Question

What's the ratio? An acid with a \(\mathrm{p} K_{\mathrm{a}}\) of 8.0 is present in a solution with a pH of \(6.0 .\) What is the ratio of the protonated to the deprotonated form of the acid?

Short answer

The ratio of protonated to deprotonated form is 100:1.

Step-by-step solution

Identify Relevant Information

We have an acid with \( \mathrm{p}K_a = 8.0 \) and the solution has a \( \mathrm{pH} = 6.0 \). We need to find the ratio of the protonated form \([HA]\) to the deprotonated form \([A^-]\).

Apply the Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation relates pH, \( \mathrm{pK}_a \), and the ratio of deprotonated to protonated forms: \[pH = \mathrm{p}K_a + \log \frac{[A^-]}{[HA]} \] Reorganize it to solve for the ratio \( \frac{[A^{-}]}{[HA]} \).

Rearrange the Equation to Find the Ratio

Rearrange the Henderson-Hasselbalch equation to:\[\log \frac{[A^-]}{[HA]} = \mathrm{pH} - \mathrm{pK}_a\] Given \( \mathrm{pH} = 6.0 \) and \( \mathrm{p}K_a = 8.0 \), compute \( 6.0 - 8.0 = -2.0 \).

Calculate the Anti-logarithm

The previous result gives:\[\log \frac{[A^-]}{[HA]} = -2.0\]Calculate the anti-logarithm to find:\[\frac{[A^-]}{[HA]} = 10^{-2} = 0.01\]

Find the Desired Ratio

The ratio of protonated to deprotonated form is the inverse of \( \frac{[A^-]}{[HA]} \):\[\frac{[HA]}{[A^-]} = \frac{1}{0.01} = 100\]

Conclusion: State the Final Result

The ratio of the protonated form to the deprotonated form of the acid in the solution is 100:1.

Key concepts

Acid-Base Chemistry

Understanding acid-base chemistry is essential for determining the behavior of acids and bases in solutions. At the core of this concept is the ability of acids to donate protons (H+) and bases to accept them.
Acids can exist in equilibrium between their protonated and deprotonated forms:

• Protonated form: This is when the acid molecule has not lost its proton. It's typically represented as HA, where H is the hydrogen that can be released.
• Deprotonated form: This is the form of the acid after it has lost a proton. It's commonly represented as A^-, indicating that it's carrying a negative charge left by the lost proton.

In solution, the balance between these forms depends on the pH of the solution and the acid's intrinsic strength, characterized by its dissociation constant, pKa.

In our example, the acid's behavior in solution will determine whether it primarily exists in its protonated form or its deprotonated form.

pH and pKa Relationship

The relationship between pH and pKa is a powerful tool for predicting how an acid behaves in a solution. The pH is a measure of how acidic or basic a solution is, while the pKa is a measure of how easily an acid can lose a proton.
To make sense of these values:

• When the pH is lower than the pKa, the protonated form (HA) is favored because the solution provides enough H+ ions.
• Conversely, when the pH is higher than the pKa, the deprotonated form (A^-) is favored. In this situation, the solution does not favor the presence of free protons, encouraging the acid to release them and become deprotonated.

Using the Henderson-Hasselbalch equation:
\[pH = pK_a + \log \frac{[A^-]}{[HA]}\]
we can predict the ratio between protonated and deprotonated forms by rearranging it, as shown in our exercise. It helps us understand how the pH shifts affect the balance between these forms.

For example, with a pH of 6 and a pKa of 8, the solution has a lower pH than the pKa, favoring the protonated form.

Protonation and Deprotonation

Protonation and deprotonation describe the process where acids lose or gain protons. This process is crucial for understanding chemical reactivity and equilibrium in solutions.

• Protonation: This occurs when a base gains a proton (H+), resulting in the formation of the protonated species.
• Deprotonation: This is when an acid loses a proton, leading to the formation of the deprotonated species.

These transformations are influenced by the pH of the solution relative to the acid's pKa. When the pH is much lower than the pKa, the environment is rich in H+ ions, encouraging protonation.

However, if the pH is significantly higher than the pKa, deprotonation occurs because the environment allows the acid to release its proton. In the example given, because the pH of 6 is lower than the pKa of 8, the environment promotes protonation, ensuring that the solution has more of the protonated form available, as evidenced by the 100:1 ratio in favor of the protonated form.