Chapter7: Statistical Data Treatment And Evaluation
Section: Chapter Questions
Problem 7.10QAP
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Using Equation 8-7 and Table 8-1 in your text, calculate the activity coefficient for Ca in a 0.01 M solution of Ca(ClO4)2?
Transcribed Image Text:TABLE 8-1 Activity coefficients for aqueous solutions at 25°C
Ionic strength (P, M)
Ion
size
0.005
(0.01
0.05
(0.001
0.1
(а, pm)
Ion
%3D
Charge = ±1
0.914
0.86
0.933
0.83
0.967
900
0.931
0.912
0.85
0.966
0.82
800
0.909
0.845
(C,Hs) CHCO,, (C,H-),N*
(O-N),C,H-O. (CH-);NH, CH,OC,H,CO
Li, CH.CO, HOC,H CO, CiC,H,CO, C,H;CH,CO,,
CH=CHCH-CO,. (CH3),CHCH,CO,, (CH;CH,),N, (C;H;),NH;
Cl-CHCO,, C1CCO, (CH;CH,);NH, (C,H;)NH
Na, CdCI, Cl0,, 10, HCO,, H;PO,, HSO, , H,AsO4 ,
Co(NH3)4(NO-), CH;CO,, C1CH,CO,, (CH3),N*,
(CH;CH-),NH, H,NCH,CO,
H,NCH-CO,H, (CH3);NH", CH;CH;NH
OH, F, SCN, OCN, HS , C1O, C10,, BrO3, IO4, MnO,,
HCO, Hcitrate, CH,NH, (CH3),NH;
K, C1, Br, I, CN, NO,, NO,
Rb , Cs, NH, TI*, Ag
0.930
0.81
0.965
700
0.929
0.907
0.835
600
0.965
0.80
0.964
0.928
0.904
0.83
0.79
500
0.928
0.902
0.82
450
0.964
0.775
0.927
0.901
0.815
0.77
400
0.964
350
0.964
0.926
0.900
0.81
0.76
300
0.964
0.925
0.899
0.805
0.755
250
0.964
0.924
0.898
0.80
0.75
Charge = ±2
0.69
Mg, Be+
CH-(CH,CH,CO,)2, (CH;CH;CH,CO,)2
Ca, Cu, Zn²*, Sn", Mn², Fe2*, Ni²*, Co, C,H¿(CO2)2,
H;C(CH-CO)2, (CH;CH;CO,)2
Sr*, Ba", Cd²*, Hg²*, S¯, S,0 , Wo , H,C(CO, )2, (CH2CO5 )2,
(CHOHCO)2
Pb, CO, So;, Moo, Co(NH3);CI²*, Fe(CN);NO² , C,0,
Hcitrate
Hg, So , S,0 , S,0, S,0 , SeO, Cro, HPO?
800
0.872
0.755
0.52
0.45
700
0.872
0.755
0.685
0.50
0.425
600
0.870
0.749
0.675
0.485
0.405
500
0.868
0.744
0.67
0.465
0.38
450
0.867
0.742
0.665
0.455
0.37
400
0.867
0.740
0.660
0.445
0.355
Charge = ±3
A13, Fe*, Cr*, Sc*, Y³*, In³+, lanthanides"
citrate-
900
0.738
0.54
0.445
0.245
0.18
500
0.728
0.51
0.405
0.18
0.115
PO, Fe(CN) , Cr(NH;),", Co(NH3),“, Co(NH3);H,O**
400
0.725
0.505
0.395
0.16
0.095
Charge = ±4
Th, Zr, Cet, Sn+
Fe(CN)
1 100
0.588
0.35
0.255
0.065
0.10
500
0.57
0.31
0.20
0.048
0.021
a. Lanthanides are elements 57-71 in the periodic table.
SOURCE: J. Kielland, J. Am. Chem. Soc. 1937, 59, 1675.
Transcribed Image Text:To interpolate a value of y, we can set up a proportion:
Unknown y interval
known x interval
Interpolation:
(8-7)
Ay
Ax
0.83 – y
20 - 16
=y = 0.766
%3D
0.83 - 0.67
20 - 10
For x = 16, our estimate of y is 0.766.
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