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During lab, a student used a Mohr pipet to add the following solutions into a 25 mL volumetric flask. They calculated the final volumes added, which are recorded below. The student then followed the directions in the manual to make the stock solution. Volumes Used to Create Stock Solution Volume 0.200 M Fe(NO3)3 (mL) Volume 0.00200 M KSCN (mL) Stock Solution 10.63 mL 1.42 mL Using the stock solution above, the student made additional dilutions, with the final volumes below. Calculate the [FeSCN2 ] in Standard 2. Report your answer in mM. Volumes Used to Create Standard Solutions Solution Volume Stock solution (mL) Volume Water (mL) Standard 1 7.43 2.53 Standard 2 4.63 5.17 Standard 3 2.66 6.58 Standard 4 0.57 9.36 Note: Reporting your answer in mM is for grading purposes only. It is not necessary during lab. Report your answer to three places after the decimal.

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Answer:

(FeSCN⁺²) = 0.11 mM

Explanation:

Fe ( NO3)3 (aq) [0.200M] + KSCN (aq) [ 0.002M] ⇒ FeSCN+2

M (Fe(NO₃)₃  = 0.200 M

V (Fe(NO₃)₃ =  10.63 mL

n (Fe(NO₃)₃ = 0.200*10.63 = 2.126 mmol

M (KSCN) =  0.00200 M

V (KSCN) = 1.42 mL

n (KSCN) =  0.00200 * 1.42 = 0.00284 mmol

Total volume = V (Fe(NO₃)₃  + V (KSCN)

                       = 10.63 + 1.42

                       = 12.05 mL

Limiting reactant = KSCN

So,

FeSCN⁺² = 0.00284 mmol

M (FeSCN⁺²) = 0.00284/12.05

                     = 0.000236 M

Excess reactant = (Fe(NO₃)₃

n(Fe(NO₃)₃ =  2.126 mmol -  0.00284 mmol

                  =2.123 mmol

For standard 2:

n (FeSCN⁺²) = 0.000236 * 4.63

                    =0.00109

V(standard 2) = 4.63 + 5.17

                       = 9.8 mL

M (FeSCN⁺²)  = 0.00109/9.8

                      = 0.000111 M = 0.11 mM

Therefore, (FeSCN⁺²) = 0.11 mM

The value of the (F-e-S-C-N⁺²) = 0.11 m-M when the student made additional dilutions.

Calculation of the value of the (F-e-S-C-N⁺²):

Since

M (F-e(N-O₃)₃  = 0.200 M

V (F-e(N-O₃)₃ =  10.63 mL

n (F-e(N-O₃)₃ = 0.200*10.63

= 2.126 mmol

M (K-S-C-N) =  0.00200 M

V (K-S-C-N) = 1.42 mL

And,

n (KS-C-N) =  0.00200 * 1.42 = 0.00284 mmol

Now

Total volume = V (F-e(N-O₃)₃  + V (K-S-C-N)

                      = 10.63 + 1.42

                      = 12.05 mL

Also, Limiting reactant = K-S-C-N

So,

F-e-S-C-N⁺² = 0.00284 mmol

M (F-e-S-C-N⁺²) = 0.00284/12.05

                    = 0.000236 M

Now

Excess reactant = (F-e(N-O₃)₃

n(F-e(N-O₃)₃ =  2.126 mmol -  0.00284 mmol

                 =2.123 mmol

Now For standard 2:

n (F-e-S-C-N⁺²) = 0.000236 * 4.63

                   =0.00109

V(standard 2) = 4.63 + 5.17

                      = 9.8 mL

M (F-e-S-C-N⁺²)  = 0.00109/9.8

                     = 0.000111 M = 0.11 mM

Therefore, (F-e-S-C-N⁺²) = 0.11 mM

Learn more about volume here: https://brainly.com/question/22695394

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