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CHM211 – Class of 2014 - Problem Set Course Instructor: Dr. Paul Piunno Due: no later than 11:10 am on November 7th, 2014 Notes: 1) Show all calculations for full marks to be awarded 2) Whenever EXCEL (or any other spreadsheet software) is used for calculations, indicate the functions used and the equations used by the software to execute the functions. Reminder: Don’t forget to include your name and student number on your answer sheets and please staple your answer sheets together.
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1. The concentration of lead in the blood of 50 children from a large school near a busy main road was determined. The sample mean was 10.12 ng·ml-1 and the standard deviation was 0.64 ng·ml-1. a) Calculate the 95% confidence interval for the mean lead concentration for all the children in the school. b) About how big should the sample have been to reduce the range of the confidence interval to 0.2 ng·ml-1 (i.e. ± 0.1 ng·ml-1)? a) 𝑥 = 10.12 ng ∙ mL!! , 𝑠 = 0.64 ng ∙ mL!! , 𝑁 = 50, t95% conf., 49 DOF = 2.01 𝑡𝑠 95% Confidence Interval = 10.12 ± ng ∙ mL!! 𝑁 2.01 0.64 = 10.12 ± ng ∙ mL!! 50 = 10.12 ± 0.18 ng ∙ mL!! [2 Marks]
Th
b) for the total uncertainly to be ± 0.1 ng·mL-1, 𝑡𝑠 0.1 ng ∙ mL!! = ng ∙ mL!! 𝑁 (2.01)(0.64) ! 𝑁= = 165 0.1
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[1 Mark]
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CHM211 – Class of 2014 - Problem Set 2. The solubility product of barium sulphate is 1.3 × 10-10, with a well-established standard deviation of 0.1 × 10-10. Calculate the 99% confidence interval for the solubility of barium sulphate in water. It can be assumed that ionic strength effects can be ignored. Assume that concentration and activity are equivalent as ionic strength effects can be ignored. BaSO4 (s) ⇌ Ba2+ (aq) + SO42- (aq) The solubility of BaSO4 is equivalent to the concentration of Ba2+ ion. Ksp = [Ba2+]·[SO42-] = 1.3 × 10-10 Let x = [Ba2+] = [SO42-] = solubility of BaSO4 x2 = 1.3 × 10-10 x = 1.1 × 10-5 M using the propagation of error equation for exponential functions, we have: 𝑠!!"
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𝑠! = 0.5 𝑥
𝐾!"
!!
𝑠! = 0.5 1.1 × 10 M
0.1 × 10!!" = 4.2 × 10!! M 1.3 × 10!!"
Confidence Interval (99%) = x ± zσ = 1.1 × 10-5 ± (2.58)(4.2 × 10-7) M = 1.1 × 10-5 ± 0.1 × 10-5 M [3 Marks]
3. The following table gives the concentration of norepinephrine (mmol per gram creatinine) in the urine of healthy volunteers in their early twenties. Male: 0.48, 0.36, 0.20, 0.55, 0.45, 0.46, 0.47, 0.23 Female: 0.35, 0.37, 0.27, 0.29
Is there any evidence that, with 95% confidence, the concentration of norepinephrine differs between the sexes?
Th
Conducting a comparison of two means. 𝑥!"#$ = 0.4, 𝑁!"#$ = 8, 𝑥!"#$%" = 0.32, 𝑁!"#$%" = 4, 𝑠!""#$% = 0.108, 𝐷. 𝑜. 𝐹. = 10 !!"#$ !!!"#$%"
𝑡!"# =
!!"#$ ∙!!"#$%"
!.!" – !.!"
(!.!"#)
!!! !∙!
= 1.21
sh
!!""#
%$!!"#$ !!!"#$%"
=
tcrit, DoF = 10, 95% Conf. = 2.23 > texp, therefore, there is no significant difference in the concentration of norepinephrine in the urine samples of the males and females obtained in this investigation. [3 Marks]
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CHM211 – Class of 2014 - Problem Set 4. The data below give the concentration of thiol (mM) in the blood lysate of the blood of two groups of volunteers, the first group being ‘normal’ and the second having rheumatoid arthritis: Normal: 1.84, 1.92, 1.94, 1.92, 1.85, 1.91, 2.07 Rheumatoid: 2.81, 4.06, 3.62, 3.27, 3.27, 3.76 (a) Can any of the data in the ‘normal’ group be rejected as an outlier with 95% confidence? (b) Do the variances of the two groups differ significantly (i.e. with 95% confidence)? a) 𝑄!"# =
!.!"!!.!" !.!"!!.!"
= 0.565
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𝑄!"#$,!"% !"#$.,!!! = 0.568 since Qexp < Qcrit, there are no outliers in the data set. [2 Marks]
!!
!.!"#
b) 𝐹!"# = !!! = !.!!"#$ = 34.0 !
Fcrit = 4.39 Since Fexp >> Fcrit, the variances of the two data sets differ significantly [2 Marks]
5. The following data give the recovery of bromide from spiked samples of vegetable matter, measured using a gas–liquid chromatographic method. The same amount of bromide was added to each specimen. Tomato: 777, 790, 759, 790, 770, 758, 764 µg·g-1
Cucumber: 782, 773, 778, 765, 789, 797, 782 µg·g-1
Th
(a) Test whether the recoveries from the two vegetables have variances that differ significantly (i.e. with 95% confidence).
(b) Test whether the mean recovery rates differ significantly (i.e. with 95% confidence). !!
!"#
a) 𝐹!"# = !!! = !"# = 1.70
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!
Fcrit = 4.28 Since Fexp < Fcrit, the variances of the two data sets do not differ significantly [2 Marks]
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CHM211 – Class of 2014 - Problem Set
b) 𝑥! = 773, 𝑁! = 7, 𝑥! = 781, 𝑁! = 7, 𝑠!""#$% = 12.1, 𝐷. 𝑜. 𝐹. = 12 𝑡!"# =
!! !!! !! !!! !!""#$% !! ∙!!
=
!!" – !"# (!".!)
!!! !∙!
= 1.24
tcrit, DoF = 12, 95% Conf. = 2.19 > texp, therefore, there is no significant difference in the mean of the recovery rates. [3 Marks] 6.
Mg(OH)2 is a metal hydroxide compound of low solubility. a) Calculate the solubility of Mg(OH)2 in deionized water. b) Calculate the solubility of Mg(OH)2 in 0.10 M KNO3.
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Note: In all cases, you may assume that the solubility of Mg(OH)2 does not appreciably affect the solution’s ionic strength. a) Ksp Mg(OH)2 = 7.1 × 10-12 (Skoog, 8th ed., p. A-7) Mg(OH)2 ⇌ Mg2+ + 2OHLet x = the solubility of Mg(OH)2 = [Mg2+] Ksp Mg(OH)2 = [Mg2+][OH-]2 = x · 2x2 = 7.1 × 10-12 4x3 = 7.1 × 10-12 x = 1.21 × 10-4 M [2 Marks]
b) In 0.10 M ionic strength solution, 𝛾!"!! = 0.44, 𝛾!" ! = 0.76 (Skoog, 8th ed., p. 274) ! !! ! = 𝛾!"!! Mg 𝐾!" !" !" ! = 𝑎!"!! ∙ 𝑎!" 𝛾!"! OH ! ! ! !! ! Mg = 𝛾!"!! ∙ 𝛾!" OH ! ! Let x equal the solubility of Mg(OH)2 = [Mg2+] ! ! ! ∙ 𝑥 ∙ 2𝑥 𝐾!" !" !" ! = 𝛾!"!! ∙ 𝛾!" 𝐾!" !" !" ! 4𝑥 ! = ! 𝛾!"!! ∙ 𝛾!" ! !
!!" !" !" ! ! !∙!!"!! ∙!!" !
Th
𝑥 =
=
!
! × !"!!"
! !.!! (!.!")!
= 1.91 × 10!! M
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[3 Marks]
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CHM211 – Class of 2014 - Problem Set 7. The concentration of Ca2+ in a well water sample was determined using a calcium ionselective electrode and a one-point standard addition. A 10.00 mL aliquot of the sample was placed in a beaker with the calcium ion-selective electrode and a reference electrode, and the measured potential was –67.04 mV. A 200 µL aliquot of a 5.00 × 10–2 M standard solution of Ca2+ was added to the 10.00 mL sample solution, after which a potential of –58.31 mV was measured. Determine the concentration of Ca2+ in the well water sample. Note that the total ionic strength of all solutions was adjusted to 100 mM prior to each measurement. It can be assumed that the ionic strength adjustment was made in such a fashion so as to not to alter the concentration of calcium in the solutions being measured, nor their volumes. The Nernst equation for the cell potential of the sample is: 𝐸!"##!"#$%& = 𝐾 +
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[1 Mark]
0.0592 log 𝑎!"!! !" !"#$%& 2
and the Nernst equation for cell potential following the standard addition is: 𝐸!"#!!"#$%&!!"#$%#&% = 𝐾 +
[1 Mark]
𝑉!"#$%& 0.0592 𝑉!"#$%#&% log ∙ 𝑎!"!! !" !"#$%& + ∙ 𝑎!"!! !" !"#$%#&% !"!!!. 2 𝑉!"#$% 𝑉!"#
%$Where Vsample, Vstandard and VTotal are the volumes of the sample solution, the quantity of standard solution added, and the total volume of the sample solution and the standard solution. The difference in the cell potential on adding the standard, ΔEcell, may be written as: Δ𝐸!"## = 𝐸!"##!"#$%&!!"#$%#&% − 𝐸!"##!"#$%&
Δ𝐸!"## =
𝑉!"#$%& 0.0592 𝑉!"#$%#&% log ∙ 𝑎!"!! !" !"#$%& + ∙ 𝑎!"!! !" !"#$%!"# !"!!!. 2 𝑉!"#$% 𝑉!"#$% 0.0592 − log 𝑎!"!! !" !"#$%& 2
Th
The above equation may be rearranged to:
[1 Mark]
𝑉!"#$%& 𝑉!"#$%#&% ∙ 𝑎!"!! !" !"#$%#&% !"!!!. 2 ∙ Δ𝐸!"## = 𝑙𝑜𝑔 + 0.0592 𝑉!"#$% 𝑉!"#$% ∙ 𝑎!"!! !" !"#$%&
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from which and the activity of calcium in the sample may be determined as follows: 2 ∙ (0.00873 V) = 𝑙𝑜𝑔 0.0592 𝑉
10.00 𝑚𝐿 0.200 𝑚𝐿 (2.00 × 10!! 𝑀) + 10.20 𝑚𝐿 10.20 𝑚𝐿 ∙ 𝑎!"!! !" !"#$%&
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CHM211 – Class of 2014 - Problem Set Note: Units of Volts (not millivolts) must be used in the above equation. The activity of the !! calcium standard was calculated as 𝑎!"!! !" !"#$%#&% !"!! !. = γ 𝐶𝑎!"#$%#&% !"! ! ! = 0.40 × –2 –2 (5.00 × 10 M) = 2.00 × 10 M, where the activity coefficient for calcium at 100 mM ionic strength was taken from the table included in our Topic 2 lecture notes. [1 Mark for correctly calculating the activity of the standard] 0.294 = 𝑙𝑜𝑔 0.9804 +
1.97 = 0.9804 +
(3.92 × 10!! 𝑀) 𝑎!"!! !" !"#$%&
(3.92 × 10!! 𝑀) 𝑎!"!! !" !"#$%&
[1 Mark]
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𝑎!"!! !" !"#$%& = 3.96 ×10!! 𝑀
𝐶𝑎 !!
!"#$%&
=
𝑎!"!! !" !"#$!"
𝛾!"!!,!""!" !"#!$ !"#$%&"!
=
3.96 ×10!! 𝑀 = 9.90 × 10!! 𝑀 0.40
[1 Mark for correctly calculating the concentration of the sample]
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Th
[6 Marks Total]
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