Henderson-Hasselbalch Equations and Buffer Solutions PDF

Henderson-Hasselbalch equations Relate pH, pKa and acid/base concentrations  For bases Using pKa of conjugate acid  Buffer solutions Solution of weak acid and its salt (conjugate base) Or a weak base and its salt (conjugate acid) E.g. acetic acid / sodium acetate AcOH we...

Henderson-Hasselbalch equations Relate pH, pKa and acid/base concentrations  For bases Using pKa of conjugate acid  Buffer solutions Solution of weak acid and its salt (conjugate base) Or a weak base and its salt (conjugate acid) E.g. acetic acid / sodium acetate AcOH weakly dissociated (pKa 4.76); AcONa fully dissociated  Buffering effect Addition of HO- neutralised by AcOH Addition of H3O+ neutralised by AcO- 0.045 mol L-1 soln of AcOH; 0.045 mol L-1 soln of AcONa  Buffering effect E.g. 2.25 × 10−3 mol AcOH and AcONa Add 1.0 × 10−4 mol HNO3 AcO- + H+ → AcOH [AcO-] = (2.25 × 10−3 − 1.0 × 10−4 ) = 2.15 × 10−3 mol [AcOH] = (2.25 × 10−3 + 1.0 × 10−4 ) = 2.35 × 10−3 mol  Buffering effect Buffering effect: pH of solution reduced by addition of acid, but not greatly reduced (from 4.76 to 4.72) 2.25 × 10−3 mol AcOH and AcONa Addition of 1.0 × 10−4 mol HNO3 HNO3 (nitric acid): very strong acid, fully dissociated into H+ and NO3- Concentration of added acid much less than concentration of buffer solution Addition of equal or greater concentrations of added acid: buffering effect would break down.  Common buffers NaH2PO4 / Na2HPO4 pH range 6-8 KH2PO4 / K2HPO4 pH range 6-8 (HOCH2)3CNH2 / (HOCH2)3CNH3+ (TRIS) – pH range 7-9 HEPES pH range 6.8 – 8.2 Worked example: TRIS / TRIS-H+ Buffer pKa TRIS-H+ 8.08 What is the TRIS/TRIS-H+ ratio at pH 8.0 ? 𝐴𝐻+ 𝐴 𝑝𝐾𝑎 = 𝑝𝐻 + log10 𝑝𝐻 − 𝑝𝐾𝑎 = log10 𝐴 𝐴𝐻+ 𝑇𝑅𝐼𝑆 𝑇𝑅𝐼𝑆 −0.08 = log10 = 0.83 𝑇𝑅𝐼𝑆 − 𝐻 + 𝑇𝑅𝐼𝑆 − 𝐻 +  Addition of acid to 0.1 M TRIS/TRIS-H+ 0.1 M = [TRIS] + [TRIS-H+] [TRIS]/[TRIS-H+] = 0.83 at pH 8.0 [TRIS] = 0.83[TRIS-H+]; 0.1 M = 1.83[TRIS-H+] [TRIS-H+] = 0.055 M; [TRIS] = 0.045 M Add 0.005 M H+ (of a strong acid) [TRIS] = 0.040 M; [TRIS-H+] = 0.060 M 𝑇𝑅𝐼𝑆 − 𝐻 + 𝑝𝐻 = 𝑝𝐾𝑎 − log10 𝑇𝑅𝐼𝑆 0.06 = 8.08 − log10 = 7.9 0.04  Addition of acid to 0.02 M TRIS/TRIS-H+ 0.02 M = [TRIS] + [TRIS-H+] [TRIS]/[TRIS-H+] = 0.83 at pH 8.0 [TRIS] = 0.009 M; [TRIS-H+] = 0.011 M Add 0.005 M H+ [TRIS] = 0.004 M; [TRIS-H+] = 0.016 M 𝑇𝑅𝐼𝑆 − 𝐻 + 0.016 𝑝𝐻 = 𝑝𝐾𝑎 − log10 = 8.08 − log10 = 7. 48 𝑇𝑅𝐼𝑆 0.004 Buffer more effective at higher concentration  Tutorial 1. Write an equation for NaHCO3 acting as an acid in water and from that, derive an appropriate Henderson-Hasselbalch equation. 2. The pKa of NaHCO3 is 10.25. How many moles of NaCO3- would be present in a solution of 1.0 moles of NaHCO3 at pH 9.50?  Tutorial Write an equation for the acid dissociation of PhCO2H (benzoic acid) in water and use the equation to derive an appropriate form of the Henderson- Hasselbalch equation. If the pKa of benzoic acid is 4.19, what is the ratio of benzoic acid to benzoate anion, i.e. [PhCO2H]/[ PhCO2−], necessary to provide a solution of pH 3.5?  Ions in solution Behaviour of ions in solutions affected by many factors Degree of ionisation, extent of solvation, ion-ion vs. ion-solvent interactions, external fields, other phenomena In pharmaceutical solutions, need to consider the effect of dissolved ions Need to consider real (rather than ideal) behaviour: i.e. activity rather than concentration  Activity of ions in solution 𝜇𝑖 = 𝜇𝑖° + 𝑅𝑇 ln 𝑎𝑖 ai = activity of component i ai = i[i] i = activity coefficient of component i 𝜇𝑖 = 𝜇𝑖° + 𝑅𝑇 ln 𝑖 + 𝑅𝑇 ln 𝛾𝑖 quantifies deviations from ideality  Activity of ions in solution 𝜇𝑖 = 𝜇𝑖° + 𝑅𝑇 ln 𝑖 + 𝑅𝑇 ln 𝛾𝑖 log10 𝛾± = −𝐴𝑧+ 𝑧− 𝐼 Debye Huckel Limiting Law A is a constant (= 0.509 for water at 25 ◦C) z+ and z- are the charges of the +ve and –ve ions ‘±’ implies mean value for both ions I is the Ionic Strength  Ionic Strength I Depends on the number of cations and anions in solutions Quantifies the ionic field generated by a system of ions in solution 1 𝐼 = ෍ 𝑖 𝑧𝑖2 2  Worked example: Ionic strength and mean activity coefficient of 0.1 M Na3PO4 Na3PO4; fully dissociated in water to 3Na+ and PO43- zNa = 1; [Na] = 3 x 0.1 M zPO4 = 3; [PO4] = 0.1 M I = ½{[(0.3M)12] + [(0.1M)32]} = 0.60 M Log10ϒ± = −(0.509)(1)(3)√(0.6) = − 1.183 ϒ± = 0.066  Mean activity of 0.1 M Na3PO4 a± = ϒ±[±] ([±] = mean ionic concentration of salt) [±] = n√{([+])n+([-])n-} n = n+ + n- For Na3PO4, [±] = 4√{(0.3)3(0.1)} = 0.228 M a± (Na3PO4) = (0.066)(0.228 M) = 0.015 M Significant difference between ionic activity and concentration Activities determine equilibria for ions in solution  Ka, Kb and Kw Note: AH and A- corresponding acid and conjugate base (or corresponding base and conjugate acid) pKa + pKb = pKw  Tutorial Determine ionic strength I, mean activity coefficient γ±, mean ionic concentration [±] and mean activity a± for a 0.05 M solution of MgCl2 in water at 25 °C. Assume MgCl2 is fully dissociated in solution.