CHEMISTRY 265
OCTOBER QUIZ 2010
[ 2] 1.
Name two strong acids, give their formula.
[ 2] 2.
Name a strong base, give the formula.
T. SMITH-PALMER
Name a weak base, give the formula.
Ammonia NH3
[2 ] 3.
ε = 1825 cm-1 M-1 for vanillin at 347 nm. If the absorbance of a solution in a 1.00 cm cell
is 0.122 at 347 nm, what is the concentration of the vanillin in the solution?
A = ε·b·c
0.122 = (1825 cm-1 M-1)(1.00 cm)(c)
0.122 = (1825 M-1) (c)
0.122 / 1825 M-1 = c
6.68 × 10-5 M = c
[ 1] 4.
Write an equation showing what happens when CO2 passes into water.
CO2 + H2O ⇌
H2CO3
H2CO3 ⇌ H+ + HCO3–
[1 ]
Why does the pH of the water change?
[ 1] 5.
Write an equation showing what happens when Na HCO3 interact with water.
HCO3 ⇌ H+ + CO32–
HCO3– + H2O ⇌ H2CO3 + OH–
[ 1]
Will the resulting solution be acidic or basic?
[ 2] 6.
Give two ways that increasing levels of CO2 in the sir can affect the environment
CHEMISTRY 265 – OCTOBER QUIZ 2010 – T. SMITH-PALMER
PAGE 2
[1]
7.
KCl is considered very soluble. A saturated solution is almost 4 M in KCl. If a saturated
KCl solution is added to concentrated solution of HCl, a precipitate may form. Why?
Common ion effect
[4]
8.
We often need to take various solids into solution so we can more easily carry out analyses
on them. Knowing that the following solids were present, what might we add or do to
make sure everything has dissolved?
(a)CaCO32–
Acid eg: HCl
(b) liver
Boiling conc sulphuric acid
[ 4] 9.
Given the Ksp of PbI2 = 7.9 × 10-10, what is its solubility?
Ksp = [Pb2+][I–]2
Let solubility of PbI2 = s;
then [I–] = 2s
[Pb2+] = s
Therefore Ksp = s·(2s)2
7.9 × 10-10 = 4s3
5.8 × 10-4 M = s
[5 ] 10.
Will a precipitate form when 25.0 mL of 0.030 IO3– mmol/mL is mixed with 35.0 mL of a
solution containing 3.0 × 10-3 mmol/mL CaCl2? Ksp for calcium iodate = 7.1 × 10-7.
Ksp = [Ca2+][IO3–]2
[IO3–] = (25 mL)/(60 mL) × 0.030 M = 0.0125
[Ca2+] = (35 mL)/(60 mL) × (3 × 10-3 M) = 0.00175
Q = 0.00175 × [0.0125]2 = 2.7 × 10-7
Q < Ksp
Therefore, no precipitate will form
[ 4] 11.
Given a solution that is 1.541 M in KNO3, how would you prepare a solution that is
0.305 M? In your explanation, include what glassware you would use. You need 19 mLs
of solution plus a little extra for rinsing the storage containers.
Will need to use a volumetric flask. So make 25 mL.
C1V1 = C2V2
V1 (1.541 M KNO3) = (0.305 M)/(1.541 M) × 25 mL
Volume of original ccn KNO3 needed = 4.95 mL
Use a graduated pipette to deliver the 9.45 mLs.
CHEMISTRY 265 – OCTOBER QUIZ 2010 – T. SMITH-PALMER
12.
PAGE 3
A mixture contains gold particles and sand particles. Some prospectors collected samples
to be analyzed. It is important that the analysis be done on a representative sample.
In fact, 4.40 % of the particles are gold and the rest are sand (silica).
[2]
(a) If a sample size is chosen where 10,000 particles will be taken, what would be the
relative standard deviation for sampling the gold particles?
𝜎 = √𝑛 ∙ 𝑝 ∙ 𝑞
𝜎 = 104 × 0.996 = 39.84
𝜎 = 6.3
6.3
sdrel for gold P = 10,000 × .004 = 15.75 %
2
= 16 %
[2]
(b) How could we decrease this uncertainty?
Take more samples
Grind them smaller
[2]
(c) What reagents would we need to add to completely dissolve our sample?
Aqua regia (HNO3 + HCl)
Hot HF
[1]
13.
What is the formula for calculating a pooled standard deviation?
2
√
∑(𝑥𝑖 − 𝑥̅1 )2 + ∑(𝑥𝑗 − 𝑥̅2 ) + ⋯
𝑛1 + 𝑛2 + ⋯ − 𝑥