Answer:
1.
was the
value calculated by the student.
2.
was the
of ethylamine value calculated by the student.
Explanation:
1.
The
value of Aspirin solution = 2.62
![pH=-\log[H^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%5BH%5E%2B%5D)
![[H^+]=10^{-2.62}=0.00240 M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D10%5E%7B-2.62%7D%3D0.00240%20M)
Moles of s asprin = 
Volume of the solution = 0.600 L
The initial concentration of Aspirin = c = 

initially
c 0 0
At equilibrium
(c-x) x x
The expression of dissociation constant :
:



was the
value calculated by the student.
2.
The
value of ethylamine = 11.87


![pOH=-\log[OH^-]](https://tex.z-dn.net/?f=pOH%3D-%5Clog%5BOH%5E-%5D)
![[OH^-]=10^{-2.13}=0.00741 M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D10%5E%7B-2.13%7D%3D0.00741%20M)
The initial concentration of ethylamine = c = 0.100 M

initially
c 0 0
At equilibrium
(c-x) x x
The expression of dissociation constant :
:



was the
of ethylamine value calculated by the student.
Answer:
The pH of solution is 2.88 .
Explanation:
The reaction is :

We know,
for this reaction is = 
Also, since volume of water is 1 L.
Therefore, molarity of solution is equal to number of moles.
Also, ![K_a=\dfrac{[CH_3COO^-][H^+]}{[CH_3COOH]}](https://tex.z-dn.net/?f=K_a%3D%5Cdfrac%7B%5BCH_3COO%5E-%5D%5BH%5E%2B%5D%7D%7B%5BCH_3COOH%5D%7D)
Let, amount of
produce is x.
So,
![K_a=\dfrac{[x][x]}{[0.1]}\\1.76\times 10^{-5}=\dfrac{[x][x]}{[0.1]}](https://tex.z-dn.net/?f=K_a%3D%5Cdfrac%7B%5Bx%5D%5Bx%5D%7D%7B%5B0.1%5D%7D%5C%5C1.76%5Ctimes%2010%5E%7B-5%7D%3D%5Cdfrac%7B%5Bx%5D%5Bx%5D%7D%7B%5B0.1%5D%7D)

We know, 
Therefore, pH of solution is 2.88 .
Hence, this is the required solution.
Answer:
for the given reaction is 130.19kJ/mol
Explanation:
To Calculate the
, we use the formula:

For the given chemical reaction:

We are given:

Now, to calculate
, we put the values in the above equation:


As, the value of
comes out to be positive, the reaction is said to be non-spontaneous reaction.