Problem 111 - Chemistry: The Central Science in SI Units, Global Edition

Question

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?

Step-by-Step Solution

Answer

The value of \(K_a\) for the parent acid HA is approximately \(3.99 \times 10^{-10}\).

Step 1: Determine the concentration of hydroxide ions OH- and the acidic salt (A-)

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-.

Step 2: Calculate the concentration of 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: \(4.71 = -\log_{10} [OH^{-}]\) \[OH^{-} = 10^{-4.71} \approx 1.94 \times 10^{-5} M\]

Step 3: Find the concentration of the acidic salt (A-)

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

Step 4: Use the equilibrium expression to find Ka

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 \(3.99 \times 10^{-10}\).

Key Concepts

pH and pOH relationship

Understanding the relationship between pH and pOH is crucial in many chemistry problems. The pH scale measures the acidity of a solution, while the pOH scale measures the basicity. These two scales are interconnected by the equation:
\[ pH + pOH = 14 \]
This means that if you know the pH of a solution, you can easily find the pOH, and vice versa. In the given exercise, the solution had a pH of 9.29. By subtracting this from 14, we find the pOH:
\[ pOH = 14 - 9.29 = 4.71 \]
Knowing this relationship helps to transition from acidity to basicity, and it's foundational for understanding how substances behave in solutions.

• pH below 7 indicates an acidic solution.
• pH above 7 indicates a basic solution.
• pH of exactly 7 indicates a neutral solution.

Hydroxide ion concentration

Hydroxide ions (\([OH^-]\)) are essential for determining a solution's basicity. In the given problem, once we found the pOH value as 4.71, we used it to find the concentration of hydroxide ions:
\[ pOH = -\log_{10} [OH^-] \]
Solving for \([OH^-]\) gives us:
\[ [OH^-] = 10^{-4.71} \approx 1.94 \times 10^{-5} \, M \]
This calculation helps us understand the extent of dissociation of the basic part of the salt in the solution. It also highlights how the logarithmic nature of the pOH scale can be used to easily switch between concepts like acidity and basicity. Deeper comprehension of this process can aid in problem-solving across various chemical contexts.

Equilibrium expression for acids

The equilibrium expression for acids is essential for finding the acid dissociation constant (\(K_a\)). It represents the balance between dissociated ions and the un-dissociated acid:
\[ HA \rightleftharpoons H^+ + A^- \]
The expression is given by:
\[ K_a = \frac{[H^+][A^-]}{[HA]} \]
For the exercise, once the concentrations of \([H^+]\) and \([A^-]\) were determined, they were substituted into the equation to find \(K_a\):

• \([H^+] = 10^{-9.29} \approx 5.13 \times 10^{-10} \text{ M}\)
• \([A^-] \approx 1.94 \times 10^{-5} \text{ M}\)
• \([HA] \approx 0.25 \text{ M}\) (adjusted for the small dissociation)

Substituting these values gives:
\[ K_a = \frac{(5.13 \times 10^{-10})(1.94 \times 10^{-5})}{0.25} \approx 3.99 \times 10^{-10} \]
This calculation of \(K_a\) reveals the strength of the acid, showing how readily it donates protons in aqueous solution. Understanding this equilibrium concept is vital for comprehending various chemical reactions and environmental conditions.