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 \(\left.\mathrm{H}_{3} \mathrm{PO}_{4}\right) ?\)

Short 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-by-step solution

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 \( \text{H}_3 \text{PO}_4 \).

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, \( K_b = \frac{K_w}{K_{a3}} = \frac{1.0 \times 10^{-14}}{4.2 \times 10^{-13}} = 2.38 \times 10^{-2} \).

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.

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: \[ 2.38 \times 10^{-2} = \frac{x^2}{1 \times 10^{-3} - x} \].

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, \([\text{OH}^-] = 4.88 \times 10^{-3} \ M\).

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.