19.02 Phosphate Buffer Capacity Buffer capacity refers to the amount of acid or

Question

19.02 Phosphate Buffer Capacity Buffer capacity refers to the amount of acid or base a buffer can "absorb" without a significant pH change. It is governed by the concentrations of the conjugate acid and base forms of the buffer. A 0.5 M buffer will require five times as much acid or base as a 0.1 M buffer for a given pH change. In this problem you begin with a buffer of known pH and concentration and calculate the new pH after a particular quantity of acid or base is added. In the laboratory you will carry out some stepwise additions of acid or base and measure the resulting pH values. Reaction: HPO4 2? + H3O+ H2PO4 ? Initial nin x nin Change -x -x +x Equilibrium nin - x small nin + x Starting with 37 mL of 0.50 M phosphate buffer, pH = 6.83, you add 4.9 mL of 1.00 M HCl. Using the Henderson Hasselbalch equation with a pK2 for phosphate of 6.64, calculate the following values to complete the ICE table.(Carry out your intermediate calculations to at least one more significant figure than needed. See "Phosphate Buffer Issues" to explain why you do not use the value of 7.21 for pK2 that is given in your textbook.) 1.) What is the composition of the buffer to begin with in terms of the concentration of the two major species? 2.) What is the millimolar quantity of the two major phosphate species? 3.) What is the millimolar quantity of H3O+ added as HCl (x in the ICE table)? 4.) What is the final millimolar quantity of HPO4 2? and H2PO4 ? at equilibrium? 5.) Calculate the new HPO4 2?/H2PO4 ? ratio, the log(HPO4 2?/H2PO4 ? ) and the new pH of the solution. (Note: You can use the molar ratio rather than the concentration ratio because both species are in the same volume.) 6.) log(HPO4 2?/H2PO4 ?) 7.) pH = 8.) Now take another 37 mL of the 0.50 M pH 6.83 buffer and add 1.3 mL of 1.00 M NaOH. Using steps similar to those above, calculate the new pH of the solution.

Explanation / Answer

The solution is at a maximum buffer capacity (the minimum dpH/dn) when pH = pKa, as can be seen algebraically: Ka = [H3O+][A-]/[HA]; [H3O+] = Ka[HA]/[A-] log [H3O+] = log Ka + log ([HA]/[A-]) pH = pKa - log([HA]/[A-]) This rearranged form of the ionization constant of an acid is sometimes called the Henderson-Hasselbalch equation. Since pH = pKa if and only if [HA] = [A-], maximum buffer capacity of a conjugate acid-base pair is always found at pH = pKa. Reasonable buffer capacity is still obtained if the ratio of [HA] to [A-] is somewhere between 0.1 and 10, but outside this range it decreases rapidly. A chemical buffer can be effective about pH = pKa +/- 1. To make up a buffer solution one would first choose a reasonable acid-base conjugate pair on the basis of its pKa value and any other chemical constraints (one would not use the HCN/CN- pair, for example, in biological systems because cyanide is toxic). One would then adjust the concentrations such that both are reasonable (say 10-3 molar or above) while the desired pH is still obtained. The conjugate acid-base pairs glycinium ion/glycine, hydrogen citrate/citrate, dihydrogen phosphate/monohydrogen phosphate, and boric acid/hydrogen borate are often used for the preparation of aqueous buffers. Example. Let us select an appropriate buffer for an aqueous solution of pH 8.00. Of the acid-base conjugate pairs commonly used for buffer preparation, boric acid/dihydrogen borate could be useful at pH 8.00 as could dihydrogen phosphate/monohydrogen phosphate. The monoprotic species hydrazine and hypochlorous acid have acid ionization constants of the right magnitude but tend to be otherwise reactive in aqueous solutions. Using the boric acid/dihydrogen borate pair, [H3O+][H2BO3-]/[H3BO3] = 5.78 x 10-10 [H2BO3-]/[H3BO3] = 5.78 x 10-10/(1.0 x 10-8) = 0.0578 Borate buffers are most conveniently prepared from solid sodium borate with addition of sufficient hydrochloric acid to obtain the correct pH. In this example that would require for each mole of sodium borate almost three moles of hydrochloric acid. Example. Let us select an appropriate buffer for an aqueous solution of pH 5.00 and calculate the equilibrium concentration ratio of acid and base forms at this pH. Then we will calculate the concentrations of acid and base forms if the total concentration of both is 0.01 molar. For a buffer of pH 5.00 one would ideally have an acid-base conjugate pair for which pKa = 5.00. No convenient such conjugate acid-base pair exists, but for acetic acid pKa = 4.756 which is reasonably close. Ka = [H3O+][CH3COO-]/[CH3COOH] = 1.75 x 10-5 [CH3COO-]/[CH3COOH] = 1.75 x 10-5/1.0 x 10-5 = 1.75 [CH3COO-] + [CH3COOH] = 0.01; 1.75[CH3COOH] + [CH3COOH] = 0.01; 2.75[CH3COOH] = 0.01; [CH3COOH] = 0.01/2.75 = 0.00364 [CH3COO-] = 0.01 - 0.00364 = 0.00636 Any procedure which yields these concentrations will produce an appropriate buffer. Example. Let us calculate the concentrations of the buffer ions in a phenol/phenolate buffer whose pH is exactly 10.50. We can then give appropriate instructions for the preparation of this buffer from phenol and sodium phenolate, assuming the total phenol/phenolate concentration is to be 0.10 molar. We begin with the acid ionization constant of the phenolate ion: Ka = 1.08 x 10-10 = [H3O+][C6H5O-]/[C6H5OH] 1.08 x 10-10/3.16 x 10-11 = 3.42 = [C6H5O-]/[C6H5OH] [C6H5O-] + [C6H5OH] = 0.10 3.42 = [C6H5O-]/(0.10 - [C6H5O-]) 0.342 - 3.42[C6H5O-] = [C6H5O-]; 0.342 = 4.42[C6H5O-] [C6H5O-] = 0.0773, [C6H5OH] = 0.0227 While any buffer solution having the correct mole ratio of 3.42 will have the correct pH, only the concentrations above would total 0.10 molar. Appropriate instructions would be to prepare a 0.100 molar solution of phenol and add to it sufficient strong base (NaOH) to amount to 0.0773 mol. This would produce the desired concentrations.

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