1) Solution # 9 contains 5 mL Fe(NO3)3 and 3mL NaSCN in total volume of 10 mL. P

Question

1) Solution # 9 contains 5 mL Fe(NO3)3 and 3mL NaSCN in total volume of 10 mL. Percent transmittance of the solution is 17.8.

2) A student plots a graph of absorbance on the y axis and concentration on the x axis, generating a regression line of y = 2795x - 0.0125. What is the concentration of the absorbing species in solution, if the percent transmittance is 42.8?

3) A solution is prepared in a 50 mL volumetric flask by mixing 3 mL of 0.00312 M NaSCN with 15mL of 0.20M Fe(NO3)3. Enough 0.25 M HNO3 is added to fill the flask to the 50 mL mark. Before equilibrium was established, what was the initial concentration of SCN- in the solution?

0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0.00E+00 Calibration Curve, FeSCN2+ y 2267.4 x 0.00074 R2 0.9945 1.00E-04 2.00E-04 3.00E-04 concentration, M 4.00E-04

Explanation / Answer

Ans. Part 1. Question is not specified- what does it seek for ?

Part 2: Absorbance of a solution can be calculated using given value of transmittance as follow-

Absorbance, A = 2 - log (% transmittance) = 2- log (T)

Or, A = 2 – log (42.8) = 2 - 1.63144 = 0.3685

Thus, absorbance of given sample = 0.3685

In the graph, Y-axis indicates absorbance and X-axis depicts concentration. That is, according to the trendline (linear regression) equation Y = 2795X – 0.0125 obtained from the graph, 1 absorbance unit (1 Y = Y) is equal to 2795 units on X-axis (concentration) minus 0.0125.

            From, Y = 2795X – 0.0125

            Or, 0.3685 = 2795X – 0.0125              ; [Y = absorbance of unknown = 0.3685]

            Or, 0.3685 + 0.0125 = 2795X

            Or, X = 0.381 / 2795 = 1.363 x 10-4

Hence, concentration of unknown solution is 1.363 x 10-4 M.

Part 3. SCN- is added in form of NaSCN. Moreover, 1 mol of NaSCN consists of 1 mol SCN-thus [NaSCN] = [SCN-].

Given,

            [NaSCN] = 0.00312 M

            Volume of NaSCN solution taken = 3.0 mL

            Final volume of solution made upto = 50.0 mL

Now, using formula: M1V1 = M2V2

M1= molarity of initial solution 1, V1= volume of initial solution 1      ;[3 ml of 0.00312 M]

M2= molarity of final solution 2, V2= volume of final solution 2         , [50 mL, M to be calculated]

Or, 0.00312 M x 3.0 mL = 50.0 mL x M2

Or, M2 = (0.00312 M x 3.0 mL) / 50.0 mL = 0.0001872 M

Thus, [NaSCN] = [SCN-] = 0.0001872 M upon final volume make up (and, before equilibrium is established).  

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