Problem 101 - Chemistry The Central Science In Si Units

Question

Salts containing the phosphate ion are added to municipal water supplies to prevent the corrosion of lead pipes. (a) Based on the \(\mathrm{pK}_{\mathrm{ad}}\) values for phosphoric acid \(\left(\mathrm{p} K_{\mathrm{at}}=7.5 \times 10^{-3}\right.\), \(\left.\mathrm{p} K_{a 2}=6.2 \times 10^{-8}, \mathrm{p} K_{a 3}=4.2 \times 10^{-13}\right)\) what is the \(\mathrm{K}_{\mathrm{b}}\) value for the \(\mathrm{PO}_{4}^{3-}\) ion? (b) What is the pH of a \(1 \times 10^{-3}\) \(M\) solution of \(\mathrm{Na}_{3} \mathrm{PO}_{4}\) (you can ignore the formation of \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) and \(\left.\mathrm{H}_{3} \mathrm{PO}_{4}\right) ?\)

Step-by-Step Solution

Answer

\( K_b = 2.38 \times 10^{-2} \); pH of \( 1 \times 10^{-3} \ M \) \( \text{Na}_3\text{PO}_4 \) is 11.69.

Step 1: Understand the Problem

We need to find the base dissociation constant \( K_b \) of \( \text{PO}_4^{3-} \) using the provided \( pK_a \) values for phosphoric acid. Then, we need to determine the pH of a \( 1 \times 10^{-3} \ M \) solution of \( \text{Na}_3 \text{PO}_4 \) while ignoring the formation of \( \text{H}_2 \text{PO}_4^- \) and \( \text{H}_3 \text{PO}_4 \).

Step 2: Calculate \( K_b \) for \( \text{PO}_4^{3-} \)

The relationship between \( K_a \) and \( K_b \) for a conjugate pair is given by \( K_w = K_a \times K_b \), where \( K_w = 1.0 \times 10^{-14} \) at 25°C. For \( \text{PO}_4^{3-} \), which is the conjugate base of \( \text{HPO}_4^{2-} \), we use \( K_{a3} = 4.2 \times 10^{-13} \). Thus, \( K_b = \frac{K_w}{K_{a3}} = \frac{1.0 \times 10^{-14}}{4.2 \times 10^{-13}} = 2.38 \times 10^{-2} \).

Step 3: Consider Dissociation of \( \text{PO}_4^{3-} \)

In solution, \( \text{PO}_4^{3-} \) will accept a proton to form \( \text{HPO}_4^{2-} \) in the equation: \[ \text{PO}_4^{3-} + \text{H}_2\text{O} \rightleftharpoons \text{OH}^- + \text{HPO}_4^{2-} \]. We assume full dissociation of \( \text{Na}_3\text{PO}_4 \) to form \( \text{PO}_4^{3-} \) ions.

Step 4: Calculate pH of the Solution

Assuming initial concentration \( c \) of \( \text{PO}_4^{3-} \) is \( 1 \times 10^{-3} \ M \). Using the equation \( K_b = [\text{OH}^-][\text{HPO}_4^{2-}] / [\text{PO}_4^{3-}] \), assume \( x \) amount dissociates to give \([\text{OH}^-] = x \) and \([\text{HPO}_4^{2-}] = x \). Substituting gives: \[ 2.38 \times 10^{-2} = \frac{x^2}{1 \times 10^{-3} - x} \].

Step 5: Solve the Quadratic Equation

Neglect \( x \) in the denominator due to low concentration and simplify: \[ x^2 = 2.38 \times 10^{-5} \] giving \( x = \sqrt{2.38 \times 10^{-5}} \approx 4.88 \times 10^{-3} \). Thus, \([\text{OH}^-] = 4.88 \times 10^{-3} \ M\).

Step 6: Determine pH from \([OH^-]\)

Find pOH using \( \text{pOH} = -\log([\text{OH}^-]) = -\log(4.88 \times 10^{-3}) \approx 2.31 \). Then calculate pH using \( \text{pH} = 14 - \text{pOH} = 14 - 2.31 = 11.69 \).

Key Concepts

pH Calculation

Understanding how to calculate pH is essential when working with acid-base equilibria, particularly in solution chemistry. pH is a measure of the acidity or basicity of a solution. It is determined by the concentration of hydrogen ions \([H^+]\) in the solution.
The formula used to calculate pH is:

• \( \text{pH} = -\log([H^+]) \)

In cases where we have a basic solution, the concentration of hydroxide ions \([OH^-]\) is often easier to measure, as demonstrated in our exercise.
For basic solutions, you calculate pOH first:

• \( \text{pOH} = -\log([OH^-]) \)

Then, you find pH with:

• \( \text{pH} = 14 - \text{pOH} \)

In the provided solution, we determined that a concentration of \(4.88 \times 10^{-3}\) M of \([OH^-]\) led to a pOH of approximately 2.31.
This was then used to find a pH of 11.69, indicating the solution's basic nature.

Conjugate Acid-Base Pairs

In acid-base chemistry, understanding conjugate acid-base pairs is crucial. These pairs consist of an acid and a base that transform into each other by the gain or loss of a proton \(H^+\).
When an acid donates a proton, it turns into its conjugate base. Conversely, when a base accepts a proton, it turns into its conjugate acid.
The relationship between an acid's dissociation constant \(K_a\) and its conjugate base's dissociation constant \(K_b\) is key to solving many equilibria problems, as seen in this exercise. The formula used is:

• \( K_w = K_a \times K_b \)
• where \( K_w \) is the ion product of water, \(1.0 \times 10^{-14} M^2\).

For \( \text{PO}_4^{3-}\), it acts as a conjugate base of \( \text{HPO}_4^{2-}\), with its \( K_a\) given as \(4.2 \times 10^{-13}\).
Using these values, the \( K_b \) was calculated as \(2.38 \times 10^{-2}\), providing insight into the base's strength.

Solution Chemistry

Solution chemistry involves the study of different solutes in a given solvent and their interactions. It is essential for understanding how substances like salts and acids behave when dissolved in water. When \( \text{Na}_3\text{PO}_4\) is dissolved, it dissociates completely to release three sodium ions \( \text{Na}^+\) and one phosphate ion \( \text{PO}_4^{3-}\).
This transformation affects the pH of the solution by altering the concentration of hydroxide ions \( \text{OH}^-\) when \( \text{PO}_4^{3-} \) acts as a base and accepts a proton from water, leading to the formation of \( \text{HPO}_4^{2-} \) and \( \text{OH}^- \).

• The dissociation interaction is key to understanding the pH adjustment proposed in the exercise, as seen by calculating \([OH^-]\) concentration.

These solutes behave according to their chemical nature and surrounding conditions, such as temperature and concentration, all crucial in equilibrium studies and solution preparation techniques.