Answer :
Explanation:
We have to calculate [tex]pK_{a}[/tex] value.
It is known that at the equivalence point concentration of acid is equal to the concentration of anion formed.
Hence, [HA] = [tex][A^{-}][/tex]
Now, relation between [tex]pK_{a}[/tex] and pH is as follows.
pH = [tex]pK_{a} + log \frac{[A^{-}]}{[HA]}[/tex]
Putting the values into the above formula as follows.
pH = [tex]pK_{a} + log \frac{[A^{-}]}{[HA]}[/tex]
4.23 = [tex]pK_{a} + log (1)[/tex] (as [HA] = [tex][A^{-}][/tex])
[tex]pK_{a}[/tex] = 4.23 (as log (1) = 0)
or, [tex]pK_{a}[/tex] = 4
Thus, we can conclude that [tex]pK_{a}[/tex] of given weak acid is 4.
A weak acid is titrated with a strong base. The pH is 4.23 at 50% T, 6.72 at 100% and 9.12 at the end point. The pKa of the weak acid is 4.23.
Let's consider the generic reaction for the titration of a weak acid (HA) with a strong base (BOH).
HA(aq) + BOH(aq) ⇒ BA(aq) + H₂O(aq)
At 50%, we have a buffer system formed by equal concentrations of the weak acid (HA) and its conjugate base (A⁻).
Given the pH is 4.23, we can calculate the pKa using the Henderson-Hasselbach equation.
[tex]pH = pKa + log [A^{-} ]/[HA] = pKa+log1=pKa\\\\pKa=pH=4.23[/tex]
A weak acid is titrated with a strong base. The pH is 4.23 at 50% T, 6.72 at 100% and 9.12 at the end point. The pKa of the weak acid is 4.23.
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