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Answer:
The degree of dissociation of acetic acid is 0.08448.
The pH of the solution is 3.72.
Explanation:
The 
The value of the dissociation constant = 
![pK_a=-\log[K_a]](https://tex.z-dn.net/?f=pK_a%3D-%5Clog%5BK_a%5D)

Initial concentration of the acetic acid = [HAc] =c = 0.00225
Degree of dissociation = α

Initially
c
At equilibrium ;
(c-cα) cα cα
The expression of dissociation constant is given as:
![K_a=\frac{[H^+][Ac^-]}{[HAc]}](https://tex.z-dn.net/?f=K_a%3D%5Cfrac%7B%5BH%5E%2B%5D%5BAc%5E-%5D%7D%7B%5BHAc%5D%7D)



Solving for α:
α = 0.08448
The degree of dissociation of acetic acid is 0.08448.
![[H^+]=c\alpha = 0.00225M\times 0.08448=0.0001901 M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3Dc%5Calpha%20%3D%200.00225M%5Ctimes%200.08448%3D0.0001901%20M)
The pH of the solution ;
![pH=-\log[H^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%5BH%5E%2B%5D)
![=-\log[0.0001901 M]=3.72](https://tex.z-dn.net/?f=%3D-%5Clog%5B0.0001901%20M%5D%3D3.72)
1.1214 mL will a 0.205-mole sample of He occupy at 3.00 atm and 200 K.
<h3>What is an ideal gas equation?</h3>
The ideal gas law (PV = nRT) relates the macroscopic properties of ideal gases. An ideal gas is a gas in which the particles (a) do not attract or repel one another and (b) take up no space (have no volume).
Using equation PV=nRT, where n is the moles and R is the gas constant. Then divide the given mass by the number of moles to get molar mass.
Given data:
P= 3.00 atm
V= ?
n=0.205 mole
R= 
T=200 K
Putting value in the given equation:


V= 1.1214 mL
Learn more about the ideal gas here:
brainly.com/question/27691721
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