Question

TIe-LaboratoTV ASSISITICI 1. A student followed the procedure of this ex KIO, solutions to the first appearance of a the procedure of this experiment to determine the solubility product of zinc(IT) 2. Solutions of Zn(NO), of known initial concentrations were titrated with 0.200M ne first appearance of a white precipitate. The following data were collected: constant for zinc(II) iodate. ne table below and, by making the appropriate plot, determine the solubility product (Use additional paper for doing the calculations required to complete the 0.0452 0.0118 [Zn(NO3)2, M initial 0.226 0101/ 0.200 0.200 [KIOslo, M titrant 0.200 0.200 100.0 100.0 Vo mL of Zn(NO3)2 100.0 100.0 13.0 18.3 V, mL of KIO, titrant 12.9 118.3 3 Vo + V, mL 112.4 112.9 Ooh .099 [Zn2+1 [103] 0.023 [Zn2+][10,12 1.06 x 10" log (Zn2+1/10,12 - 3.98 VZn²+] 1 + V (Zn2+] log Ksp. Ksp of Zn(IO3)2 –

Answer 1

" A student followed the procedure of this experiment to de the Ky of zinc (El iodate, an (703) solution of Zn(NO3)2 of known concentration were titrated with 0.200 m kios solutions, to the first apperence of a white precipitate the following data were collected. 0.0452 0.0114 (2n (NO3)2] andal m 0.226 0.101 0.200 0. 200 0.200 1000 [k103] M Vinitial of Zn(NO3)2 ML of thoz added mL 100.0 0.200 100.0 1219 100.0 18.3 13.0 1120.4 113.0 118.3 0.010 0.040 0090 0.200 0.022 0.023 0.031 9.68x106 At endpoint veotal mi [2n2+] .M (9031).m [20²] [103] ² log (C 2n2") (103)) izn247 14 Jizn2+] 4.36x105 2012 x105 0.023 1.06x104 - 3.97 0.309 -5:02 -4.67 -4.36 0.091 0.231 0.167

Graph using equation 7.11, equal to y. intercept. log kepubble top of an (103)2 = 3.42 x 106 found in which of the ion product Ksp ? why? question is closest to the the titration volumes for the first three sample in question I do not vary greatly the fitration volume for the last sample. however, is considerably larger than that of the first samples. why is this to be expected? the table is filled with the following calculations. Total = u zn (NO3)2 + vk 103 (zn+2] = (zn (No3) 2) intial * vzn (NO3)2 lutotal [103-] = [ki 03] initial * v K103 ( utatal [zn +2]* (103 -72 109 Cr2n +2] * [103-73) r (2n+2])i + V (2n+2] A plot of the log data is made ((2n+2] * (103-73) verses rizn+2] /1+ Vizn+2).

-3,97 Log [Zn+2]*[1-1^2 V[Zn+2] /1 + 1 Log [Zn+2] *[1- 0,309 0,231 -4,36 0,167 -4,67 0,091 -5,02 1 0 0,05 0,1 0,15 0,2 0,25 0,3 0,35 -Log (Zn+2]*[O-]^2 y = 4,8193x-5,4664 - Lineal (Log (Zn+2]* [l- 1^2)

and you get the cut with the yaris, in this case - 5.ubby the antilogarithm is calculated and you have ksp = Antilog (-5,4664) : 3.42x tó