CHEM II Long Island University Chemistry Exam Practice

Exam 2For this exam you may only use what is provided in this assignment. The periodic table, equations, and
constants are provided along with the exam. No books notes or class work may be used during the
exam. The internet nor another person cannot be used to obtain answers. The work completed on the
exam must be your own work.
Answer all the following questions in the order provided. Write the number for each question in the
upper left corner. If you are skipping a question indicate the number in the upper left corner and leave
space in case, you decide to return to the problem.
When answering the question show all the relevant work. All dimensional analysis must be shown
properly. Partial credit will be given where appropriate.
By submitting your exam, via blackboard, you affirm that you have followed the exam rules and are
doing all the work on the exam by yourself, with no help from any outside source.
1. A solution was prepared by dissolving 0.03456 grams of Mg(OH)2 in 550 mL of distilled water at
25.0℃. (12 points)
a. Write out the balanced dissociation equation for magnesium hydroxide.
b. Determine the concentration of Hydroxide ions in the solution.
c. Calculate the pH of the solution.
d. If HBr were added to the solution would the pH increase, decrease or remain the same.
Explain your answer. Assume the temperature and volume of solution was constant.
2. Determine the pH at the equivalence point when 45.0 mL of 0.231 M HNO2 is mixed with 0.183
M KOH. Ka for HNO2 is 4.5 x 10-4. Your answer must include a completed ICE box and assumption
check. (12 points)
3. A buffer was prepared by mixing 135.0 grams of H2S and 218.2 grams of KSH in a 1.5 L vessel at
25.0℃. Ka for H2S 9.5 x 10-8 (12 points)
a. Determine the moles of acid in the buffer.
b. Determine the moles of base in the buffer
c. What is the pH for this buffer at 25.0℃?
d. Using an ICE box determine the quantity of moles present for H2S and KSH after the addition
of 0.045 mol of HI to the buffer.
e. Determine the pH of the buffer after the addition of 0.045 mol of HI.
(Assume the volume of the buffer does not change with the addition of HI)
4. For each of the following equations identify the acid, base, conjugate acid, and conjugate base.
(4 points)
a. K2HAsO3(aq) + CBr3COOK(aq) ⇌ K3AsO3(aq) + CBr3COOH(aq)
b. C6H5OH(aq) + C5H5NH2(aq) ⇋ C5H5NH3+(aq) + C6H5O-(aq)
5. Determine the pOH of a 1.567 M of Na2SO3 solution. (Ka for HSO3- is 6.3 x 10-8) (10 points)
6. Explain the following acid trends. (6 points)
a. CH3CH2CBr2COOH is a stronger acid than CH3CH2CH2COOH
b. H2O is a weaker acid than H2Se
7. A chemist mixed 68.5 mL of 0.724 M HNO3 and 95.0mL of 0.0045 M KOH in a reaction vessel.
(12 points)
a. Determine the moles of acid.
b. Determine the moles of base.
c. Determine the pH of the solution.
8. Calculate the pOH for a 0.452 M C2H5NH3+ solution at 25.0℃. Ka for C2H5NH3+ is 1.79 x 10-11
Your answer must include a completed ICE box and assumption check. (10 points)
9. A 0.263 M solution of a weak acid has a pH of 2.38 at 25.0℃. (8 points)
a. Write out the dissociation equation for the weak acid, HA.
b. Complete the ICE box determine the equilibrium concentrations of all species present at
equilibrium.
c. Determine the % dissociation for the acid.
10. A chemist mixed 32.0 mL of 4.28 M HBr and 45.0 mL of 3.75 M ClO2- at 25.0℃. Kb = 9.09 x 10-13
(10 points)
a. Determine the moles of acid.
b. Determine the moles of base.
c. Determine the pH of the mixture. Your answer must include a completed ICE box.
11. Predict whether the following aqueous solutions will be acidic, basic, or neutral. Explain your
answer using a balanced chemical equation. (4 points)
a. NaBrO2
b. NaBr
Equations and Constants
𝐽
𝑘𝐽
R = 8.314 𝑚𝑜𝑙 𝐾
R = 8.314 x 10-3 𝑚𝑜𝑙 𝐾
Kw = 1.0 x 10-14
F = 96,500 𝑚𝑜𝑙 𝑒 −
1 Ci = 3.7 x 1010 Bq
NA = 6.022 x 1023
𝐶
Rate = kN
[𝐴]
ln[𝐴] 𝑡 = -kt
0
t1/2 =
1
0.693
𝑘
= kt +
[𝐴]𝑡
1
[𝐴]0
1
t1/2 = 𝑘[𝐴]
lnk = (−
𝒌
ln(𝒌𝟏 ) =
0
𝐸𝑎
1
) (𝑇) + lnA
𝑅
𝑬𝒂
𝟐
𝑹
𝟏
𝟏
𝟐
𝑻𝟏
(𝑻 −
)
𝑁
ln𝑁𝑡 = -kt
0
Kp = Kc(RT)∆n
pH = -log[H3O+]
pH = -log[OH-]
pH + pOH = 14.00
Kw = [H3O+][OH-]
Kw = KaKb
pKa = -logKa
pKb = -logKb
pKa + pKb = 14.00
pH = pKa + log
[𝑐𝑜𝑛𝑗𝑢𝑔𝑎𝑡𝑒 𝑏𝑎𝑠𝑒]
[𝑎𝑐𝑖𝑑]
𝑎𝑡𝑚 𝐿
R = 0.08206 𝑚𝑜𝑙 𝐾
𝐽
F = 96,500 𝑉 𝑚𝑜𝑙 𝑒 −
% dissociation of an acid =
% dissociation of a base =
[𝐻 + ]𝑒
[𝐻𝐴]𝑖
[𝑂𝐻 − ]𝑒
[𝐴− ]𝑖
x 100%
x 100%
ax2 + bx + c = 0
𝑥=
−𝑏 ± √𝑏 2 − 4𝑎𝑐
2𝑎
ΔS°rxn = ΣnS° (products) – ΣmS°(reactants)
ΔH°rxn = ΣnH° (products) – ΣmH°(reactants)
ΔG°rxn = ΣnG° (products) – ΣmG°(reactants)
∆G = ∆H – T∆S
∆G° = -RTlnK
E°cell = E°cathode – E°anode
∆G° = -nFE°cell
∆Suni = ∆Ssys + ∆Ssurr
∆Ssurr =
E°cell =
−∆𝐻𝑟𝑥𝑛
𝑇
0.0257𝑉
𝑛
Ecell = E°cell –
𝑙𝑛𝐾
0.0257𝑉
𝑛
∆G = ∆G° + RTlnQ
I=
𝑞
𝑡
𝑙𝑛𝑄