Review Questions From Chem 112 and Chem 115

7/17/2019 Review Questions From Chem 112 and Chem 115

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Review questions from Chem 112 and Chem 115 (Optional Material):

The following questions are from old Chem 112 / 115 courses and review some of the basic

background material you’ve already been exposed to. Completing this material is entirely

optional, and answers are provided. We will not review this in class.

1. Write the formulae for the following: (a) rubidium nitrite (b) potassium sulfide (c) sodiumhydrogen sulfide (d) magnesium phosphate (e) calcium hydrogen phosphate (f) potassium

dihydrogen phosphate (g) iodine heptafluoride (h) ammonium sulfate (i) silver perchlorate (j)

boron trichloride.

Ans. (a) RbNO2 (b) K 2S (c) NaHS (d) Mg3(PO4)2 (e) CaHPO4

(f) KH2PO4 (g) IF7 (h) (NH4)2SO4 (i) AgClO4 (j) BCl3

2. Identify each of the following elements: (a) a halogen whose anion contains 36 electrons (e-)

(b) a radioactive noble gas with 86 protons (c) a Group 6 element whose anion contains 36 e- (d)

an alkali metal cation which contains 36 e- (e) a Group 4 cation which contains 80 electrons

Ans. (a) Br (b) Rn (c) Se (d) Rb+ (e) Pb

2+

3. Calculate the number of C, H, and O atoms in 1.50 g of glucose (C6H12O6).

Ans. Molecular Mass of glucose = (6(12.01) + 12(1.008) + 6(16.00)) g mol-1

= 180.156 g mol-

1. Thus, moles of glucose = 1.50 g / 180.156 g mol

-1 = 8.326 x 10

-3 mol. Thus, moles of C atoms =

6(8.326 x 10-3

) mol = 0.04996 mol. Since each mol of atoms contains 6.022 x 1023

atoms, numberof C atoms = 6.022 x 10

23 x 0.04996 = 3.009 x 10

22atoms. Sample contains 3.01 x 10

22 C atoms

(and, of course, O atoms). It contains exactly twice as many H atoms: 6.02 x 1022

atoms.

4. Consider the reaction: KCN(aq) + HCl(aq) => KCl(aq) + HCN(g).

If 0.140 g KCN is dissolved in water and treated with an excess of HCl(aq), what mass of

HCN results, assuming complete reaction?

Ans. F. Mass KCN = (39.10 + 12.01 + 14.01) g mol-1

= 65.12 g mol-1

. Thus, moles of KCN =

0.140 g / 65.12 g mol-1

= 2.150 x 10-3

mol. This also = moles HCN produced (look at theequation). Molecular mass HCN = (1.008 + 12.01 + 14.01) g mol

-1 = 27.028 g mol

-1. Thus, mass

HCN produced = 2.150 x 10-3

mol x 27.028 g mol-1

= 0.0581 g.

If you understand this thoroughly, do it in one step:

mass HCN = 0.140 g x 27.028 g mol-1

/ 65.12 g mol-1

= 0.0581 g.



5. Consider the reaction: 2 NO(g) + O2(g) => 2 NO2(g).If 0.886 mol of NO is mixed with 0.503 mol of oxygen, which is the limiting reagent? And howmany moles of NO2 are produced, assuming complete reaction?

Ans. 0.886 mol of NO will make 0.886 mol NO2. 0.503 mol of O2 will make 1.006 mol NO2.

So NO is limiting, and 0.886 mol NO2 produced.

6. Calculate the volume in mL of a 0.270 mol L-1

solution of NaCl required to provide 2.14 g of

NaCl.

Ans. Molar mass of NaCl = ( 22.99 + 35.45 ) g mol-1

= 58.44 g mol-1

.

Thus, moles NaCl desired = (mass NaCl desired) / (molar mass) = 2.14 g / 58.44 g mol-

1= 3.662 x 10

-2 mol.

Thus, vol of soln required (L) = (moles NaCl required) / (molarity of soln) = 3.662 x 10-2

mol /

0.270 mol L-1

= 0.1356 L = 136 mL.

7. How would you prepare 60.0 mL of 0.200 mol L-1

nitric acid from a stock solution of 4.00

mol L-1

nitric acid?

Ans. Consider the new solution: moles HNO3 = (molarity of soln) x (vol soln) = 0.200

mol L-1

x 0.0600 L = 1.20 x 10-2

mol.

This, of course, is the same in the sample taken of the stock solution. Thus, for the stock

solution:

vol required = (moles HNO3 required) / (molarity of soln) = 1.20 x 10-2

mol / 4.00 mol L-

1= 3.00 x 10

-3 L.

Put 3.00 mL of the stock soln into a flask, then dilute with water to a total volume of 60.0

mL.

8. A sample of methane gas is heated from 25oC to 88

oC at constant pressure. The initial volume

is 36.4 L; what is the final volume?

Ans. First, convert temperatures to K: T1 = (25+273) K = 298 K, T2 = (88 + 273) K = 361 K.

Now, since n and P are constant, V1/T1 = V2/T2 => V2 = V1 (T2/T1) = 36.4 L x 361 K / 298K = 44.1 L.

9. The formation of ammonium chloride from ammonia and gaseous hydrogen chloride is a rare

example of a reaction of two gases producing a solid. What mass of the solid is produced when73.0 g ammonia are mixed with an equal mass of HCl(g)? What gas remains after reaction?

What is its volume, measured at 14.0oC and 752 mm Hg?

Ans. A limiting reagent problem. BCE: NH3(g) + HCl(g)-----> NH4Cl(s)

Now moles NH3 = mass NH3 / molar mass NH3 = 73.0 g / 17.034 g mol-1

= 4.286 mol. This could

make 4.286 mol NH4Cl (look at the equation).



Similarly, moles HCl = mass HCl / molar mass HCl = 73.0 g / 36.458 g mol-1

= 2.002 mol. This

could make 2.002 mol NH4Cl (look at the equation).Thus, HCl is limiting, and 2.002 mol NH4Cl is formed.

Mass NH4Cl = moles NH4Cl x molar mass NH4Cl = 2.002 mol x 53.492 g mol-1

= 107 g.

The gas remaining is ammonia, NH3. Since there were 4.286 mol NH3 initially, and 2.002 mol

were used in reaction (look at the equation), 2.284 mol remain. Use the Ideal Gas Equation tofind the volume:

PV = nRT V = nRT/P = 2.284 mol x 0.0821 L atm mol-1

K -1

x (14 + 273) K / ((752/760)

atm) = 54.4 L.

10. Write Lewis dot symbols for the following atoms and ions (a) I (b) I- (c) S (d) S

2- (e) P (f) P

3-

(g) Na (h) Na+

(i) Mg (j) Mg2+

(k) Al (l) Al3+

(m) Pb (n) Pb2+

Ans.

11. Write Lewis structures for the following species. Include all likely resonance forms, andshow formal charges. (a) HCO2

-(C is central atom) (b) CH2 NO2

- (H atoms attached to C, C

attached to N, O atoms attached to N).

Ans. (formal charges in [ ])

Note that the last two structures would make fairly minor contributions to the actual structure,

since there is a [-1] charge on a C atom, rather than the much more electronegative O atom.

12. Predict the shape of the following ions: (a) ammonium (b) NH2- (c) carbonate (d) IBr 2

- (e)

IBr 4- (f) AlH4

+ (g) hydronium (h) BeF4

2-

Ans. (a) 8 e-. Complete octet round N => 4 single bonds (to H atoms). 4 e- pairs round N =>

tetrahedral electron pair distribution. Since all e- pairs are bonds, shape is also tetrahedral.

(b) Complete octet round N => 2 single bonds (to H atoms), two lone pairs. 4 e- pairs round N

=> tetrahedral electron pair distribution. Since two are lone pairs, shape (which does not includelone pairs) is angular or bent at 109

o (the tetrahedral angle).

(c) 22 e-. Three single bonds from central C to O atom, complete octets round O atoms =>

incomplete octet on C atom. Shift two e- from any one O atom into a double bond. (There areresonance forms here, but since these only involve pi electrons moving they are irrelevant to this

discussion.) Central atom has three e- pairs (pi bonds "don't count"), all bonds, so structure is

planar triangular (or trigonal).



(d) 22e-. Complete octets round Br atoms => 6 e- left, put on central I atom. 5e- pairs => trigonal

bipyramidal electron pair distribution. Lone pairs go equatorial => shape is linear.

(e) 36e-. Complete octets round Br atoms => 4e- left, put on central I atom. 6 e- pairs =>

tetrahedral electron pair distribution. Lone pairs go trans => square planar shape.

(f) A misprint. There is no such species. It was intended to be AlH4-, which has the same bonding

pattern (and hence, shape) as ammonium ion.(g) Isoelectronic with ammonium; however, one electron pair is now a lone pair. Hence the

shape is triangular pyramidal.

(h) 32 e-. Complete octets round F atoms => 4 e- pairs round central Be atom. Tetrahedral.

13. For the reaction: H2(g) + CO2(g) => H2O(g) + CO(g) , K = 0.534 at 700oC.

Calculate the moles of H2 present at equilibrium after 0.300 mol of CO and 0.300 mol of H2O are

heated to 700oC in a 10.0 L container.

Ans. K = [H2O] [CO] / ( [H2] [CO2] ) = 0.534.

[H2O], mol L

-

[CO], mol L

-

[H2], mol L

-

[CO2], mol L

-

Initial 3.00 x 10

- 3.00 x 10

- 0 0

-x -x +x +x

Equilibrium (3.00 x 10-

- x) (3.00 x 10-

- x) x x

K = (3.00 x 10-2

- x)2 / x

2 = 0.534 . (This is a perfect square; square root both sides and

solve for x ) => x = 1.733 x 10-2

. Thus, at equilibrium, there are 0.173 mol H2 present.

14. Give the formulae for the conjugate acid and the conjugate base of (a) water (b) hydroxide

ion (c) ammonia.

Ans. (Conjugate acid, followed by conjugate base) (a) H3O+, OH- (b) H2O, O2- (c) NH3

+, NH2-

15. Calculate the pH of (a) 1.0 x 10-3

mol L-1

HCl (b) 0.76 mol L-1

KOH.

Ans. (a) Complete ionization (strong acid), so [H3O+] = 1.0 x 10

-3 M => pH = 3.00.

(b) Complete ionization (strong base), so [OH-] = 0.76 M => [H3O

+] = K w / [H3O

+] = 1.0 x 10

-14

/0.76 = 1.32 x10-14

M => pH = 13.88.

16. Classify each of the following as a strong or weak acid: (a) nitric acid (b) hydrofluoric acid

(c) sulfuric acid (d) HSO4- (e) carbonic acid (f) HCO3

- (g) hydrochloric acid (h) HCN(aq) (i)

nitrous acid.

Ans. If you haven't already, memorize the list of six strong acids: hydrochloric, hydrobromic,

hydroiodic, nitric, sulfuric and perchloric . All other common acids are weak. (a), (c), (g) are

strong, others weak.

17. Which of the following solutions has the highest pH? Assume all are 0.40 mol L-1

. (a) formicacid (pK a = 3.77) (b) perchloric acid (c) acetic acid (pK a = 4.74).



Ans. Since all three acids have the same concentration, the solution with the highest pH (i.e.,

lowest [H3O+]) will be the one containing the weakest acid. HClO4 is a strong acid. Considering

the remaining two, acetic acid is the weaker (higher pK a value). The acetic acid solution will

have the highest pH.

18. The K a for benzoic acid is 6.5 x 10

-5

. Calculate the pH of a 0.10 mol L

-1

solution.

Ans. Using the approximate formula, [H3O+] = (K a. CHA) = (6.5 x 10

-5x 0.10 ) = 2.55 x 10

-3

M.

Since [H3O+] is << CHA, this result is valid. Thus, pH = -log10(2.55 x 10

-3) = 2.59

19. What is the molarity of a solution of ammonia (K b = 1.8 x 10-5

) whose pH is 11.22?

Ans. pOH = 14.00 - 11.22 = 2.78 => [OH-] = 10

-pOH = 1.66 x 10

-3.

The formula [OH-] = (K b. CB) is valid since the given [OH

-] << CB.

Square to [OH-]

2 = K b.CB => CB = [OH

-]

2 / K b = (1.66 x 10

-3)

2 / 1.8 x 10

-5 = 0.15 M.

20. Calculate the pH of a buffer solution containing 0.15 mol L-1

ammonia and 0.35 mol L-1

ammonium chloride. K a for the ammonium ion = 5.6 x 10-10

.

Ans. K a for NH4+ = 5.6 x 10

-10 => pK a = 9.252.

Use the Henderson -Hasselbalch (H-H) Equation: pH = pK a + log ([NH3] / [NH4+] ) = 9.252 + log

(0.15 / 0.35) = 9.252 - 0.368 = 8.88.

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Review Questions From Chem 112 and Chem 115