Question A6.3: For the countercurrent extraction in Example A6.2, calculate...
For the countercurrent extraction in Example A6.2, calculate the recovery and separation factor for solute A if the contents of tubes 85-99 are pooled together.
From Example A6.2 we know that after 100 steps of the countercurrent extraction, solute A is normally distributed about tube 90 with a standard deviation of 3. To determine the fraction of solute in tubes 85-99, we use the single-sided normal distribution in Appendix 1A to determine the fraction of solute in tubes 0-84 and in tube 100 . The fraction of solute A in tube 100 is determined by calculating the deviation z (see Chapter 4 )
z=\frac{r-\mu}{\sigma}=\frac{99-90}{3}=3
and using the table in Appendix 1A to determine the corresponding fraction. For z=3 this corresponds to 0.135 \% of solute A. To determine the fraction of solute \mathrm{A} in tubes 0-84, we again calculate the deviation
z=\frac{r-\mu}{\sigma}=\frac{85-90}{3}=-1.67
From Appendix 1A we find that 4.75 \% of solute A is present in tubes 0-84. Solute A’s recovery, therefore, is
100 \%-4.75 \%-0.135 \% \approx 95 \%
To calculate the separation factor, we must determine the recovery of solute \mathrm{B} in tubes 85-99. This is determined by calculating the fraction of solute B in tubes 85-100 and subtracting the fraction of solute B in tube 100 . By calculating z and using Appendix 1A, we find that approximately 10.6 \% of solute \mathrm{B} is in tubes 85-100, and that essentially no solute \mathrm{B} is in tube 100 . The separation factor, S_{\mathrm{B}, \mathrm{A}} therefore, is
S_{\mathrm{B}, \mathrm{A}}=\frac{R_{\mathrm{B}}}{R_{\mathrm{A}}}=\frac{10.6}{95}=0.112
| Appendix 1A | ||||||||||
| Single-Sided Normal Distribution^a | ||||||||||
| u | 0.00 | 0.01 | 0.02 | 0.03 | 0.04 | 0.05 | 0.06 | 0.07 | 0.08 | 0.09 |
| 0.0 | 0.5000 | 0.4960 | 0.4920 | 0.4880 | 0.4840 | 0.4801 | 0.4761 | 0.4721 | 0.4681 | 0.4641 |
| 0.1 | 0.4602 | 0.4562 | 0.4522 | 0.4483 | 0.4443 | 0.4404 | 0.4365 | 0.4325 | 0.4286 | 0.4247 |
| 0.2 | 0.4207 | 0.4168 | 0.4129 | 0.4090 | 0.4052 | 0.4013 | 0.3974 | 0.3936 | 0.3897 | 0.3859 |
| 0.3 | 0.3821 | 0.3783 | 0.3745 | 0.3707 | 0.3669 | 0.3632 | 0.3594 | 0.3557 | 0.3520 | 0.3483 |
| 0.4 | 0.3446 | 0.3409 | 0.3372 | 0.3336 | 0.3300 | 0.3264 | 0.3228 | 0.3192 | 0.3156 | 0.3121 |
| 0.5 | 0.3085 | 0.3050 | 0.3015 | 0.2981 | 0.2946 | 0.2912 | 0.2877 | 0.2843 | 0.2810 | 0.2776 |
| 0.6 | 0.2743 | 0.2709 | 0.2676 | 0.2643 | 0.2611 | 0.2578 | 0.2546 | 0.2514 | 0.2483 | 0.2451 |
| 0.7 | 0.2420 | 0.2389 | 0.2358 | 0.2327 | 0.2296 | 0.2266 | 0.2236 | 0.2206 | 0.2177 | 0.2148 |
| 0.8 | 0.2119 | 0.2090 | 0.2061 | 0.2033 | 0.2005 | 0.1977 | 0.1949 | 0.1922 | 0.1894 | 0.1867 |
| 0.9 | 0.1841 | 0.1814 | 0.1788 | 0.1762 | 0.1736 | 0.1711 | 0.1685 | 0.1660 | 0.1635 | 0.1611 |
| 1.0 | 0.1587 | 0.1562 | 0.1539 | 0.1515 | 0.1492 | 0.1469 | 0.1446 | 0.1423 | 0.1401 | 0.1379 |
| 1.1 | 0.1357 | 0.1335 | 0.1314 | 0.1292 | 0.1271 | 0.1251 | 0.1230 | 0.1210 | 0.1190 | 0.1170 |
| 1.2 | 0.1151 | 0.1131 | 0.1112 | 0.1093 | 0.1075 | 0.1056 | 0.1038 | 0.1020 | 0.1003 | 0.0985 |
| 1.3 | 0.0968 | 0.0951 | 0.0934 | 0.0918 | 0.0901 | 0.0885 | 0.0869 | 0.0853 | 0.0838 | 0.0823 |
| 1.4 | 0.0808 | 0.0793 | 0.0778 | 0.0764 | 0.0749 | 0.0735 | 0.0721 | 0.0708 | 0.0694 | 0.0681 |
| 1.5 | 0.0668 | 0.0655 | 0.0643 | 0.0630 | 0.0618 | 0.0606 | 0.0594 | 0.0582 | 0.0571 | 0.0559 |
| 1.6 | 0.0548 | 0.0537 | 0.0526 | 0.0516 | 0.0505 | 0.0495 | 0.0485 | 0.0475 | 0.0465 | 0.0455 |
| 1.7 | 0.0446 | 0.0436 | 0.0427 | 0.0418 | 0.0409 | 0.0401 | 0.0392 | 0.0384 | 0.0375 | 0.0367 |
| 1.8 | 0.0359 | 0.0351 | 0.0344 | 0.0336 | 0.0329 | 0.0322 | 0.0314 | 0.0307 | 0.0301 | 0.0294 |
| 1.9 | 0.0287 | 0.0281 | 0.0274 | 0.0268 | 0.0262 | 0.0256 | 0.0250 | 0.0244 | 0.0239 | 0.0253 |
| 2.0 | 0.0228 | 0.0222 | 0.0217 | 0.0212 | 0.0207 | 0.0202 | 0.0197 | 0.0192 | 0.0188 | 0.0183 |
| 2.1 | 0.0179 | 0.0174 | 0.0170 | 0.0166 | 0.0162 | 0.0158 | 0.0154 | 0.0150 | 0.0146 | 0.0143 |
| 2.2 | 0.0139 | 0.0136 | 0.0132 | 0.0129 | 0.0125 | 0.0122 | 0.0119 | 0.0116 | 0.0113 | 0.0110 |
| 2.3 | 0.0107 | 0.0104 | 0.0102 | 0.00964 | 0.00914 | 0.00866 | ||||
| 2.4 | 0.00820 | 0.00776 | 0.00734 | 0.00695 | 0.00657 | |||||
| 2.5 | 0.00621 | 0.00587 | 0.00554 | 0.00523 | 0.00494 | |||||
| 2.6 | 0.00466 | 0.00440 | 0.00415 | 0.00391 | 0.00368 | |||||
| 2.7 | 0.00347 | 0.00326 | 0.00307 | 0.00289 | 0.00272 | |||||
| 2.8 | 0.00256 | 0.00240 | 0.00226 | 0.00212 | 0.00199 | |||||
| 2.9 | 0.00187 | 0.00175 | 0.00164 | 0.00154 | 0.00144 | |||||
| 3.0 | 0.00135 | |||||||||
| 3.1 | 0.000968 | |||||||||
| 3.2 | 0.000687 | |||||||||
| Single-Sided Normal Distribution^a—continued | ||||||||||
| u | 0.00 | 0.01 | 0.02 | 0.03 | 0.04 | 0.05 | 0.06 | 0.07 | 0.08 | 0.09 |
| 3.3 | 0.000483 | |||||||||
| 3.4 | 0.000337 | |||||||||
| 3.5 | 0.000233 | |||||||||
| 3.6 | 0.000159 | |||||||||
| 3.7 | 0.000108 | |||||||||
| 3.8 | 0.0000723 | |||||||||
| 3.9 | 0.0000481 | |||||||||
| 4.0 | 0.0000317 | |||||||||
| 4.1 | 0.0000207 | |||||||||
| 4.2 | 0.0000133 | |||||||||
| 4.3 | 0.00000854 | |||||||||
| 4.4 | 0.00000541 | |||||||||
| 4.5 | 0.00000340 | |||||||||
| 4.6 | 0.00000211 | |||||||||
| 4.7 | 0.00000130 | |||||||||
| 4.8 | 0.000000793 | |||||||||
| 4.9 | 0.000000479 | |||||||||
| 5.0 | 0.000000287 | |||||||||
^aThis table gives the proportion, P, of the area under a normal distribution curve that lies to the right of the deviation z, where z is defined as
z = (X – µ)/σ
For example, the proportion of the area under a normal distribution curve that lies to the right of a deviation of 0.04 is 0.4840, or 48.40%. The area to the left of the deviation is given as 1 – P. Thus, 51.60% of the area under the normal distribution curve lies to the left of a deviation of 0.04. When the deviation is negative, the values in the table give the proportion of the area under the normal distribution curve that lies to the left of z; therefore, 48.40% of the area lies to the left, and 51.60% of the area lies to the right of a deviation of –0.04.