A \(0.25 M\) solution of a salt NaA has \(\mathrm{pH}=9.29 .\) What is the value of \(K_{a}\) for the parent acid HA?

To find the concentration of hydroxide ions, we first need to determine the pOH value. Since the pH and pOH values are related by the equation: \(pH + pOH = 14\), we can calculate pOH as follows: pOH = 14 - pH Now, we can calculate the concentration of hydroxide ions, OH-, using the definition of pOH: \(pOH = -\log_{10} [OH^{-}]\) Finally, we know that the concentration of the acidic salt A- is the same as the concentration of the hydroxide ions OH-.

Using the pH value provided in the problem, we can determine the pOH value: pOH = 14 - 9.29 = 4.71 Now, we can find the [OH-] concentration: \(4.71 = -\log_{10} [OH^{-}]\) \[OH^{-} = 10^{-4.71} \approx 1.94 \times 10^{-5} M\]

Since the concentration of the acidic salt, A-, is equal to the concentration of OH-: \[A^{-} = 1.94 \times 10^{-5} M\]

We can now use the equilibrium expression for the dissociation of the parent acid, HA, into A- and H+: \[HA \rightleftharpoons H^{+} + A^{-}\] \[K_{a} = \frac{[H^{+}][A^{-}]}{[HA]}\] Since we have a 0.25 M solution of NaA and the concentration of A- is 1.94 x 10^-5 M, we can find the concentration of the parent acid, HA: \[HA = 0.25 M - 1.94 \times 10^{-5} M \approx 0.25 M\] We also know that the pH = 9.29, so the concentration of H+ ions can be found: \[pH = -\log_{10} [H^{+}]\] \[H^{+} = 10^{-9.29} \approx 5.13 \times 10^{-10} M\] Now, we can substitute these values into the equilibrium expression to find Ka: \[K_{a} = \frac{(5.13 \times 10^{-10})(1.94 \times 10^{-5})}{0.25}\] \[K_{a} \approx 3.99 \times 10^{-10}\] So the value of Ka for the parent acid HA is approximately \(3.99 \times 10^{-10}\).