Question

When I was a boy, Uncle Wilbur let me watch as he analyzed the iron content of runoff from his banana ranch. A 25.0-mL sample was acidified with nitric acid and treated with excess KSCN to form a red complex. (KSCN itself is colorless.) The solution was then diluted to 100.0 mL and put in a variable-path length cell. For comparison, a 10.0-mL reference sample of 6.74 times 10^-4 M Fe^3+ was treated with HNO_3 and KSCN and diluted to 50.0 mL, The reference was placed in a cell with a 1.00-cm light path. The runoff sample exhibited the same absorbance as the reference when the path length of the runoff cell was 2.41 cm. What was the concentration of iron in Uncle Wilbur's runoff?

Answer 1

When 25.0-mL sample was acidified with nitric acid and treated with excess KSCN to form a red complex, the concentration of iron in Uncle Wilbur's runoff sample is 1.62 x 10⁻³ M.

Step-by-step explanation:

In this question, Uncle Wilbur's runoff sample and a reference sample were both treated with nitric acid and KSCN to form a red complex.

The absorbance of the reference sample was measured using a cell with a 1.00-cm light path.

The runoff sample exhibited the same absorbance, but with a path length of 2.41 cm.

By comparing the path lengths and absorbances, we can calculate the concentration of iron in the runoff sample.

We know that the absorbance of a solution is directly proportional to the concentration of the absorbing substance, as well as the path length of the cell.

In this case, the path length of the runoff cell is 2.41 cm, which is 2.41 times longer than the reference cell.

Therefore, the concentration of iron in the runoff sample is also 2.41 times higher than the reference concentration.

If we assume the concentration of iron in the reference sample is 6.74 x 10⁻³⁴ M, we can calculate the concentration of iron in the runoff sample as follows:

Concentration of iron in the runoff sample = Concentration of iron in the reference sample x Path length of the runoff cell / Path length of the reference cell

Concentration of iron in the runoff sample = 6.74 x 10⁻⁴ M x 2.41 cm / 1.00 cm

Concentration of iron in the runoff sample = 1.62 x 10⁻³ M

So therefore the concentration of iron is 1.62 x 10⁻³ M

Answer 2

C = 2.24x10⁻⁴ M

Step-by-step explanation:

The concentration of iron in Uncle Wilbur's runoff can be calculated using Beer-Lambert law:

`A = \epsilon*C*l` (1)

Where:

A: is the absorbance of the compound

ε: is the molar absorptivity of the compound

C: is the concentration of the compound

l: is the optical path length

Since the runoff sample exhibited the same absorbance as the reference sample, we can find the concentration using equation (1):

`\epsilon*C_(1)*l_(1) = \epsilon*C_(2)*l_(2)` (2)

Where:

Subscripst 1 and 2 refer to Uncle Wilbur's runoff and to reference sample, respectively.

l₁ = 2.41 cm

l₂ = 1.00 cm

We can find C₂ as follows:

`C_(2) = (C_(2i)*V_(i))/(V_(f))` (3)

Where:

`C_(2i)`: is the initial concentration of the reference sample = 6.74x10⁻⁴ M

`V_(i)`: is the initial volume = 10.0 mL

`V_(f)`: is the final volume = 50.0 mL

`C_(2) = (6.74 \cdot 10^(-4) M*10.0 mL)/(50.0 mL) = 1.35 \cdot 10^(-4) M`

Now, we can find C₁ using equation (2):

`C_(1) = (C_(2)*l_(2))/(l_(1)) = (1.35 \cdot 10^(-4) M*1.00 cm)/(2.41 cm) = 5.60 \cdot 10^(-5) M`

Finally, since the runoff solution was diluted to 100.0 mL, the initial concentration can be calculated using equation (3) for
`C_(1i)`:

`C_(1i) = (C_(1)*V_(f))/(V_(i)) = (5.60 \cdot 10^(-5) M*100.0 mL)/(25.0 mL) = 2.24 \cdot 10^(-4) M`

Therefore, the concentration of iron in Uncle Wilbur's runoff is 2.24x10⁻⁴ M.

I hope it helps you!