CHEM*2400/2480 Summer 2004 Assignment 7 ANSWERS

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1 CHEM*4/48 Summer 4 Assignment 7 ANSWERS 1. This is a buffer question. We simply need to find the ratio of the base-to-acid concentrations and use the Henderson-Hasselbalch equation. What makes this problem a little more interesting is that it is a diprotic acid. While using the equation for either proton 1 or would work, the data will be easily obtainable for only one instance. A buffer is most useful when it is within ±1 ph unit of the pk a value. Since the target ph is 5.15, we will use the pk a which is closest to that value, in this case, pk a. We first need to know the amount of H M added g n ^H Mh = 14.6 g mol =.43 mol We could determine the formal concentration from this, but we are really only interested in the concentration ratio which is identical to the mole ratio, since they are both in the same volume. So we won t go that unnecessary step. The required ratio comes from the Henderson-Hasselbalch equation. n ^M ph = pk a + log - h n HM - ^ h n ^M 5.15 = log - h n HM - ^ h n ^M log - h n HM - = =-.546 ^ h n ^M - h = =.84 n ^HM - h We start with H M and add base until we convert it all to HM - : this requires.43 mol of NaOH. We then add a little more base to convert some of the HM - to M -, sufficient to achieve the ratio calculated above. n ^M - h =.84 n ^HM - h n^hm - h =.43 - n^m - h Ñ n ^M - h =.84 _.43 - n^m - hi = n^m - h 1.84 n ^M - h =.1 n ^M - h = =.951mol This is the amount of base needed beyond the conversion of H M to HM -. The total amount of added base must therefore be n ^OH - h = =.55 mol But we want to know the volume of NaOH solution to add. V ^OH - h =.75 M.55 mol =.7 L = 7. ml. The molar mass of each compound is different because of the counter ions. In the case of the leucine-hydrochloride, an HCl is added to the leucine. In the case of the sodium leucinate, a proton is removed and replaced by a Na ion. The molar masses are then given by:

2 MM(L-HCl) = g/mol MM(L) = g/mol MM(NaL) = g/mol Since the weighed mass is the same in each case, the number of moles employed must be different. We find 1. g n ^L - HClh = g =.596 mol mol 1. g n ^Lh = g =.76 mol mol 1. g n ^NaLh = g =.653 mol mol The leucine hydrochloride introduces the fully acidic form of the molecule into solution. The sodium leucinate delivers the fully basic form. Leucine itself is the intermediate form. In the case of L-HCl and NaL, we use the quadratic solution for the ph while we need to use the intermediate form equation in the case of the leucine. (a) leucine hydrochloride: F ^L - HClh =.596 mol =.596 M.1 L (b) leucine: F ^Lh = 6 H = 6 H = ph = 6.51 ph = K a1 + K a1 + 4K a1 F =.145 M.76 mol =.76 M.1 L K a1 K a F + K a1 K w K a1 + F = 8.9 # 1-7 M alternate method as check: ph = 1 _ pk a1 + pk ai = 6.38 (c) sodium leucinate: F ^NaLh =.653 mol =.653 M.1 L K b1 = K w K a = 1.1 # # 1-1 = 5.64 # OH = poh =.73 -K b1 + K b1 + 4K b1 F =.189 M ph = 14 - poh = = This is just a direct application of the fractional composition equations. We need only find the [H + ] and then work out the value for each species. Remember that

3 these are fractional compositions, so that the results are not in terms of concentrations but rather are numbers that range from to 1. ph = H = 1 - ph = = 3.98 # 1-8 M 6 a H 3 A = H 3 = 7.6 # 1 6 H 3 + K a16 H K a1 K a6 H + K a1 K a K a3 a H A = K a16 H = 1.5 # 1-6 H 3 + K a16 H K a1 K a6 H + K a1 K a K a3 a HA = K a1 K a6 H = H 3 + K a16 H + K a1 K a6 H + K a1 K a K a3 a A = K a1 K a K a3 =.8 # H 3 + K a16 H K a1 K a6 H + K a1 K a K a3 Concentration is found by multiplying the formal concentration by each fractional composition. We have [H 3 A] = 7.6 x 1-9 (.1 mm) = 7.6 x 1-1 mm= 7.6 x 1-13 M [H A - ] = 1.5 x 1-3 (.1 mm) = 1.5 x 1-4 mm= 1.5 x 1-7 M [HA - ] =.996 (.1 mm) = 9.96 x 1 - mm= 9.96 x 1-5 M [A 3- ] =.8 x 1-3 (.1 mm) =.8 x 1-4 mm=.8 x 1-7 M 4. The iterative procedure proceeds as follows: (a) Use the formal concentration as [HM - ] in the intermediate equation to calculate a value for [H + ]. (b) Use these two values for [HM - ] and [H + ] in the two equilibrium expressions (K a1 and K a ) to find values for [H M] and [A - ]. (c) Find a new approximation for [HM - ] i+1 = F - [H M] i - [A - ] i. (d) Rinse and repeat (iterate through steps b and c) until [HM - ] is sufficiently stable, meaning [HM - ] i - [HM - ] i-1 /[HM - ] i <.1? The approximation is simply the value for [H + ] found in the first pass of this routine. (i) Approximate ph =.651 Exact Results (about 5 iterations): [H + ] = 1.9 mm ph =.717 [H M] = 1.1 mm [HM - ] = 5.84 mm [M - ] = 3.4 mm (ii) Approximate ph = 5.5 Exact Results (about 1 iterations): [H + ] = 3.16 x 1-6 M ph = [H M] = 1.94 x 1-3 M [HM - ] = 6.13 x 1-3 M [M - ] = 1.94 x 1-3 M (iii) Approximate ph = 6.5 Exact Results (about 5 iterations): [H + ] = 3.16 x 1-7 M ph = 6.5

4 [H M] =.97 x 1-4 M [HM - ] = 9.41 x 1-3 M [M - ] =.98 x 1-4 M You will note how the error was only.5 ph unit in the worst case, where the pk a difference was only a factor of 1 and the ph was very low. In the other cases, the approximate result was accurate to a precision greater than the input data would justify. This clearly confirms the validity of the approximation method in almost all circumstances. However, it should also be noted that the assumption that there is insignificant amount of dissociation is not true and even in the best case, the amount of dissociation is still about 6%; but this turns out to not have a big effect on the ph. 5. Here is the procedure for calculating points on a titration curve, involving a diprotic acid. Use the quadratic equation with K a1 for the starting ph. Use the quadratic equation with K b1 for the second equivalence point ph. Use the intermediate equation for the first equivalence point ph. Use the Henderson-Hasselbalch equation for all points between the equivalence points (K a1 for the first buffer region and K a for the second). Just consider dilution of the titrant beyond the last equivalence point. Remember throughout to adjust any formal concentration for dilution effects to that point in the titration. Also, throughout the buffer regions, you only need the ratio of the concentrations which is identical to the mole ratio; using moles rather than concentrations always eliminates some unnecessary calculation steps and nicely accounts for dilution effects at the same time. From the initial concentrations and volumes, it is easy to see that, since the titrant has twice the concentration of the analyte, that the first equivalence point will be found at a volume equal to half the original analyte volume, namely 5. ml. The second equivalence point is then necessarily at 5. ml of added titrant. Taking the titration to 6. ml goes out 1. ml beyond the last equivalence point. V =. ml 6 H = n ^H Mh =.5 M #.5 L = 1.5 # 1-3 mol -K a1 + K a1 + 4K a1 F = 4.76 # 1-3 M ph =.3 V = 5. ml n ^OH - h =.5 M #.5 L =.5 # 1-4 mol n ^HM - h = n ^OH - h =.5 # 1-4 mol n ^H Mh = 1.5 # # 1-4 = 1. # 1-3 mol 6 HM ph = pk a1 + log =.95 + log.5 # H M@ 1. # 1-3 =.348 The pattern continues through the first buffer region:

5 V = 1. ml n ^OH - h =.5 M #.1 L = 5. # 1-4 mol n ^HM - h = n^oh - h = 5. # 1-4 mol n ^H Mh = 1.5 # # 1-4 = 7.5 # 1-4 mol 6 HM ph = pk a1 + log =.95 + log 5. # H M@ 7.5 # 1-4 =.774 And so on V = 15. ml n ^OH - h =.5 M #.15 L = 7.5 # 1-4 mol n ^HM - h = n ^OH - h = 7.5 # 1-4 mol n ^H Mh = 1.5 # # 1-4 = 5. # 1-4 mol 6 HM ph = pk a1 + log =.95 + log 7.5 # H M@ 5. # 1-4 = 3.16 and again V =. ml n ^OH - h =.5 M #. L = 1. # 1-3 mol n ^HM - h = n ^OH - h = 1. # 1-3 mol n ^H Mh = 1.5 # # 1-3 =.5 # 1-4 mol 6 HM ph = pk a1 + log =.95 + log 1. # H M@.5 # 1-4 = 3.55 Another data point that we did not hit, but is easy to calculate is the 1/ E.P., where the ph = pk a. Hence at V = 1.5 ml ph = pk a1 =.95 The next point is the first equivalence point where we use the intermediate form equation to find the ph. V = 5. ml F =.5 M # = M 6 H = K a1 K a F + K a1 K w = 6.4 # 1 K a1 + F - 5 M ph = Now we move into the second buffer region. We need to find the ratio of HM - and M - and we use pk a in the buffer equation. V = 3. ml n ^OH - h =.5 M #.5 L =.5 # 1-4 mol beyond 1st E.P. n ^M - h = n ^OH - h =.5 # 1-4 mol beyond 1st E.P. n ^HM - h = 1.5 # # 1-4 = 1. # 1-3 mol 6 M ph = pk a + log HM - = log.5 # # 1-3 = 4.86 And without belabouring the point, the next three values are V = 35. ml ph = 5.3

6 V = 4. ml ph = V = 45. ml ph = 6.1 Also, the ph at the 1/ E.P. point is easily given again. V = = 37.5 ml ph = pk a = 5.48 Now we are at the second equivalence point. We first need to turn K a into K b1 and then use the quadratic solution for [OH - ] which in turn will give the ph. V = 5. ml F =.5 M # =.15 M K b1 = K w K = 1.1 # 1-14 a 3.9 # 1-6 =.59 # H -K b1 + K = b1 + 4K b1 F = 5.69 # 1-6 M poh = 5.45 ph = 14 - poh = 14 = 5.45 = The last two points only need to account for dilution of the excess added strong titrant. Remember that you will be directly calculating the poh which then needs to be turned into the ph. V = 55. ml 6 OH =.5 M # b l =.38 # 1-3 M poh =.63 ph = = V = 6. ml 6 OH =.5 M # b 1. l = # 1-3 M poh =.34 ph = = We now gather together all these data points and put them on a graph of added titrant volume vs. ph to create the titration curve. Volume Titrant (ml) ph

7 And the titration curve looks like Titration of Phthalic Acid ph Volume Added Titrant (ml) Note how the first E.P. is barely noticeable (at 5. ml) but the second E.P. is quite distinct and could be used to identify the titration endpoint. Under these conditions, your calculations would clearly suggest that if you were to experimentally employ this titration, you should not spend any effort in identifying the 1 st E.P. but focus only on the nd E.P. 6. Here we need only to code in the correct equations from the back of chapter 1. I give here the titration curves but will also make available for download the Excel file used. (a) A weak, monoprotic base titrated with a strong acid

8 Titrate a Weak Base with Strong Acid ph Volume Added Titrant (ml) (b) A weak, diprotic acid titrated with a strong base. Titration of Diprotic Acid with Strong Base ph Volume Added Titrant (ml) (c) Titration of a weak triprotic acid with a strong base.

9 Titrate Triprotic Weak Acid with Strong Base ph Volume Added Titrant (ml) 7. Here is the graph of the data. You will note that most measurements were made at intervals of 5. ml added titrant but near the endpoint, where the ph was changing quickly, the steps are 1 ml in size. This is, of course, to give better accuracy. But using these two techniques, we do not have to hit the equivalence point right on. First of all, here is the graph of the titration. Titration of Monoprotic Unknown Acid ph Volume of Added Titrant (ml)

10 Clearly we will be finding the endpoint somewhere in the vicinity of 37 or 38 ml. But to get an accurate measure we take the second derivative of the data. We need to find where the data crosses zero. The first graph is the second derivative of the entire data set. nd Deriv 4 3 Second Derivative Volume Added Titrant (ml) The data suggests the endpoint is just above 37 ml. We will zoom in on the graph between 37 and 38 ml.

11 nd Deriv Second Derivative Volume Added Titrant (ml) It is quite clear that the zero-crossing is at ml. Now, this endpoint depends upon the data points just around the equivalence point. The next procedure determines the endpoint but uses almost all of the titration data. (b) The Gran plot.

12 Gran Plot.1.1 V(base)*1^-pH Volume Added Titrant This uses the Gran function on all of the data. Clearly the data beyond the equivalence point and the first couple of data points near the beginning are not valid. If we leave them out and then fit the data to a linear least squares straight line, we find

13 Abbreviated Gran Plot y = -.344x The equation of the line can be used to find the value of the x-axis when extrapolated to the zero-crossing. We just want to find the value of x when y =. We find y = =-.344 x x = = ml You can see that both are valid ways to identify the endpoint in a titration. From here you could continue to calculate the unknown concentration.