Final Examination Form IB

James Richard Fromm


1. Name, using correct modern chemical nomenclature, each of the compounds whose formula is given. Give the correct molecular formula for each of the named compounds; if no molecular formula exists, give the correct empirical formula. a. MnO2; b, CoSO3.5H2O; c, K4Fe(CN)6; d, NaHCO3; e, Ba3(PO4)2; f, lithium tetrahydroxochromate(III); g, sodium hypochlorite; h, iron(III) oxide; I, periodic acid; j, magnesium chlorate

2. Write out the requested item below in as simple a form as possible:
.a. the equilibrium constant for the ionization of carbonate ion as a base:
b. the complete atomic electronic structure of a fluorine atom:
c. the solubility product of silver phosphate:
d. the numeric value of the solubility product of silver phosphate, with molar units:
. the stability constant of the hexamminecobalt(III) ion:
f. the gas-phase equilibrium constant for the oxidation of sulfur dioxide to sulfur trioxide by molecular oxygen:
g. the form of the equilibrium constant for the ionization of benzoic acid, C6H5COOH, as an acid:
h. the Lewis structure of sulfur trioxide, SO3:
i. the numeric value of the ionization constant of benzoate ion, C6H5COO-, in molar units:
j. the Lewis structure of sulfate ion:

3. A sample of mass 1.4796 g of an organic compound containing carbon, hydrogen, fluorine, and probably oxygen was analyzed by combustion methods. The analysis produced values for carbon (as 1.6687 g CO2) and hydrogen (as 0.5123 g H2O); oxygen is not determinable by combustion methods. A separate analytical method gave a value (corrected to the original sample mass) of 0.3602 g fluorine (as F). Calculate the correct empirical formula for this compound.
Empirical formula is ______________________________.

4. When zinc metal is placed in dilute hydrochloric acid, a quantitative reaction takes place which produces hydrogen gas and zinc(II) ions. A chemist placed a sample of 98.93 mg zinc metal in excess hydrochloric acid in an experiment so constructed that all of the evolved hydrogen gas was collected over water at 22oC in an inverted buret. The volume of gas collected in the buret was 378.94 mL. The atmospheric pressure as measured by a barometer was 746.3 torr on that day. Calculate a value for the universal gas constant R from these data.
The value of R is ________________________ with units of . . . . . ..

5. The reaction between molecular nitrogen and molecular hydrogen to produce gaseous ammonia, called the Haber process, is used to produce virtually all industrial nitrogen compounds and fertilizers.
a. Calculate the amount of heat evolved or absorbed when one mole of nitrogen gas reacts to form ammonia at 25oC. The heat which is _______________ (evolved, absorbed) in this process is __________ kJ/mol N2(g).
b. The value for the equilibrium constant for the Haber process at 25oC (which is not the usual temperature at which the process operates industrially) is 5.961 x 10+5 atm-2. In a reactor operated at 80 atm which is initially fed with the stoichiometric mixture of 3 H2(g) to 1 N2(g), compute the equilibrium partial pressures of hydrogen, nitrogen, and ammonia.
Hydrogen: ____________ atm. Nitrogen: ____________ atm. Ammonia: _____________ atm.

6. A student needed to determine the concentration and pKa value of a weak acid prepared by organic synthesis. A sample of 3.874 g of the acid was dissolved in a 250 mL volumetric flask. Several 50 mL aliquots taken from this flask required a mean volume of 41.37 mL of 0.1463 molar standardized NaOH to reach the cresol purple end point. The weak acid is believed to be monoprotic.
a. Calculate the molar mass of the weak acid, in g/mol acid:
b. I probably (could, could not) determine whether or not this acid was monoprotic using only an indicator during the titration. I probably (could, could not) determine whether or not this acid was monprotic using only a pH meter during the titration.
c. The pKa of this acid would be the pH on the titration curve when the volume of NaOH solution added was ______________ mL.
d. The color of the cresol purple indicator should be _______________ at the start of the titration; at and beyond the endpoint its color should be _______________.

7. A chemist prepared a solution as a primary standard by dissolving 29.73 g of potassium dihydrogen phosphate in 500 mL of solution. To this solution she then added 47.62 mL of standard 1.00 molar hydrochloric acid solution to prepare a buffer.
a. Calculate the pH of the buffer she prepared. The pH of the buffer she prepared is ____________________.
b. Her klutz of a lab partner came in the following morning and knocked the bottle of buffer mixture off the desk. After cleaning up the mess, he decided to conceal his clumsiness by preparing a new buffer for her. Unfortunately for her (and maybe for him, when she finds out) he misread the lab notebook and added 47.62 mL of the standard 1.00 molar sodium hydroxide solution instead of the hydrochloric acid. The solution he prepared (is, is not) a buffer and its pH will be ____________________, as calculated below.

8. a. The molar solubility of barium fluoride in water was measured and found to be 0.0182 molar at 25oC. The solubility product of BaF2 at this temperature must then be ________________ with nominal units of _______________ as calculated below:
b. When the temperature increases the molar solubility of BaF2 will probably (increase, decrease).
c. When NaF, which is soluble in water, is added to a saturated solution of BaF2, one would expect the molar solubility of BaF2 to (increase, decrease) (greatly, slightly, not at all) as a consequence of the (common, diverse) ion effect.
d. When NaBr, which is soluble in water, is added to a saturated solution of BaF2, one would expect the molar solubility of BaF2 to change (greatly, slightly, not at all) as a consequence of the (positive, negative, common, diverse) ion effect.

9. For each of the molecules below, the carbon-oxygen bond lengths and strengths are the same for all carbon-oxygen bonds within each molecule. However, the bond lengths and strengths differ between the molecules. The molecules are: CO(g), CO2(g), CO32-(aq), and C2H5OH(l).
a. Draw the complete Lewis structure of each of these four molecules.
b. As the strength of bonding between two atoms in a molecule increases, the distance between those atoms decreases; shorter bonds are stronger. Rank these four molecules in order of bond strength.
1. (strongest, shortest):
2. :
3. :
4. (longest, weakest):

10. In the Bohr model of hydrogen, the energies of the electrons in orbits are given by the relationship E = -2.178 x 10-18 J (Z2/n2). The symbol Z stands for the nuclear charge or atomic number (one, for hydrogen) and n is the orbit number (in quantum theory, the principal quantum number). The lowest possible principal quantum number is 1 for any atom. A chemist wishes to see how well the Bohr model fits the element lithium.
a. Calculate the energy of the lowest Bohr orbit in lithium in J/molecule and in kJ/mole. The energy is ____________________ J/molecule which is ____________________kJ/mole.
b. Draw a sketch of the lowest two Bohr orbits of lithium. On that sketch show how a photon of electromagnetic radiation could remove an electron from the lowest Bohr orbit (n = 1) to the next lowest Bohr orbit (n = 2).
c. Use the relationship DE = hn to calculate the frequency of electromagnetic radiation which would move one electron from the n = 1 orbit to the n = 2 orbit of lithium. The frequency will be ____________________ Hz.


1. a, manganese dioxide or manganese(IV) oxide; b, cobalt(II) sulfite pentahydrate; c, potassium hexacyanoferrate(II) or potassium ferrocyanide; d, sodium hydrogen carbonate; e, barium phosphate; f, Li[Cr(OH)4]; g, NaClO; h, Fe2O3; i, HIO4; j, Mg(ClO3)2.

2. a, Kb = [OH-][HCO3-]/[CO32-]; b, 1s22s22p5; c, Ksp = [Ag+]3[PO43-]; d, Ksp = 1 10-19 mol4/L4; e, Kstab = [Co(NH3)63+]/[Co3+][NH3]6; f, Kp = p2(SO3)/p2(SO2)p(O2); g, Ka = [H3O+][C6H5COO-]/[C6H5COOH]; h, ::=S( O:::)=O:: with 24 electrons; i, 1.5 10-10; j, S( O:::)4

3. The empirical formula is obtained from the ratios of the moles of elements present in the compound.
F: 0.3602 g F/(18.998 g F/mol F) = 18.960 mmol F
C: molar mass of CO2 is 12.011 + 2(15.999) = 44.009 g/mol
1.6687 g CO2/(44009 g CO2/mol CO2) = 37.917 mmol CO2
37.917 mmol CO2 corresponds to 37.917 mmol C
H: molar mass of H2O is 2(1.008) + 15.999 = 18.015 g/mol
0.5123 g H2O/(18.015 g H2O/mol H2O) = 28.437 mmol H2O
28.437 mmol H2O corresponds to 2 28.437 = 56.875 mmol H
O: oxygen can be obtained only by difference, so the mass of all of the others must be calculated.
mass F = 18.960 mmol x 18.998 g/mol = 360.2 mg
mass C = 37.917 mmol x 12.01 g/mol = 455.4 mg
mass H = 56.875 mmol x 1.008 g/mol = 57.3 mg
The total mass is therefore 872.9 mg
1.4796 g - 0.8729 g = 0.6067 g O
0.6067 g O/(15.998 g O/mol O) = 37.923 mmol O
The value for F is the smallest, making the mole ratio F: 1.000; C: 2.000; H: 3.000; O: 2.000. The empirical formula is C2H3O2F, which would correspond to the molecular formula of fluoroacetic acid, CH2FCOOH.

4. The universal gas constant R appears in the ideal gas law pV = nRT. The experimental data must give us values for p, V, n, and T in order to establish a value for R.
T: temperature is 22oC which is 22 + 273.15 = 295.15 K
V: volume was measured as 378.94 mL or 0.03794 dm3
p: the atmospheric pressure is 746.3 torr, but the gas was collected over water. The vapor pressure of water at 22oC is 19.8 torr. The actual pressure of the gas is then 746.3 - 19.8 = 726.5 torr. Since 760 torr is 101.3 kPa, this pressure is 96.83 kPa.
n: the overall balanced reaction is Zn + 2H+ --> Zn2+ + H2(g), so each mole of zinc produces one mole of molecular hydrogen. There are 98.93 mg/(65.38 g/mol) = 1.513 mmol zinc and so in a quantitative reaction there must be 1.513 mmol H2.
R: Since R = pV/nT = (96.83 kPa)(0.03794 dm3)/(1.513 mmol)(295.15 K), the measured value of R is 8.227 kPa dm3 mol-1K-1. This is significantly higher than the theoretical value of 8.314. Some reasons for this might be the use of impure zinc (zinc with some ZnO as well) or solubility of hydrogen in water (probably negligible).

5. a. The required tabulated standard molar enthalpies of formation are: N2(g), zero; H2(g), zero; NH3(g), -46 kJ/mol. The reaction is N2(g) + 3H2(g) rarrow.gif (63 bytes)> 2NH3(g), so DHoR = 2(-46 kJ) - 1(0) - 3(0) = -92 kJ for the reaction of one mole of nitrogen to give two moles of ammonia. A negative enthalpy change corresponds to the evolution of heat from a chemical reaction. The heat which is evolved is -92 kJ/mol N2(g).
b. For the reaction N2(g) + 3H2(g) larrow.GIF (55 bytes)rarrow.gif (63 bytes) 2NH3, the gas-phase equilibrium constant is Kp = p2(NH3)/p(N2)p3(H2) = 5.961 x 10+5 atm-1. Since the reaction begins with a stoichiometric 3H2:1N2 ratio, and the formation of ammonia removes the gases in that stoichioometric ratio, it continues as equilibrium is reached. So since p(H2) = 3p(N2),
Kp = p2(NH3)/27 p4(N2) = 5.961 x 10+5 atm-2
(27)(5.961 x 10+5 atm-2) p4(N2) = p2(NH3)
4011.8 atm-1 p2(N2) = p(NH3)
The equilibrium constant is large, so the partial pressure of the product, ammonia, should be relatively high while that of nitrogen should be relatively low. The total pressure of 80 atm would then be mostly ammonia; if all of it were, then
4011.8 atm-1 p2(N2) = 80 atm
p2(N2) = 0.01994 atm2, p(N2) = 0.141 atm and p(H2) is three times that or 0.424 atm.
A more precise method would involve the quadratic equation:
p(N2) + p(H2) + p(NH3) = 80 atm
4p(N2) + p(NH3) = 80 atm
4p(N2) + 4011.8 atm-1p2(N2) = 80 atm
p(N2) + 1003.0 atm-1p2(N2) = 20 atm
but it gives almost the same answers, since most of the gas present is indeed ammonia.

6. a. 41.37 mL x 0.1463 molar = 6.0524 mmol NaOH
This titrated 50 mL/250 mL or 1/5 of the acid, so there must have been 30.262 mmol of acid present. The molar mass if the acid is monoprotic is then: 3873 mg/30.262 mmol = 127.98 g/mol
b. You probably could not determine whether or not the acid was monoprotic using an indicator titration since only one color change would appear, certainly with cresol purple which has only one color change. You probably could determine whether or not the acid was monoprotic using a pH meter over the course of the titration because both plateaus would probably appear at significantly different pH values.
c. The half-titrated point, which in this problem would correspond to addition of 20.69 mL of NaOH, is the point at which the solution pH would equal the pKa value for the acid.
d. The indicator cresol purple changes from yellow to purple as it changes from acid form to base form. The indicator would be yellow (acidic) at the start of the titration and purple (basic) at and after the end point of the titration.

7. a. The moles of HCl added are (47.62 mL)(1.00 molar) or 47.62 mmol. The amount of potassium dihydrogen phosphate present originally can be calculated from the molar mass of KH2PO4:
39.0983 x 1 = 39.0983 g
1.008 x 2 = 2.016 g
30.9738 x 1 = 30.9738 g
15.9994 x 4 = 63.9976 g
Therefore the molar mass is 136.0857 g/mol. g/mol
29.73 g/(136.0857 g/mol) = 0.2185 mol KH2PO4. The added HCl reacts with the dihydrogen phosphate base:
Before reaction, 218.5 mmol H2PO4-, 47.6 mmol HCl, 0 mmol H3PO4
After reaction, 170.9 mmol H2PO4-, 0 mmol HCl, 47.6 mmol H3PO4
This is a buffer using K1 of phosphoric acid:
K1 = 7.5 x 10-3 = [H2O+][H2PO4-]/[H3PO4]
[H3O+] = 7.5 x 10-3 (47.6/170.9) = 2.089 x 10-3, pH = 2.68
b. The klutz weighs out the same amount of KH2PO4, 0.2185 moles, but addes NaOH instead. The potassium dihydrogen phosphate now reacts as an acid:
Before reaction, 218.5 mmol H2PO4-, 47.6 mmol NaOH, 0 mmol HPO42-
After reaction, 170.9 mmol H2PO4-, 0 mmol NaOH, 47.6 mmol HPO42-
This is a buffer using K2 of phosphoric acid:
K2 = 6.2 x 10-8 = [H3O+][HPO42-]/[H2PO4-]
[[H3O+]= 6.2 x 10-8 [H2PO4-]/[HPO42-]
[[H3O+]- 6.2 x 10-8 (170.9/47.6) = 2.22 x 10-7, pH = 6.65
I predict a short life expectancy for the klutz!

8. a. Ksp = [Ba2+][F-]2, BaF2(s) --> Ba2+ + 2F-, [F-] = 2[Ba2+]
Ksp = [Ba2+](2[Ba2+])2 = 4 [Ba2+]3 = 4(0.0182)3
Ksp = 2.41 x 10-5 mol3/L3
b. The solubility of most ionic salts in water will increase with increasing temperature.
c. When NaF is added to a saturated solution of BaF2, one would expect the molar solubility of BaF2 to decrease greatly as a consequence of the common ion effect.
d. When NaBr is added to a saturated solution of BaF2, one would expect the molar solubility of BaF2 to change slightly as a consequence of the diverse ion effect.

9. a,:C:::O: with ten electrons; b, ::O=C=O:: with sixteen electrons; c, [C( O:::)2=O::]2- with twenty-four electrons; d, H3C CH2 C O::(-H) with sixteen electrons.
b. The bonds are strongest in CO (triple bond), next strongest in CO2 (double bond), next strongest in CO32- (single and double bonds - resonance here), and weakest in C2H5OH (single bond only).

10 a. E = -2.178 x 10-18 J (Z2/n2)
For lithium Z is three and for its lowest orbit n is one. Then
E = -2.178 x 10-18 J (9/1) = 1.960 x 10-17 J/molecule
E = -1.960 x 10-17 J/molecule x 6.022 x 10+23 molecule/mol
E = 1.180 x 10+7 J/mol = 11,800 kJ/mol
b. The drawing should show two concentric circles surrounding the central nucleus. The photon, most easily represented by hn, is absorbed and the energy of the photon moves an electron from the inner orbit circle to the outer orbit circle.
c. For n = 1 the calculation above yields 1.960 x 10-17 J. For n = 2 the value is 1/22 or 1/4 of this, or 0.490 x 10-17 J. The difference, 1.470 x 10-17 J, is the value of DE. Since DE = hn, n = 1.470 x 10-17 J/6.626 x 10-34 J s
n = 2.219 x 10+16 s-1 = 2.219 x 10+16 Hz
Since c = nl, the wavelength is 2.998 x 10+8 m s-1/2.219 x 10+16 s-1 which is 13.5 x 10-9 m (13.5 nm). This is in the far ultraviolet to low-energy X-ray region of the electromagnetic spectrum.

Copyright 1997 James R. Fromm