Question

Write the equilibrium constant expression for the experiment you will be studying this week. 2. If the equilibrium values of Fe3+, SCN- and FeSCN2+ are 9.5 x 10-4 M, 3.6 x 10-4 M and 5.7 x 10-5 M respectively, what is the value of Kc? 3. Write the general form of the dilution equation. 4. A solution is prepared by adding 18 mL of 0.200 M Fe(NO3)3 and 2 mL of 0.0020 M KSCN. Calculate the initial concentrations of Fe3+ and SCNin the solution.

Answer 1

Answer:

Kc = 166.7

[Fe³⁺] =  0.18 M

[SCN⁻] = 2×10⁻⁴ M

Explanation:

In the reaction of Fe³⁺ and SCN⁻, it is formed a complex:

Fe³⁺  +  SCN⁻  ⇄  FeSCN²⁺             Kc

Let's make the expression for Kc →  [FeSCN²⁺] / [Fe³⁺] . [SCN⁻]

5.7×10⁻⁵ / 9.5×10⁻⁴. 3.6×10⁻⁴  = 166.7

We determine the mmoles, we add from each reactant:

18 ml . 0.2M = 3.6 mmoles of Fe³⁺

2 ml . 0.002M = 4×10⁻³ mmoles of SCN⁻

General form of the dilution equation is:

Concentrated [C] . Concentrated Volume = Diluted [C] . Diluted Volume

Total volume = 20mL

[Fe³⁺]: 3.6 mmoles /20mL = 0.18 M

[SCN⁻]: 4×10⁻³ mmoles /20mL = 2×10⁻⁴ M

Answer 2

The value should be 1.67 x 10^2

The initial concentration should be 0.18 M and 2.0 x 10^(-4) M

Calculation of the value and initial concentration:

The value is

= 5.7 x 10^(-5)/(9.5 x 10^(-4) x 3.6 x 10^(-4))

= 167

= 1.67 x 10^2

we know that

Initial moles  = volume x concentration

So,

= 18/1000 x 0.200

= 0.0036 mol

Now

Initial moles  = volume x concentration

= 2/1000 x 0.0020

= 4.0 x 10^(-6) mol

So,

Total volume should be

= 18 + 2

= 20 mL

= 0.02 L

Now

Initial concentration   

= moles /total volume

= 0.0036/0.02

= 0.18 M

Now

Initial concentration

= moles  /total volume

= 4.0 x 10^(-6)/0.02

= 2.0 x 10^(-4) M

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