substitution of equ (10. 17a. 17b) into eq. d to gives Ordinarily eq (18) would
ID: 490203 • Letter: S
Question
substitution of equ (10. 17a. 17b) into eq. d to gives Ordinarily eq (18) would be sufficient to determine a single measurement of is a K, from A+rt and use would be available from independent measurements known However, the determination of E in this case is not orward parameter, usually I cm. (an) depends on the unknown um because the concentration of [FescNP is required. constant for reaction (10). so another approach Expanding the right-side denominator in eq. (18) gives (9a) Because the under conditions where [Fe'' [scN lo, and the reaction experiment will be done stoichiometry assures [scN llFescNP net, it can be concluded that [Fe "h Auvet, and the that lo applied to the right- approximation that [Fe' lo-A4evet [Fe' lo can be safely side denominator i eq. (19a) to give 09b) Following a series of algebraic manipulations multiply by clear fraction [Fe" lo ISCN le divide by [Fes JolsCN lo iFes JalscN le Equation (19e) has the straight-line form, y mx b, so plotting as x as y and [Fe" le[SCNExplanation / Answer
[Fe(NO3)3]=0.1 M
[KSCN]=0.002 M in 10 ml
For
(1) M1V1=M2V2
So [Fe3+]
V1=1 ml
V2=10+1=11ml
0.1*0.001=M2 *0.011
[Fe3+]0=0.0091 M
[SCN-]=0.002*0.01/0.011=0.001818 M
Same way calculated for another data.
V added V in lit A [Fe3+] [SCN-] x y 1 0.001 0.04 0.009091 0.001818 26.4 2420 2 0.002 0.101 0.016667 0.001833 61.15091 3305.455 3 0.003 0.165 0.023077 0.001846 96.525 3872.917 4 0.004 0.212 0.028571 0.001857 121.5738 3995.385 5 0.005 0.257 0.033333 0.001867 145.3886 4130.357 6 0.006 0.294 0.0375 0.001875 164.64 4181.333 7 0.007 0.325 0.041176 0.001882 180.5491 4193.08 8 0.008 0.351 0.044444 0.001889 193.721 4181.029 9 0.009 0.36 0.047368 0.001895 197.6 4011.111 10 0.01 0.385 0.05 0.0019 210.3316 4052.632
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