2. HARRIS’s Ch. 10-12 Supplement Zumdahl’s Chapter 15

3. Supplementary Content  Harris Chapter 10 Buffer Capacity,  Ionic Strength and Henderson–Hasselbalch  Harris Chapter 11 Diprotics Intermediate form Buffers Fractional Composition  Harris Chapter 12 Titration accuracy limit Diprotic titration Indicated indicators Spreadsheet Titration Curves Monoprotic Diprotic Triprotic

4. Buffer Capacity,   Buffers resist pH change only while both forms present in significant quantity.  We want slope of pH with C base to be low!  Alternately, dC b /dpH should be high and represents , the capacity to resist.   = (ln 10) SA / (S+A) while H–H viable. Limits:  min =2.303 [smaller],  max =1.152 [S=A]

5. Henderson–Hasselbalch  pH = pK a + log( a S / a A )  pK a + log([S]/[A])  a Y =  Y  [Y] = thermodynamic “activity”  Y = “activity coefficient,” a correction factor, well known for dilute ionic solutions. (Ch. 8)  For Y z±, ion of radius  (pm) in a solution of many ions and  = ½  c i z i 2 is ionic strength, log  Y = – 0.51z 2  ½ / [ 1 + (   ½ / 305) ]

6. Diprotic Intermediate Form H 2 A  HA – + H + ( abbreviated below as A  B + H) HA –  A 2– + H + ( abbreviated below as B  C + H) HA – is the intermediate (amphiprotic) form.  HA – dominates at 1 st equivalence point, where  ~ zero! World’s worst buffer.  Also, near equivalence, K w is important (to all conjugates), and must be included. How?

7. Explicit Water Equilibrium  If both H + and OH – are important, they must be included in MHA’s charge balance: [H + ] + [M + ] = [HA – ] + 2[A 2– ] + [OH – ] or H + F  H + (A + B + C) = B + 2C + K w /H H = C – A + K w /H = K 2 B/H – HB/K 1 + K w /H H 2 = K 2 B – H 2 B/K 1 + K w H 2 (B/K 1 + 1) = K 2 B + K w

8. Solution from B = HA – H = [ (K 1 K 2 B + K 1 K w ) / (B + K 2 ) ] ½  But if weak [B] 0 = F, then B eq  F >> K 2 H  [ (K 1 K 2 F + K 1 K w ) / F ] ½  And if K w << F K 2 H ~ [ K 1 K 2 F / F ] ½ = (K 1 K 2 ) ½ pH ~ ½ (pK 1 + pK 2 ) Like pH = pK a at V ½ for monoprotic titration.

9. Diprotic Buffer Solutions  Both Henderson–Hasselbalch eqns apply! pH = pK 1 + log( [HA – ] / [H 2 A] ) pH = pK 2 + log( [A 2– ] / [HA – ] ) same pH!  Make buffer of A & B or B & C, using the appropriate pK; the other H–H equation then governs the third species (not added).  What if we add amounts of A, B, and C? 

10. Fractional Composition,   For monoprotic,  (A – ) = % diss. / 100% (wow) Since [HA] = [H + ][A – ]/K a = [H + ](F–[HA])/K a, [HA] = [H + ] F / ( [H + ] + K a )  HA = [HA] / F = [H + ] / ( [H + ] + K a )  A¯ = 1 –  HA = K a / ( [H + ] + K a )  For diprotic,  (H 2 A) +  (HA – ) +  (A 2– ) = 1 and after we agonize over mass balance,

11. Diprotic Fractional Composition   H 2 A = [H + ] 2 / D   HA¯ = K 1 [H + ] / D   A²¯ = K 1 K 2 / D  D = [H + ] 2 + K 1 [H + ] + K 1 K 2 gives the max  HA¯ where [H + ] = (K 1 K 2 ) ½ and its max value is K 1 ½ / (K 1 ½ + 2 K 2 ½ )

12. Titration Accuracy Limits  Besides indicators and eyeballs, endpoint accuracy depends upon high sensitivity of pH to V titrant. Here are things to avoid: Very weak acids have low pH sensitivity there. Very dilute solutions are similarly insensitive. Very strong analytes may be sensitive but imply refilling burettes, increasing reading errors.

13. Diprotic Titrations  Deal with two buffer regions and equivalence points associated with the K a1 and K a2. While buffer formulae (H–H) are identical, equivalence point formulae are not. pH eq2  14 + ½ log[ F K w /K a2 ] (same as monoprotic) pH eq1  ½ ( pK a1 + pK a2 ) Assuming the curve shows clear equivalence points.

14. Choosing Indicators  Paradoxical as it may seem, the proper indicator has its best buffer region, pH = pK i where the analyte has its equivalence point, e.g. pH  14 + ½ log( F K w / K a ). At such a pH, the indicator has equal amounts of acid and conjugate base where the analyte has pure conjugate.

15. Spreadsheet Titration Curves  Assume C a and C b known; plot V b vs. pH. Because it’s easier than pH vs. V b ! [H + ] + [Na + ] = [A – ] + [OH – ] [Na + ] = C b V b / V total = C b V b / ( V a + V b ) [A – ] =  A¯ F a =  A¯ C a V a / ( V a + V b )   C b V b / ( C a V a )  = {  A¯ – ([H + ]–[OH – ])/C a } / {1+([H + ]–[OH – ])/C b }

16. Only Somewhat Twisted  = {  A¯ – ([H + ]–[OH – ])/C a } / {1+([H + ]–[OH – ])/C b } All known if pH is known.  A¯ = K a / ( [H + ] + K a ) [OH – ] = K w / [H + ]  Therefore V b =  C a V a / C b  as a function of pH. Excel doesn’t care which is the X or Y axis.

17. “To Diprotica and Beyond”  For weak polyprotic acids and strong bases:  X = 1 – ( [H + ] – [OH – ] ) / C a X  m = {  A¯ – ([H + ]–[OH – ])/C a } X  d = {  HA¯ + 2  A²¯ – ([H + ]–[OH – ])/C a } X  t = {  H 2 A¯ +2  HA²¯ +3  A³¯ – ([H + ]–[OH – ])/C a }  D for triprotic fractional compositions: D = [H + ]³ + K 1 [H + ]² + K 1 K 2 [H + ] + K 1 K 2 K 3