Cuối kỳ hóa phân tích (bản tiếng anh )

Question 1:
Arrange the steps in the order that will allow you to determine the
concentration of the base solution:
+ Write a balance equation for the reaction between the analyte and
the titrant
+ Calculate the number of moles of titrant using the volume of titrant
required and the concentration of titrant
+ Calculate the number of moles of analyte using the stoichiometric
coefficients of the equation
+ Calculate the concentration of the analyte using the number of moles
of analyte and the volume of analyte titrated

Question 2:


Write the balance titration equation:
Fe3+ + Cu+  Fe2+ + Cu2+
Complete the two half-reaction that occur at the Pt indicator
electrode. Write the half-reactions as reductions:
Half reaction of Copper : Cu2+ + e-  Cu+ (Estd = 0.161V)
Half reaction of Iron : Fe3+ + e-  Fe2+ (Estd = 0.767V)
Select the two equations given in the pictures below that can be used
to determine the cell voltage at different points in the titration. E of
the Ag/AgCl electrode is 0.197V
Equation 1: E = ECu2+/Cu+ - 0.0592 log – ESCE
= 0.161 – 0.0592 log – 0.197 (V)
Equation 2: E = EFe3+/Fe2+ - 0.0592 log – ESCE
= 0.767 – 0.0592 log – 0.197 (V)
Calculate the value of E for the cell after each of the given volume of
the 1.50 ml Cu++ titrant have been added:
Initial moles of Fe3+ = 0.02 x 100 ml = 2 mmol

Initial moles of Cu+ = 0.1 x 1.5 ml = 0.15 mmol
After reaction:
[Fe2+] =


[Fe3+] =
Ecell = 0.767 – 0.0592 log – ESCE
= 0.767 – 0.0592 log – 0.197 = 0.635 V
Question 3:
Electricity is the flow of electrons. The question relate to how
electricity is quantified.
Electron are charged particles. The amount of charge that passes per
unit time is called current.
The driving force for the electron (i.e., the reason they are flowing in
the first place is measured by potential(thế năng)
Charge is measured in coulombs (C)
Current is measured in amperes (A)

Question 4:
A solution of Cu+ is undergoing a redox titration by the addition of
Ce4+. The standard reduction half-reactions and potentials are given:


Select a suitable indicator to observe the end point:
Which color change would you observe the end point?
Cu+ + Ce4+  Cu2+ + Ce3+
The two half cell:
Ce4+ +e-  Ce3+ Estd = 1.72V
Cu2+ + e-  Cu+ Estd = 0.161V
Cu+  Cu2+ + e- (Anode reaction)

Ce4+ + e-  Ce3+ (Cathode reaction)
Ecell = Eanode – Ecathode = 1.72 – 0.161 = 1.559V
 The suitable indicator is 5,6 – dimethylphenanthroline
 The color turns from yellow-green to red.

Question 5:
Match each of the electrical circuit terms with their definition


Electrical potential: The work that can be done to move an electric
charge from one point to another (volts)
Electrical charge(E): the amount of positive or negative particles
(coulombs)
Current: the quantity of charge moving through a circuit each second
(amperes)
Question 6:
End point: The point in the titration when the added amount of
standard reagent is equal to the amount of analyte being titrated.
Equivalence point: The point in the titration when a change in the
analyte solution is observed, indicating equivalency.
Analyte: A solution of unknown concentration
Indicator: It is added to the analyte solution and aids in the observation
of the completion of the reaction.
Titrant: The standard reagent of known concentration that is added
from a buret to the analyte solution.
Molarity: A common concentration unit – moles per liter
Titration: A volumetric analysis technique used to determine unknown
concentration of a solution by using the known concentration of other.

Question 7:

Making the table about the indicator and their pH


Analyzing the curve and observe what pH range that the pH changes
rapidly  Equivalence point  Compare it with the table  Suitable
indicator.
Question 8:
Which of the factors would have a major effect (greater than 0.1 pH
units) on the pH range for the end point of an acid-base titration
+ Titrant concentration
+ Strength of acid or base
Question 9:
When a weak acid (HA) is titrated with a strong base, such as NaOH
what species are present in the weak acid solution before the titration
is started ?
 HA, H+ (H3O+) , A- , H2O

Question 10:
Dilute solution with a same weak acid (pKa = 4.5) ranging in
concentration from 2 x 10^-6 to 2 x 10^-2 M (0.002 to 20mM) are


titrated with a strong base that is five times more concentrated than
the acid.
At low concentration of a weak acid, how does the concentration of
the acid affect the shape of the titration curve?
 Answer : The lower the concentration of the acid, the sharper the
inflection at the equivalence point
More information:
+ The shape depends on the pKa and the concentration of acid

Question 11:
Consider a monoprotic weak acid (HA) that is titrated with a strong
base
What is the relationship between the strength of the weak acid and
the pH of the solution at the equivalence point?
 Answer: The weaker the acid, the higher the pH at the
equivalence point
Explanation: According to Bronsted lowry concept : Stronger the acid ,
weaker will be it's conjugate base. Or
Weaker the acid, stronger will be it's conjugate base.

Question 12:


Identify the equilibrium shift that takes place after each reaction is
cooled in a refrigerator in which direction does the equilibrium of this
reaction shift due to cooling?
2H2S + SO2
Answer: ∆H is given in negative form. So, it is an exothermic reaction. With
decrease in temperature, equilibrium shifts in product side according to Lechateliers principle.  Toward the products
∆H is negative  The right shift is exothermic  an exothermic reaction
So : Increasing the temperature  Shift to left  Toward the reactants
Decreasing the temperature  Shift to right  Toward the products

Question 13:

A particular gas-phase reaction has an equilibrium constant of Kp =
0.30. A mixture is prepared where all the reactants and products are
in their standard states.
Which direction will the reaction proceed?

We know that Kp =
If Kp > 1  The reaction is favored  It will proceed toward products that
is right
Oppositely, it will proceed toward reactants that is left
 Qp > Kp (0.3) The reaction will proceed to the left
Question 14:


In the Fajans titration of Hg2(2+) , NaCl is added to produce the
precipitate Hg2Cl2. The end point of the titration is detected with
bromophenol blue.
What charge do you expect the precipitate to have after the
equivalence point ?
Answer: Before equivalence mercurous ion will be in the solution. So formed precipitate
of mercurous chloride will adsorb excess of mercurous ion so solution will be
positively charged.
In Fazan's methid,
Before the equivalance point of titration, the precipitate is negatively charged since
solution consists of excess Cl- ions which are adsorbed ob the precipitate, giving
negative charge to precipitate.
But after the equivalance point of titration, Hg2+ ions are in excess, so the surface of
precipitate becomes positively charged with adsorbed layer of excess Hg 2+ ions.

Question 15:


Methyl orange is an indicator used for detecting the pH of acidic
solutions. The red, protonated form of the indicator (Hln) dissociates
into a yellow, deprotonated form (In-) as shown in the reaction. The
pKa for methyl orange is 3.4

Hln(aq)
Red

Yellow

Identifying the color of a solution containing methyl orange indicator
at pH 2
 The color will be red due to the acidic solution(pH = 2)
If pH = 3.4  pKa  Equivalence point  Orange
If pH = 6  Basic solution  Yellow
Question 16+17:
Both liquid bleach, ClO- and iodate,IO3- , can react with iodide ions, IIn a reaction with iodide ions, liquid bleach is best classified as ?
Reaction of ClO- with IClO- + I- ----> Cl- + IOThis reaction occur in presence of OH- as catalyst.
Iis oxidized to IO- and ClO- is reduced to Cl- .

 Liquiđ bleach is best classified as oxidizing agent
Reaction of IO3- with IIO3- + 5 I- + 6H+ ---> 3 I2 + 3 H2O
IO3- is reduced to I2 while I- is oxidized to I2 .

 Iodate ion is best classified as oxidizing agent
Question 18:


Equal volumes of two different weak acids are titrated with 0.20M
NaOH, resulting in the following titration curves.
Which curve corresponds to the titration of the more concentrated
weak acid solution?

To determine the curve of more and less concentrated weak acid we need to look at titration
curves carefully.

When the titration is of more concentrated weak acid with strong base the pH at its equivalence
point is equal to 7 while in less concentrated weak acid and strong base titration pH at
equivalence point will be greater than 7 (as base is strong).
In the blue curve it is clearly visible that pH at equivalence point is nearly equal to 7 while
in red curve pH is greater than 7(between 8 - 10).
So we can say blue curve corresponds to titration of more concentrated weak acid
solution.


Question 19:
Cell membranes contain channels that allow ions to cross the
phospholipid bilayer. A particular K+ channel carries a current of 1.61
pA. How many K+ ions, N, pass through the channel n 1.65 m/s ?
Hint : Current is the rate of charge flow. Consider how much charge a
single K+ ion possesses.
Charge on 1 K+ ion = 1.6 x 10^-19 C
Total charge passed = Q = It = (1.61 x 10^-12) x (1.65 x 10^-3) = 2.6565 x
10^-15 C
Number of K+ ion = N = Q/e =

Question 20:
Simply living, an 87.5 kg human being will consume approximately 20.0
mol of O2 per day. To provide energy for the human, the O2 is reduced
to H2O during food oxidation by the reaction
O2 + 4H+ + 4ea) Determine the current (I) generated by the human per day. In this
case, the current is defined as the flow of electrons (e-) to O2,
from the food the human consumes
b) Reduction of O2 can occur by a donation of electrons from
nicotinamide adenine dinucleotide (NADH). When this occurs, the
electron experience a potential drop of 1.00 V. Determine the

power output(P) from the 87.5kg human


ANSWER
a) O2 + 4H+ + 4e20

80

Moles of electron = n(e) = 20 x 4 = 80 moles
Q(charge) = n(e) x F(Faraday constant) = 80 x 96485 = 7718800 C
= nNF
Current(I) = = 89.3 A
b) Power(P) = U(potential drop) x I (Current) = 1 x 89.3 = 89.3 W

Question 21:
During the titration of Cl- with Ag+, the precipitate is in equilibrium
with the ion pair AgCl(aq). Determine the concentration of AgCl(aq)


during the precipitation titration. The formation constant for the ion
pair is 4.9 x 10^-4 and the Ksp for AgCl is 1.8 x 10^-10
Ksp = [Ag+][Cl-] = x*x = 1.8 x 10^-10
 X = 1.34 x 10^-5 = [Ag+] = [Cl-]
Ag+ + Cl- AgCl
Kf =
 [AgCl] = 8.82 x 10^-14 M
Question 22:
A 2.108g sample of a solid mixture containing only potassium
carbonate (MM = 138.2058 g/mol) and potassium bicarbonate (MM =
100.1154 g/mol) is dissolved in distilled water. A volume of 35.79 mL

of a 0.744 M HCl standard solution is required to titrate the mixture to
a bromocresol green end point. Calculate the weight percent of
potassium carbonate and potassium bicarbonate in the mixture.

Moles of HCl = 0.02663 moles
K2CO3 + 2HCl  2KCl + CO2 + H2O
X

2x


KHCO3 + HCl  KCl + CO2 + H2O
Y

y

x
2x + y = 0.02663
138.2058x + 100.1154y = 2.108
 x = 0.009 ; y = 0.00863
 Mass of K2CO3 = 1.244g
 Mass of KHCO3 = 0.864g
Weight percent of K2CO3 =
Weight percent of KHCO3 = 100% – 59.01% = 40.99%
Question 23:

a) Au + 3HNO3 + 4HCl  HAuCl4 + 3NO2 + 3H2O
b) In this reaction, the oxidation state of Cl- does not change
 It does not participate in the redox process(oxidation-reduction)
 Hence, HCl supplies chloride ions to form a complex ion with the

oxidized gold
Question 24:
You perform an electrochemical reaction in which 0.800 mol of Cu+
are reduced to solid Cu


a) How many coulombs of charge are transferred ?
b) How many electrons are in this amount of charge ?
a)
Cu+  Cu + e0.8

0.8

Q = n x F = 0.8 x 96486 = 77188 C
b)
Moles of electrons = 0.8 x 6.022 x 10^-23 = 4.8176 x 10^-23
Number of electrons in this amount of charge = Q/n(e) = 1.602 x 10^27
Question 25:
What is the concentration of a 50.8 mL solution of HBr that is
completely titrated by 29.00 ml of a 0.200 M NaOH solution ?
Moles of NaOH = 0.0058 moles
Moles of HBr = Moles of NaOH = 0.0058 moles
 The concentration of a 50.8 mL solution of HBr = 0.114 M
Question 26:
Balance the redox reaction by inserting the appropriate coefficients
Fe3+ + NO2-  Fe2+ + NO3-


Adding water : Fe3+ + NO2- + H2O  Fe2+ + NO3Adding H+ : 2Fe3+ + NO2- + H2O  2Fe2+ + NO3- + 2H+
 This is balanced chemical equation in acidic medium

Question 27:
Enter the solubility product expression for Al(OH)3 (s)
Al3+ + 3OH-  Al(OH)3
The solubility product = Ksp = [Al3+]
Question 28:
Enter the solubility expression for Mg3(PO4)2
3Mg2+ + 2PO43-  Mg3(PO4)2
Ksp =