c) Calculate the [SCN ] in solution A, where 5.8 mL of 1.0103 M NaSCN is diluted
ID: 526168 • Letter: C
Question
c) Calculate the [SCN ] in solution A, where 5.8 mL of 1.0103 M NaSCN is diluted to 60.0 mL with 0.1 M Fe(NO3)3. WebAssign will check your answer for the correct number of significant figures.
(d) Calculate the [FeSCN2+] in the solutions prepared by diluting volumes of solution A in part c with 0.1 M Fe(NO3)3 to a final volume of 1.0101 mL. (Use the rounded value you calculated for part (c) to calculate your answers to part (d).)
I keep getting the first part wrong, why? And how should I set up the second part, formula wise?
Lab 11 Spectroscopic D x VA Lab 11 Spectroscopic D x C The solutions ed Are Calculate the in al [SCN X C www.webassign ne 15815516 Q3 t/web/Student/Assignment-Responses/submit?dep (c) Calculate the CSCN n solution A, where 5.8 mL of 1.0 x 10 M NascN is diluted to 60.0 mL with 0 1 M Fe(NO3)3. 0.00008814 x M 4.0 (d) Calculate the CFescN2+] in the solutions prepared by diluting volumes of solution A in part c with o.1 M Fe (NO3)3 to a final volume of 1.0x101 mL. (Use the rounded value you calculated for part (c) to calculate your answers to part (d).) (i) 1.0 mL (ii) 3.0 mL 4.0 (iii) 5.0 mL 4.0 (iv) 7.0 m 4.0 (v) 9.0 mL. 4.0 Additional Materials LA Spectroscopic Determination of an Equilibrium Constant 07 PM Search Windows 5/7/2017Explanation / Answer
Use the dilution equation:
M1*V1 = M2*V2 where M1 = 1.0*10-3 M; V1 = 5.8 mL; V2 = 60.0 mL.
Plug in values:
(1.0*10-3 M)*(5.8 mL) = M2*(60.0 mL)
===> M2 = (5.8*1.0*10-3)/(60.0) M = 9.667*10-5 M 9.7*10-5 M
The concentration of SCN- in solution A = 9.7*10-5 M (ans).
d) Write down the equation for the reaction as
Fe3+ (aq) + SCN- (aq) -------> [Fe(SCN)]2+
Note that we use 0.1 M Fe(NO3)3 and 9.7*10-5 M SCN- (from part C). Therefore, [Fe(SCN)2+] is decided by [SCN-], since [SCN-] is the limiting reactant.
(i) Take the first example; volume of SCN- taken = 1.0 mL and volume of the solution = 1.0*101 mL = 10.0 mL.
Use the dilution law:
(9.7*10-5 M)*(1.0 mL) = (10.0 mL)*[SCN-]
===> [SCN-] = (9.7*10-5*1.0)/(10.0) M = 9.7*10-6 M.
Since, we have [SCN-] = 9.7*10-6 M, therefore, [Fe(SCN)2+] = 9.7*10-6 M (ans).
Fill up the following table:
#
Vol. of 9.7*10-5 M SCN- (mL)
Total vol. of solution (mL)
[SCN-] final (M)
[Fe(SCN)2+]
i
1.0
10.0
9.7*10-6
9.7*10-6
ii
3.0
10.0
2.91*10-5
2.91*10-5
iii
5.0
10.0
4.85*10-5
4.85*10-5
iv
7.0
10.0
6.79*10-5
6.79*10-5
v
9.0
10.0
8.73*10-5
8.73*10-5
#
Vol. of 9.7*10-5 M SCN- (mL)
Total vol. of solution (mL)
[SCN-] final (M)
[Fe(SCN)2+]
i
1.0
10.0
9.7*10-6
9.7*10-6
ii
3.0
10.0
2.91*10-5
2.91*10-5
iii
5.0
10.0
4.85*10-5
4.85*10-5
iv
7.0
10.0
6.79*10-5
6.79*10-5
v
9.0
10.0
8.73*10-5
8.73*10-5
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