1) where the acidic (HIn) but not the basic form (In-) of the indicator strongly absorbs radiation and the wavelength (�2) where the basic but not the acid form strongly absorbs radiation. Method 1: • Prepare a set of BTB solutions at different pH (Refer to the manual). • Measure the absorbance of each solution at the two wavelengths. • Plot Absorbance versus pH at each wavelength. • Determine pKa: It corresponds to the pH where the absorbance of each form is half of its maximum absorbance. Determination of pKa of BTB 11 A plot of absorbance as a function of pH for an indicator. Determination of pKa of BTB (continued) Method 2: The ratio [In-]/[HIn] can be obtained from spectrophotometric measurements which are made at two wavelengths (�1) and (�2). If Beer's law is obeyed, the absorbance at �1 and �2 are A (λ1) = ε (HIn, λ1) b [HIn] A (λ2) = ε (In-, λ2) b [In-] where A is absorbance, � is molar absorptivity and b is cell pathlength. [In-] [HIn] pH = pKa + log 12 At any pH, the total concentration (CT) of both forms of the indicator is constant and is the sum of the individual concentrations of each species: CT = [HIn] + [In-] In highly acid solutions (lowest pH), CT = [HIn] In highly basic solutions (highest pH) CT= [In-] By substitution, taking ratios and rearrangements, we can write The plot of pH versus log [In-]/[HIn] yields a straight line of an intercept equal to pKa A plot of pH as a function of log([In-]/[HIn]).