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alexira
4 months ago
15

The standard curve was made by spectrophotographic analysis of equilibrated iron(III) thiocyanate solutions of known concentrati

on. You are asked to analyze a Fe(SCN)2+ solution with an unknown concentration and an absorbance value of 0.410 . The slope-intercept form of the equation of the line is y=4541.6x+0.0461 . The unknown was analyzed on the same instrument as the standard curve solutions at the same temperature. What is the Fe3+ concentration of the unknown solution?
Chemistry
1 answer:
VMariaS [2.9K]4 months ago
6 0

Answer:

The molar concentration of Fe³⁺ in the unknown solution is 8.01x10⁻⁵M.

Explanation:

When creating a calibration curve in spectrophotometric analysis, you apply Lambert-Beer’s law, which indicates that the absorbance of a compound is directly related to its concentration:

A = E*l*C

Where A stands for absorbance, E is the molar absorption coefficient, l is the path length, and C represents the molar concentration

Using the line equation, you find:

y = 4541.6X + 0.0461

Where Y is the absorbance and X refers to the concentration - we will presume the concentration is expressed in molarity-

Given that the absorbance of the unknown is 0.410:

0.410 = 4541.6X + 0.0461

X = 8.01x10⁻⁵M

The molar concentration of Fe³⁺ in the unknown solution is 8.01x10⁻⁵M.

You might be interested in
(a) The original value of the reaction quotient, Qc, for the reaction of H2(g) and I2(g) to form HI(g) (before any reactions tak
KiRa [2933]

Response:

Here's my calculation

Clarification:

Assume the starting concentrations of H₂ and I₂ are 0.030 and 0.015 mol·L⁻¹, respectively.

We need to determine the initial concentration of HI.

1. We will need a chemical equation with concentrations, so let's compile all the information in one location.

H₂ + I₂ ⇌ 2HI

I/mol·L⁻¹: 0.30 0.15 x

2. Calculate the concentration of HI

Q_{\text{c}} = \dfrac{\text{[HI]}^{2}} {\text{[H$_{2}$][I$_{2}$]}} =\dfrac{x^{2}}{0.30 \times 0.15} =  5.56\\\\x^{2} = 0.30 \times 0.15 \times 5.56 = 0.250\\x = \sqrt{0.250} = \textbf{0.50 mol/L}\\\text{The initial concentration of HI is $\large \boxed{\textbf{0.50 mol/L}}$}

3. Plot the initial values

The graph below visualizes the initial concentrations as plotted on the vertical axis.

7 0
4 months ago
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