Essential laboratory data are recorded (Table 1) for the determination of the so

Question

Essential laboratory data are recorded (Table 1) for the determination of the solubility-product constant for silver chromate (Ag_2CrO_4) solutions. Ag_2CrO_4(s) harr 2 Ag^+ + CrO_4^2- Confirm that the above data obey Beer-Lambert law by plotting Absorbance (Y axis) versus concentration (X axis) (i.e. A = abc, where a is absorptivity, c is concentration and b the path length is 1 cm) Use A = abc to calculate the molar absorptivity (a) for each standard solution Calculate the mean value for the four standards Determine [CrO_4^2-] for the sample solutions (use the calibration curve or c=A/ab) Calculate [Ag^+] for the sample solutions. Calculate the Ksp for Ag_2CrO_4 and complete table 2.

Explanation / Answer

b) molar absorptivity = absorbance/concentration

Thus,

For std 1, molar absorptivity = 0.091/2.4 x 10^-5 = 3792 M-1.cm-1

For std 2, molar absorptivity = 0.510/1.2 x 10^-4 = 4250 M-1.cm-1

For std 3, molar absorptivity = 0.990/2.4 x 10^-4 = 4125 M-1.cm-1

For std 4, molar absorptivity = 1.460/3.6 x 10^-4 = 4056 M-1.cm-1

Mean molar absorptivity = 4056 M-1.cm-1

c) concentration of [CrO4^2-] in,

Sample #1, [CrO4^2-] = 1.270/4056 = 3.13 x 10^-4 M

Sample #2, [CrO4^2-] = 1.310/4056 = 3.23 x 10^-4 M

Sample #3, [CrO4^2-] = 1.850/4056 = 4.56 x 10^-4 M

d) [Ag+] = 2[CrO4^2-]

So,

Sample #1, [Ag+] = 2 x 3.13 x 10^-4 = 6.26 x 10^-4 M

Sample #2, [Ag+] = 2 x 3.23 x 10^-4 = 6.46 x 10^-4 M

Sample #3, [Ag+] = 2 x 4.56 x 10^-4 = 9.12 x 10^-4 M

e) Ksp = [Ag+]^2.[CrO4^2-]

Sample #1, Ksp = (6.26 x 10^-4)^2.(3.13 x 10^-4) = 1.23 x 10^-10

Sample #2, Ksp = (6.46 x 10^-4)^2.(3.23 x 10^-4) = 1.35 x 10^-10

Sample #3, Ksp = (9.12 x 10^-4)^2.(4.56 x 10^-4) = 3.79 x 10^-10

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