Question

The Henderson-Hasselbalch equation can be written as $\mathrm{pH}=\mathrm{p}{K}_{\mathrm{a}}-\log (\frac{1}{\alpha }-1)$ where $\alpha =\frac{\left[{\mathrm{A}}^{-}\right]}{\left[{\mathrm{A}}^{-}\right]+[\mathrm{HA}]}$ Thus, the degree of ionization $(\alpha )$ of an acid can be determined if both the $\mathrm{pH}$ of the solution and the $\mathrm{p}{K}_{\mathrm{a}}$ of the acid are known. (a) Use this equation to plot the pH versus the degree of ionization for the second ionization constant of phosphoric acid $(K_{\mathrm{a}}=6.3\times {10}^{-8})$ (b) If $\mathrm{pH}=\mathrm{p}{K}_{\mathrm{a}}$ what is the degree of ionization? (c) If the solution had a pH of 6.0 what would the value of $\alpha$ be?

Short answer

For the phosphoric acid with $Ka=6.3\times {10}^{-}8$, the $pKa=7.20$. By plotting pH against degree of ionization, it will show a negative slope and intercept at $pKa$. When pH = pKa, the degree of ionization is 0.50. If the solution has a pH of 6.0, the degree of ionization (alpha) is 0.7937.

Step-by-step solution

Determine the pKa.

The second ionization constant of phosphoric acid is given as: $K_a=6.3\times {10}^{-8}$. To convert $K_a$ to $pKa$, take the negative logarithm of the $K_a$ (i.e. $pKa=-\log (K_a)$). This gives $pKa=-\log (6.3\times {10}^{-8})=7.20$.

Express pH in terms of pKa and alpha.

The Henderson–Hasselbalch equation can be rewritten to express $pH$ in terms of $pKa$ and $\alpha$. Essentially, this equation is linear with slope $=-1$ and intercept $=pKa$. This allows you to plot the relationship between $pH$ and $\alpha$.

Calculate Degree of Ionization When pH = pKa.

When $pH=pKa$, substituting into Henderson-Hasselbalch equation, $-\log (\frac{1}{\alpha }-1)=0$. Solving for alpha, we will get: $\alpha =\frac{1}{2}$. This degree of ionization is the value of alpha when $pH=pKa$.

Calculate Degree of Ionization for a Given pH.

Given pH of 6.0, we substitute $pH=6.0$ and $pKa=7.20$ into the Henderson-Hasselbalch equation. Solving for $\alpha$, we will get $\alpha$ as the degree of ionization for a $pH$ of 6.0. $pH=7.20-\log (\frac{1}{\alpha }-1)$ can be rearranged as: $\alpha =\frac{1}{{10}^{(7.20-6.0)}+1}$, hence $\alpha =\frac{1}{1.26}=0.7937$.

Key concepts

Degree of Ionization

The degree of ionization, represented by $\alpha$, refers to the fraction of an acid that ionizes in a solution. It is a critical concept that helps us understand how much of the acid converts into ions. This conversion is essential in determining the acidity or alkalinity of a solution.To calculate $\alpha$, use the formula:$$\alpha =\frac{[A^{-}]}{[A^{-}]+[HA]}$$Here, $[A^{-}]$ is the concentration of the ionized form, and $[HA]$ is the concentration of the non-ionized acid. When the degree of ionization is high, the acid is strong, meaning most of it dissociates into ions. However, if $\alpha$ is low, the acid is weak, indicating that only a small portion dissociates. Understanding $\alpha$ helps in predicting the behavior of acids in different solutions, making it a foundational concept in chemistry.

pKa Calculation

The $pKa$ calculation is an essential tool in chemistry, used to describe the acidity of a substance. It is derived from the acid dissociation constant $K_a$ and is calculated as follows:$$pKa=-\log (K_a)$$This formula involves taking the negative logarithm of the $K_a$ value.Understanding $pKa$ helps in determining the strength of an acid. A low $pKa$ indicates a strong acid, which ionizes easily, while a higher $pKa$ suggests a weaker acid.In this exercise, we calculated the $pKa$ for phosphoric acid's second ionization constant, which is given as $6.3\times {10}^{-8}$.The calculation gives:$$pKa=-\log (6.3\times {10}^{-8})=7.20$$Knowing how to calculate $pKa$ is crucial for plotting pH relationships and understanding acid ionization behaviors.

Phosphoric Acid Ionization

Phosphoric acid is a compound that can ionize in multiple steps, each with its own ionization constant. The second ionization constant is particularly interesting because it shows how a substance can change state as it interacts with its environment.In the context of this exercise, the second ionization constant of phosphoric acid $(K_a=6.3\times {10}^{-8})$ was used to determine how it behaves at specific pH levels.Using the Henderson-Hasselbalch equation:$$\mathrm{pH}=\mathrm{p}Ka-\log (\frac{1}{\alpha }-1)$$We can assess the degree of ionization $(\alpha )$ at different pH levels. For example:

• At $\mathrm{pH}=6.0$, $\alpha$ was calculated to be 0.7937, indicating that a significant portion of phosphoric acid ionizes.
• When $\mathrm{pH}=\mathrm{p}Ka$, $\alpha =0.5$, showing that half of the acid is ionized.

Understanding phosphoric acid ionization provides insight into how acids behave under different conditions, which is pivotal for fields like biochemistry, pharmacy, and environmental science.