Question

Concentration of blue dye Absorbance a. Calculate the slope and y-intercept for the line of best fit. Show your work and include units. (1.6810 0.201) (4.0x10L, 6.510) 8.0 x 10 0.100 1.6 x 100 .201 1.3 x 165, 2.4 x 10 0.302 3.2 x 10 0.403 0.510 moY₂-Y, (0.510-0.201) 0.309 Xxx (4.0x104-1.6x10on) - 24010" Ymyib b: 0.2-01 -(1.3 x U'M (1.6 X 10M buy-m+ b 0.201- 0.204 slope: 1.3x105 m y-intercept: -0.007 4.0 x 10 b. Using the above values, write the equation for the best fit line on the absorbance (A) vs. concentration (c) graph. Equation: A-(1.3x105) (c)-0.007 3. At 635 nm, the absorbance of an unknown blue-orange dye solution was measured to be 0.360. Using your equation of A vs. c, calculate the concentration of the blue dye component of this solution.

please help with parts a. and #3

Answer 1

You cannot just pickup any two points from multiple given points to calculate the slope.

It requires application of least square method to create a best fit line.

We first have to calculate the mean of both X and it's corresponding Y values.

$\therefore X_{MEAN} = \frac{\sum_{i=1}^{n} x_i}{n}$

$\therefore X_{MEAN} = \frac{(0.8+1.6+2.4+3.2+4.0) \times 10^{-6}}{5}$

$\therefore X_{MEAN} = \frac{12 \times 10^{-6}}{5}$

$\therefore X_{MEAN} =2.4 \times 10^{-6}$

Similarly $\therefore Y_{MEAN} = \frac{0.100+0.201+0.302+0.403+0.510}{5}$

$\therefore Y_{MEAN} = \frac{1.516}{5}$

$\therefore Y_{MEAN} = 0.303$

Now, Slope (m) will be given by : -

$\therefore m = \frac{\sum_{i = 1}^{n} (x_i - X_{MEAN}) (y_i - Y_{MEAN})}{\sum_{i=1}^{n}(x_i - X_{MEAN})^2}$

$(x_1 - X_{MEAN}) (y_1 - Y_{MEAN}) = (0.8 - 2.4)\times 10^{-6} \times(0.100 - 0.303)$

$= (-1.6)\times 10^{-6} \times (-0.203) = 0.325 \times 10^{-6}$

Similarly $(x_2 - X_{MEAN}) (y_2 - Y_{MEAN}) = 0.082 \times 10^{-6}$

$(x_3 - X_{MEAN}) (y_3 - Y_{MEAN}) = 0$

$(x_4 - X_{MEAN}) (y_4 - Y_{MEAN}) = 0.080 \times 10^{-6}$

$(x_5 - X_{MEAN}) (y_5 - Y_{MEAN}) = 0.331 \times 10^{-6}$

Now $(x_1 - X_{MEAN})^2 = ((0.8 - 2.4)\times 10^{-6})^2 = 2.56 \times 10^{-12}$

Similarly $(x_2 - X_{MEAN})^2 = ((1.6 - 2.4)\times 10^{-6})^2 = 0.64 \times 10^{-12}$

$(x_3 - X_{MEAN})^2 = ((2.4 - 2.4)\times 10^{-6})^2 = 0$

$(x_4 - X_{MEAN})^2 = ((3.2 - 2.4)\times 10^{-6})^2 = 0.64 \times 10^{-12}$

$(x_5 - X_{MEAN})^2 = ((4.0 - 2.4)\times 10^{-6})^2 = 2.56 \times 10^{-12}$

Putting all the values in slope equation : -

$\therefore m = \frac{0.818 \times 10^{6}}{6.4}$

$\therefore m = 1.278 \times 10^{5}$

By equation of straight line y = mx + c

where c is the y-intercept

$\therefore y = (1.278 \times 10^{5})x+c$

$\therefore c = y - (1.278 \times 10^{5})x$

Putting x and y values of first point : -

$\therefore c = 0.100 - (1.278 \times 10^{5})(8.0 \times 10^{-7})$

$\therefore c = 0.100 - (0.102)$

$\therefore c = -0.002$

$\therefore A = (1.278 \times 10^{5})C-0.002$

Question #3

A = 0.360

Putting in the derived equation above

$\therefore C = \frac{A+0.002}{1.278 \times 10^{5}}$

$\therefore C = \frac{0.360+0.002}{1.278 \times 10^{5}}$

$\therefore C = 2.83 \times 10^{-6}$

$C = 2.8 \times 10^{-6}$ (answer to one decimal point)