CHEMISTRY 31

Fall, 2017 - Dixon

Homework Set 1.3 Solutions

Ch. 4: 14 [from 1.2] 22, 32, 34a

14. A trainee in a medical lab will be released to work on her own when her results agree with those of an experienced worker when her results agree with those of an experienced worker at the 95% confidence level. Results for a blood urea nitrogen analysis are shown below.

Trainee: mean = 14.57 mg/dL, S = 0.53 mg/dL, n = 6 samples

Experienced mean = 13.95 mg/dL, S = 0.42 mg/dL, n = 5 samples

worker

a) What does the abbreviation dL stand for? deciliter

b) Should the trainee be released to work alone?

To test if there is a significant difference, we can do i) an F test to see if the standard deviations are the same or if the trainee has significantly worse precision and ii) case 2 t-test to see if there is a significant difference in the blood values

i) F = S²_trainee/S²_Expworker = (0.53/0.42)² = 1.59 F_table (n₂ – 1 = 5, n₁ – 1= 4) = 6.26, so the difference in standard deviations is not significant at 95%.

ii) t_calc = |14.57 mg/dL – 13.95 mg/dL|[n₁n₂/(n₁ + n₂)]^0.5/S_pooled where S_pooled = 0.484 (see text for equation)

t_calc = 0.62[6∙5/11]^0.5/(0.484) = 2.11 and t_table = 2.262. Since t_table > t_calc, the difference in standard deviations is not significant at 95%

The trainee can be released to work alone.

22. A standard Reference Material is certified to contain 94.6 ppm of an organic contaminant in soil. Your analysis gives values of 98.6, 98.4, 97.2, 94.6, and 96.2 ppm. Do your results differ from the expected result at the 95% confidence level? If you made one more measurement and found 94.5, would your conclusion change?

Original Set: mean = 97.00 ppm, standard deviation = 1.66 ppm.

Calculated t value = (std. value – mean value)·(n)^0.5/(std. dev.) = (94.6 – 97.0) ·(n)^0.5/1.66 = 3.24. The table t value (n-1 = 4 and 95%) = 2.78.

Since calculated t > table t value, yes, there is a significant bias.

With one additional value of 94.5: mean = 96.58, standard deviation = 1.80

Calculated t value = 2.70, Table t value = 2.57.

Calculated t value is still larger than the table value, so, there still is a significant bias. No change in the result.

32. A calibration curve based on n = 10 known points was used to measure the protein in an unknown. The results were protein = 15.2₂ (+0.4₆) mg, where the standard uncertainty is u_x = 0.46 mg. Find the 90% and 99% confidence intervals for protein in the unknown.

Confidence interval: x + tu_x, where t is found from n-2 degrees of freedom ( = 8)

For 90%, t = 1.860 and for 99%, t = 3.355

For 90%, confidence interval = 15.2₂ + (1.86)(0.4₆) mg = 15.2₂ (+0.8₆) mg (or 15.2 + 0.9)

For 99%, confidence interval = 15.2₂ + (3.355)(0.4₆) mg = 15.₂ (+1.₅) mg (or 15 + 2)

34a) The linear calibration curve in Figure 4-13 is y = 0.0163₀ (+0.0002₂)x + 0.004₇ (+0.002₆) with s_y = 0.005₉. Find the quantity of unknown protein that gives a measured absorbance of 0.264 when a blank has an absorbance of 0.095.

Sample – blank absorbance = y = 0.169

x = (y – b)/m = (0.169 – 0.0047)/(0.0163) = 10.1 mg

Ch. 6: 4, 6, 8, 15-17, 19, 23, 25, 33 [postponing to set 2.1]

4. Write the expression for the equilibrium constant for each of the following reactions. Write the pressure of a gas, X, as P_X.

a) 3Ag⁺(aq) + PO₄^3-(aq) ↔ Ag₃PO₄(s)

K = 1/([Ag⁺]³[PO₄^3-])

b) C₆H₆(l) + 15/2O₂(g) ↔ 3H₂O(l) + 6CO₂(g)

K = P(CO₂)⁶/P(O₂)^15/2

6. From the equations 1) HOCl ↔ H⁺ + OCl^- (K = 3.0 x 10^-8) and 2) HOCl + OBr^- ↔ HOBr + OCl^- (K = 15), find the value of K for: HOBr ↔ H⁺ + OBr^-.

All species are aqueous.

HOBr ↔ H⁺ + OBr^- = (HOBr + OCl^- ↔ HOCl + OBr^-) + (HOCl ↔ H⁺ + OCl^-)

K = K₁/K₂ = 3.0 x 10^-8/15 = 2.0 x 10^-9.

8. For the reaction HCO₃^- ↔ H⁺ + CO₃^2-, ΔG° = +59kJ/mol at 298.15K. Find the value of K for the reaction.

K = exp(-ΔG°/RT) = exp(-59000 J mol^-1/(8.314 J mol^-1 K^-1*298.15 K)) = 5 x 10^-11.