Consider that you have two solutions of acetic acid, Ka = 1.8x10-5, one solution that is 1.59 M and another solution that is 0.1
1 answer:
Answer:
The ratio of percent dissociation of the 1.59 M solution to the 0.186 M solution: 0.343 : 1.
Explanation:
Initially c
At equilibrium c-cα cα cα
Dissociation constant of an acetic acid:
Degree of dissociation = α
Dissociation constant of an acid is given by:
1) Concentration of acid = c = [HAc] = 1.59 M
Degree of dissociation = α
Percentage of dissociation = 0.3359%
2) Concentration of acid = c' = [HAc] = 0.186 M
Degree of dissociation = α'
Percentage of dissociation = 0.9789%
The ratio of percent dissociation of the 1.59 M solution to the 0.186 M solution:
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Answer:
Value of equilibrium constant is 0.0888
Explanation:
Both and
are gaseous. Hence equilibrium constant depends upon partial pressures of
and
.
Initially no and
were present.
Hence mole fraction of and
at equilibrium can be calculated from coefficient of
and
in balanced equation.
Mole fraction of = (number of moles of
)/(total number of moles of
and
) =
Mole fraction of = (number of moles of
)/(total number of moles of
and
) =
Let's assume both and
behaves ideally.
Therefore partial pressure of ,
and P_{CO_{2}}= x_{CO_{2}}.P_{total}
Where x represents mole fraction
So,
So,