When a diprotic acid is titrated with a strong base, and the Ka1 and Ka2 are significantly different, then the pH vs. volume plo
t of the titration will have Group of answer choices
1 answer:
Complete question is;
When a diprotic acid is titrated with a strong base, and the Ka1 and Ka2 are significantly different, then the pH vs. volume plot of the titration will have
a. a pH of 7 at the equivalence point.
b. two equivalence points below 7.
c. no equivalence point.
d. one equivalence point.
e. two distinct equivalence points
Answer:
Option E - Two Distinct Equivalence points
Explanation:
I've attached a sample diprotic acid titration curve.
In diprotic acids, the titration curves assists us to calculate the Ka1 and Ka2 of the acid. Thus, the pH at the half - first equivalence point in the titration will be equal to the pKa1 of the acid while the pH at the half - second equivalence point in a titration is equal to the pKa2 of the acid.
Thus, it is clear that there are two distinct equivalence points.
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Answer:
Explanation:
From the given information;
The chemical reaction can be well presented as follows:
⇄
Now, K is known to be the equilibrium constant and it can be represented in terms of each constituent activity:
i.e
However, since we are dealing with liquids solutions;
since the activity of
is equivalent to 1
Hence, under standard conditions(i.e at a pressure of 1 bar)
(b)
From the CRC Handbook, we are meant to determine the value of the Gibb free energy by applying the formula:
Thus, for this reaction; the Gibbs frree energy = -68 kJ/mol
(c)
Le's recall that:
At equilibrium, the instantaneous free energy is usually zero &
Q(reaction quotient) is equivalent to K(equilibrium constant)
So;
(d)
The direction by which the reaction will proceed can be determined if we can know the value of Q(reaction quotient).
This is because;
If Q < K, then the reaction will proceed in the right direction towards the products.
However, if Q > K , then the reaction goes to the left direction. i.e to the reactants.
So;
Since we are dealing with liquids;
Q = 1
Since Q < K; Then, the reaction proceeds in the right direction.
Hence, SO2 as well O2 will combine to yield SO3, then condensation will take place to form liquid.