1. Use the slope and intercept of data from the fit in Figure 1 with the value o

Question

1. Use the slope and intercept of data from the fit in Figure 1 with the value of absorbance for your unknown to calculate the ferrous iron concentration (mol/L) in your unknown after dilution to 250 mL. Use the standard deviations of the slope and intercept to calculate the 95% confidence interval for the unknown concentration.

Slope = 1.022 x 10^4

Intercept = -0.011

Absorbance = 0.228

The problem I'm having is taking into account that our unknown solution was diluted twice. Here is the hint from our instructor:

Your unknown solution was put into a 250 volumetric and filled to the mark. This 250-mL solution is the unknown solution whose concentration you are to calculate and report.

Don't forget that you analyzed this solution after diluting it 4x: you took 25 mL and diluted to 100. So you'll need to account for this dilution when you calculate the concentration of your unknown solution.

Explanation / Answer

Ans. Part A: In the graph, Y-axis indicates absorbance and X-axis depicts concentration. That is, according to the trendline (linear regression) equation y = (1.022 x 104)X – 0.011 obtained from the graph, 1 absorbance unit (1 Y = Y) is equal to (1.022 x 104) units on X-axis (concentration) minus 0.011.

Now,

            Given, absorbance of unknown = 0.228

Concentration of unknown is calculated as follows-

y = (1.022 x 104)X – 0.011              ; [y = abs. of unknown = 0.228 ; X = [unknown]

or, 0.228 = (1.022 x 104)X – 0.011

or, 0.228 + 0.011 = (1.022 x 104)X

or, X = 0.239 / (1.022 x 104) = 2.338 x 10-4

Hence, concentration of unknown = 2.338 x 10-4 M = 2.338 x 10-4 mol/L

Part B: Tracking back to original (undiluted) concentration of the unknown.

As mentioned, the original unknown is diluted twice as follows-

            I. 25 mL of original unknown is diluted to 100 mL. Let this solution be labeled I.

            II. 100 mL (i.e. solution I) is further diluted upto 250 mL.

Total dilution factor = Initial volume of original unknown / final volume made upto

                                    = 25 mL / 250 mL

                                    = 1/ 10

That is, through dilution I and II, the original unknown solution is diluted 10 times or dilution factor is (1/10).

Now, we have-

            Concentration of diluted unknown = 2.338 x 10-4 M

            Dilution factor (from original to diluted unknown) = 1/ 10

So,

            Concentration of original unknown = concentration of diluted unknown / dilution factor

            Or, [Original unknown] = 2.338 x 10-4 M / (1/10)

                                                = 2.338 x 10-3 M

Hence, the concertation of undiluted (original) unknown sample = 2.338 x 10-3 M or 2.338 x 10-3 mol/ L.

Note: Molarity, M = mol/L

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