A lab technician makes three successive 100 to 1000 mL dilutions of a 0.970 M st

Question

A lab technician makes three successive 100 to 1000 mL dilutions of a 0.970 M stock solution of copper(II) chloride, then adds 3.00mL to a test tube initially containing 7.00 mL, of an solution. What the molarity of copper(II) ions in the test tube? A calibration curve was prepared for the measurement of the concentration of the complex ion FeSCN, Find the slope, y-interrupt, and correlation coefficient of the linear least squares line best-fit line). I would recommended practicing is Excel is do this? Relying the calibration carve of question 2. Find the concentration of FeSCN in a unknown sample with an absorbance of 0.21.

Explanation / Answer

Ans. 1. Given,

Stock [CuCl2] = 0.970 M

3 successive dilution of 10: 100 = 1: 10 are made.

So, Total dilution = (1/10) x (1/10) x (1/10) = 1/ 1000

Final concentration of CuCl2 = [CuCl2] of stock x total dilution

                                    = 0.970 M x (1/ 1000)

                                    = 9.70 x 10-4 M

Volume of final dilution added to ammonia solution = 3.0 mL

Volume of ammonia solution in tube = 7.0 mL

Final volume of reaction mixture = 3.0 mL (diluted CuCl2 soln.) + 7.0 mL (NH3 soln)

                                                            = 10.0 mL

Now, Using                            C1V1 = C2V2                        - equation 1

C1= Concentration, and V1= volume of std. solution 1     ;( 9.70 x 10-4 M CuCl2)

C2= Concentration, and V2 = Vol. of final solution 2                 ;( Reaction mixture)

Putting the values in equation 1-

            9.70 x 10-4 M x 3.0 mL = C2 x 10.0 mL

            Or, C2 = (9.70 x 10-4 M x 3.0 mL) / 10.0 mL = 2.91 x 10-4 M

Therefore, molarity of CuCl2 in test tube = 2.91 x 10-4 M

#3. Given,

Trendline equation is-

                                    y = 4766.7x + 0         ; in form of Y = mX + C

                                    where, m = 4766.7 = slope ; c = 0 = intercept

The calibration curve, the trendline equation or linear regression equation is in form of “Y = mX + C”.

In the graph, Y-axis indicates absorbance and X-axis depicts concentration. That is, according to the trendline (linear regression) equation y = 4766.7x + 0, one absorbance unit (1 Y = Y) is equal to 47667 units on X-axis (concentration) plus zero. Now,

Putting, y = 0.21 in trendline equation-

            0.21 = 4766.7 x + 0

            Or, x = 0.21 / 4766.7 = 4.41 x 10-5

Therefore, concentration of the unknown = x = 4.41 x 10-5 M

Note: Please put the same unit of concentration as given to you if it other than “M”. the question does not mention the unit of concentration, so it’s assumed to be molar.

The calculation is done using the information given about the curve. If the linear regression slope and intercept are not correct, deviation will be observed from calculated concentration.

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