Ost-lar 1. Write out the balanced equation for the reaction of 1 mole of NaOH with 1 mole of H;PO4: 2. Write out the balanced equation for the reaction of 2 moles of NaOH with 1 mole of H3PO4: 3. Write out the balanced equation for the reaction of 3 moles of NaOH with 1 mole of H3PO4:
Basics in Organic Reactions Mechanisms
In organic chemistry, the mechanism of an organic reaction is defined as a complete step-by-step explanation of how a reaction of organic compounds happens. A completely detailed mechanism would relate the first structure of the reactants with the last structure of the products and would represent changes in structure and energy all through the reaction step.
Heterolytic Bond Breaking
Heterolytic bond breaking is also known as heterolysis or heterolytic fission or ionic fission. It is defined as breaking of a covalent bond between two different atoms in which one atom gains both of the shared pair of electrons. The atom that gains both electrons is more electronegative than the other atom in covalent bond. The energy needed for heterolytic fission is called as heterolytic bond dissociation energy.
Polar Aprotic Solvent
Solvents that are chemically polar in nature and are not capable of hydrogen bonding (implying that a hydrogen atom directly linked with an electronegative atom is not found) are referred to as polar aprotic solvents. Some commonly used polar aprotic solvents are acetone, DMF, acetonitrile, DMSO, etc.
Oxygen Nucleophiles
Oxygen being an electron rich species with a lone pair electron, can act as a good nucleophile. Typically, oxygen nucleophiles can be found in these compounds- water, hydroxides and alcohols.
Carbon Nucleophiles
We are aware that carbon belongs to group IV and hence does not possess any lone pair of electrons. Implying that neutral carbon is not a nucleophile then how is carbon going to be nucleophilic? The answer to this is that when a carbon atom is attached to a metal (can be seen in the case of organometallic compounds), the metal atom develops a partial positive charge and carbon develops a partial negative charge, hence making carbon nucleophilic.
![Post-lab Questions
1. Write out the balanced equation for the reaction of 1 mole of NaOH with 1 mole of H3PO4:
2. Write out the balanced equation for the reaction of 2 moles of NaOH with 1 mole of H3PO4:
3. Write out the balanced equation for the reaction of 3 moles of NaOH with 1 mole of H3PO4:
4. Given similar concentrations, the stronger acid corresponds to the lower pH. Comment on the
relative strengths of the acids H3PO4 and H2PO4. H2PO,²-
Some NOTES: Read very carefully
For a Diprotic and Triprotic Ka’s (& see YOUR class notes & textbook)
In determining the quantity of the acid or the molarity of the acid, we are normally just
interested in the final equivalence point. In a pH titration plot, this is determined by
finding the point of inflection on the final area where we see a significant rise in pH (This
can be approximated by determining the midpoint). For a diprotic acid, the equivalence
point corresponds to the addition of exactly 1/2 the volume of NaOH required to reach
the final equivalence point. At the the 1/2 way point in the 1st titration, the concentration
of H2X(aq) remaining in the solution is equal to 1/2 the initial concentration of H2X. The
concentration of NaHX(aq) produced is also numerically equal to 1/2 the initial
concentration of H2X, such that
H2X(aq) + H2O(1) = HX¯ + H3O*
K, = [H;Oʻ][HX] / [H,X]
using the concentrations that we know for H2X and HX (= NaHX) at the 1/2 way point
we get
[H;O*] = K, {1/2[H2X]initial / [1/2H2X]initial}
[H3O*] = Ka
From a diprotic acid titration graph we can determine the pH at this point since
pH=-log[H3O*], and thus obtain the Ka for this equilibrium. Since this is a polyprotic
acid, this corresponds to Kal. From the pH at the midpoint of the second titration, Ka2
can be determined.
For a triprotic acid, the other two equivalence points should correspond to 1/3 and 2/3
of the volume of the base required to reach the final inflexion and thus one can still
determine the Ka values, even when they are not obvious [or outside the pH instrument
range].
5. Given this information, and using the first equivalence point, calculate the exact molarity of
phosphoric acid.
7
6. Determine the three Ka values for phosphoric acid. How do these numbers compare to the
actual values (theory) given earlier ?
Report [DUE 8pm, 4/12/21]
• Cover page
• Aims/Objectives
• R&D (including 3 graphs and 6 post-lab Q’s)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F85915cf5-de07-402b-a483-54987fecbc19%2Ff40bd23f-b3e6-471f-8750-b4a59488f415%2Fhq31rk_processed.jpeg&w=3840&q=75)
![Calculations and Graph [via Excel]:
You need to fully interpret the pH titration curve, making note of the inflections, equivalence
points, buffer regions, where pH = pKa. You will need to make use of “basic calculus' in order to
accurately measure the volumes at the end-points, and thus determine the exact molarity of the
phosphoric acid.
Sample Data [Iprovided below]:
Note - Since the NaOH and the H3PO4 solutions are about the same strength, 0.100 M, the ratio
of the volumes reacted should be the same as the ratio of the moles reacted. (note: 10.00 ml acid
was used). For pH titration - You will graph this data in Excel.
Volume
pH
Volume
pH
NaOH (ml)
NaOH (ml)
2.52
19
7.86
1
2.56
20
8.19
2
2.61
21
8.80
3
2.69
22
10.33
4
2.76
23
10.96
5
2.85
24
11.21
2.95
25
11.42
3.09
26
11.58
8
3.28
27
11.65
9
3.54
28
11.72
10
4.19
29
11.80
11
6.04
30
11.88
12
6.52
31
11.92
13
6.77
32
11.98
14
7.01
33
12.03
15
7.14
34
12.07
16
7.32
35
12.10
17
7.46
36
12.12
18
7.64
Graph: Plot in Excel pH (ordinate, ie, y-axis) vs. volume of NaOH (abscissa, ie., x-axis). Show
the data points, and then overlay the points with a best-fit polynomial line. Indicate on the plots.
a) the buffer region(s),
b) the region(s) where pH = pKa,
c) the end point(s),
d) the region where OH is in excess, and
e) the major species present in solution in each region. (acid, conjugate base, etc)
5
Using Excel, also construct first-derivative & second derivative plots for the
titration. A first-derivative plot is constructed by plotting (pH2 – pH1)/(V2 – V1) vs. volume of
titrant used. V1 and V2 are two successive titration volumes and pH1 and pH2 are the two
corresponding pH values. The second derivative is computed from the first derivative: The
second derivative is then the 'difference of the differences' vs. volume of titrant used. ie. You
need to plot:
(1) pH against volume of the titrant [you've just done this]
(2) ApH / AV against volume of the titrant
(3) A’pH /AV²
against volume of the titrant
The phosphoric acid content can then be identified using the interpolated volume at the
equivalence point of first and second derivative curve. The first derivative plots will have peaks
where the original graphs have inflection points (i.e. the end points of the titrations). This
equivalence point will be the x-intercepts for the second derivative.
DO YOU UNDERSTAND THIS, really really understand
this ??????](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F85915cf5-de07-402b-a483-54987fecbc19%2Ff40bd23f-b3e6-471f-8750-b4a59488f415%2F38hwn7_processed.jpeg&w=3840&q=75)
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