The acid dissociation constant is 1.3 × 10^-3.
<h3>What is acid-dissociation constant?</h3>
The acid-dissociation constant is a constant that shows the extent of dissociation of an acid in solution. We have to set up the reaction equation as shown below;
Let the acid be HA;
HA + H2O ⇄ H3O^+ + A^-
since the pH of the solution is 2.57 then;
[H3O^+] = Antilog(-pH) = Antilog(-2.57) = 2.7 × 10^-3
We can see that; [H3O^+] = [A^-] so;
Ka = (2.7 × 10^-3)^2/(5.5 × 10^–3)
Ka = 1.3 × 10^-3
Learn more about acid-dissociation constant: brainly.com/question/9728159
According to what is known about chemical equilibrium and Le Chatelier's principle, when you increase the amount of the reactants, the reaction will be moved to the products, this is because, the most reactants we have the most products we can produce.
From the given choices, the one that goes according to this reason is the third one: The volume of water vapor increases.
For equal moles of gas, temperature can be calculated from ideal gas equation as follows:
P×V=n×R×T ...... (1)
Initial volume, temperature and pressure of gas is 3.25 L, 297.5 K and 2.4 atm respectively.
2.4 atm ×3.25 L=n×R×297.5 K
Rearranging,
n\times R=0.0262 atm L/K
Similarly at final pressure and volume from equation (1),
1.5 atm ×4.25 L=n×R×T
Putting the value of n×R in above equation,
1.5 atm ×4.25 L=0.0262 (atm L/K)×T
Thus, T=243.32 K
