Question

Calculate [OH \(^{-}\) ] and pH in a solution in which dihydrogen phosphate ion, \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\), is \(0.335 M\) and hydrogen phosphate ion, \(\mathrm{HPO}_{4}{ }^{2-}\), is (a) \(0.335 \mathrm{M}\) (b) \(0.100 \mathrm{M}\) (c) \(0.0750 \mathrm{M}\) (d) \(0.0300 \mathrm{M}\)

Short answer

Answer: \(K_{a} \times K_{b} = K_{w}\), where K\(_w\) is the ion product of water (1 × 10\(^{-14}\) at 25°C).

Step-by-step solution

Write down the equilibrium equations

We have two equilibrium equations in this exercise. The first is the dissociation of the dihydrogen phosphate ion: \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-} \rightleftharpoons \mathrm{H}^{+} + \mathrm{HPO}_{4}^{2-}\) The second is the reaction of hydrogen phosphate ion with water: \(\mathrm{HPO}_{4}^{2-} + \mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{OH}^{-} + \mathrm{H}_{2} \mathrm{PO}_{4}^{-}\)

Write the expression for the equilibrium constants

In this step, write the expressions for the equilibrium constant of acid dissociation (K\(_a\)) and base dissociation (K\(_b\)). For the first equation: \(K_{a} = \frac{[\mathrm{H}^{+}][\mathrm{HPO}_{4}^{2-}]}{[\mathrm{H}_{2} \mathrm{PO}_{4}^{-}]}\) For the second equation: \(K_{b} = \frac{[\mathrm{OH}^{-}][\mathrm{H}_{2} \mathrm{PO}_{4}^{-}]}{[\mathrm{HPO}_{4}^{2-}]}\)

Use the relationships between K\(_a\), K\(_b\) and K\(_w\)

The relationship between the equilibrium constants of acid and base dissociation is as follows: \(K_{a} \times K_{b} = K_{w}\) where K\(_w\) is the ion product of water, which is \(1 \times 10^{-14}\) at 25°C.

Calculate [OH\(^{-}\)] and [H\(^{+}\)]

In this step, we will use the given concentrations of H\(_{2}\)PO\(_{4}^{-}\) and HPO\(_{4}^{2-}\) to find the concentrations of OH\(^{-}\) and H\(^{+}\) using K\(_b\) equation and substituting K\(_a\) values. For H\(_{2}\)PO\(_{4}^{-}\), K\(_{a}\) is 6.2 × 10\(^{-8}\) and K\(_{b}\) is 1.62 × 10\(^{-7}\). Using the given concentrations for each case--(a) 0.335 M, (b) 0.100 M, (c) 0.0750 M, and (d) 0.0300 M--calculate the corresponding [OH\(^{-}\)] and [H\(^{+}\)] concentrations.

Calculate pH

Since we have the [H\(^{+}\)] concentration for each case, we can calculate the pH using the formula: \(\mathrm{pH}=-\log [\mathrm{H}^{+}]\) Calculate the pH for each case.

Calculate pOH

Finally, calculate the pOH for each case using the [OH\(^{-}\)] concentration and the formula: \(\mathrm{pOH} = -\log [\mathrm{OH}^{-}]\) The final results are the [OH\(^{-}\)] concentration and pH values for each case (a to d).

Key concepts

Acid Dissociation Constant (Ka)

Understanding the acid dissociation constant, commonly represented as \(K_a\), is essential when studying chemical equilibria involving acids. This value gives us an idea of how readily an acid will donate a proton (hydrogen ion, \(H^+\)) in an aqueous solution. The higher the \(K_a\), the stronger the acid is, indicating a greater tendency to lose its proton. In the provided exercise, the dissociation of dihydrogen phosphate ion \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) is considered:

\[ \mathrm{H}_{2} \mathrm{PO}_{4}^{-} \rightleftharpoons \mathrm{H}^{+} + \mathrm{HPO}_{4}^{2-} \]
We can express \(K_a\) mathematically as the ratio of the product of the concentrations of the products to the concentration of the reactants, without including water, as it is the solvent:

\[K_{a} = \frac{[\mathrm{H}^{+}][\mathrm{HPO}_{4}^{2-}]}{[\mathrm{H}_{2} \mathrm{PO}_{4}^{-}]}\]
Through \(K_a\), we can predict the direction of the equilibrium and calculate the concentration of hydrogen ions, which in turn helps us calculate the pH of the solution.

Base Dissociation Constant (Kb)

In parallel to \(K_a\), the base dissociation constant \(K_b\) tells us how readily a base will accept a proton, forming its conjugate acid, in an aqueous solution. The \(K_b\) of a substance is a numerical value that indicates its strength as a base. The value of \(K_b\) is crucial for calculating the equilibrium concentration of hydroxide ions \(\mathrm{OH}^{-}\) in the solution. For the hydrogen phosphate ion, the equilibrium with water is represented as:

\[\mathrm{HPO}_{4}^{2-} + \mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{OH}^{-} + \mathrm{H}_{2} \mathrm{PO}_{4}^{-}\]
The \(K_b\) expression for this reaction is given by:

\[K_{b} = \frac{[\mathrm{OH}^{-}][\mathrm{H}_{2} \mathrm{PO}_{4}^{-}]}{[\mathrm{HPO}_{4}^{2-}]}\]
By knowing the \(K_b\), we can determine the concentration of \(\mathrm{OH}^{-}\) in a solution, which then allows for the calculation of pOH, and subsequently, the pH.

pH Calculation

The pH scale is a measure of the acidity or basicity of a solution, calculated from the concentration of hydrogen ions \( [H^+] \). It's important for students to understand that pH is logarithmic, which means a small change in pH corresponds to a large change in hydrogen ion concentration. The pH can be found using the formula:

\[\mathrm{pH} = -\log [\mathrm{H}^{+}]\]
Estimating pH requires knowing the concentration of \(H^+\) ions, which may involve solving chemical equilibria calculations like in the exercise we've discussed. By practicing pH calculations, students become proficient in translating between the concentration of \(H^+\) and the pH value, which is vital in understanding chemical reactivity and the behavior of solutions in various scientific and industrial contexts.

pOH Calculation

Similarly, pOH is a measure of the basicity of a solution, associated with hydroxide ion concentration \( [OH^-] \). pOH is also a logarithmic scale and is related to pH through the relation \(pH + pOH = 14\), at 25°C, which is based on the self-ionization constant of water \(K_w\). The pOH is calculated using the formula:

\[\mathrm{pOH} = -\log [\mathrm{OH}^{-}]\]
Understanding pOH is equally important as understanding pH because it provides complete insight into the acid-base character of solutions. It helps students appreciate the symmetric nature of proton donor (acidic) and proton acceptor (basic) strength. When solving problems, such as those in the exercise, calculating pOH serves as a practical step towards comprehending the complexities of acid-base equilibria.