Problem 8 - Chemical Principles

Question

Calculate the \(\mathrm{pH}\) and \(\mathrm{pOH}\) of (a) a solution that is \(0.17 \mathrm{M}\) \(\mathrm{Na}_{2} \mathrm{HPO}_{4}(\mathrm{aq})\) and \(0.25 \mathrm{M} \mathrm{Na}_{3} \mathrm{PO}_{4}(\mathrm{aq})\); (b) a solution that is \(0.66 \mathrm{M}\) \(\mathrm{Na}_{2} \mathrm{HPO}_{4}(\mathrm{aq})\) and \(0.42 \mathrm{M} \mathrm{Na}_{3} \mathrm{PO}_{4}(\mathrm{aq})\); (c) a solution that is \(0.12 \mathrm{M}\) \(\mathrm{Na}_{2} \mathrm{HPO}_{4}(\mathrm{aq})\) and \(0.12 \mathrm{M} \mathrm{Na}_{3} \mathrm{PO}_{4}(\mathrm{aq})\).

Step-by-Step Solution

Answer

Solve for [H+] using the equilibrium constant K3 in steps 3-5 for each solution, calculate pH = -log([H+]), and then pOH = 14 - pH. For (a) and (b) use initial concentrations to find equilibrium concentrations; for (c) the initial concentrations are the same, so no net reaction occurs and [H+] can be calculated directly.

Step 1: Understand the Problem

We are calculating the pH and pOH of solutions containing mixtures of disodium hydrogen phosphate (Na2HPO4) and trisodium phosphate (Na3PO4). Both are salts that hydrolyze to form basic solutions. The pH is determined by the concentration of hydroxide ions (OH-), which we'll calculate using an equilibrium expression involving the phosphate equilibrium constants.

Step 2: Determine the Relevant Equilibrium Constants

The disodium hydrogen phosphate (Na2HPO4) acts as a weak acid (with a corresponding K2 value) and trisodium phosphate (Na3PO4) produces hydroxide ions in solution. The overall equilibrium reaction can be represented as HPO42- + H2O ↔ PO43- + H+. We need to use the third ionization constant of phosphoric acid (K3), since it represents the equilibrium of HPO42- losing a proton to form PO43-. The value of K3 is approximately 4.7 x 10^-13.

Step 3: Set Up the Equilibrium Concentration Table for Solution (a)

For the reaction HPO42- (from Na2HPO4) + H2O ↔ PO43- (from Na3PO4) + H+, let the change in concentration of HPO42- be -x, PO43- be +x, and H+ be +x. Initial concentrations are 0.17 M and 0.25 M respectively. At equilibrium, we'll have (0.17-x) M of HPO42- and (0.25+x) M of PO43-.

Step 4: Write the Equilibrium Expression and Solve for x [Solution (a)]

The equilibrium expression is K3 = [PO43-][H+]/[HPO42-]. Substituting the equilibrium concentrations we have K3 = [(0.25+x)(x)]/[(0.17-x)]. Assuming that x is small relative to the initial concentrations, the equation simplifies to K3 ≈ (0.25*x)/(0.17), we can solve for x, which represents [H+].

Step 5: Calculate the pH and pOH for Solution (a)

Once x is calculated, it represents [H+], and pH = -log([H+]). Because pH + pOH = 14, pOH can be found by subtracting the calculated pH from 14.

Step 6: Repeat Steps 3-5 for Solution (b) and (c) with Their Initial Concentrations

For solution (b) use initial concentrations of Na2HPO4 and Na3PO4 as 0.66 M and 0.42 M, respectively, to find the equilibrium concentrations. For solution (c), use initial concentrations of 0.12 M for both Na2HPO4 and Na3PO4.

Step 7: Analyze the Symmetry in Solution (c)

In solution (c), since the initial concentrations of Na2HPO4 and Na3PO4 are the same, the reaction is symmetric with respect to HPO42- and PO43-. Therefore, no net reaction occurs, and [H+] remains equal to x. This simplifies the calculations for solution (c).

Step 8: Compile the Results for All Solutions

After performing calculations for each solution (a, b, c), compile the pH and pOH values obtained from the equilibrium expressions and pH equations. These results give the final values of pH and pOH for each solution.

Key Concepts

pOH Calculation

Understanding pOH calculation is vital for students who are delving into the behavior of solutions in chemistry. When considering a solution's basicity, the pOH is as important as pH, standing for the negative logarithm of the hydroxide ion concentration. To calculate pOH, one needs to find the concentration of hydroxide ions (OH^-) which typically comes from water's self-ionization or from the ionization of bases in solution.

For instance, in the absence of any acidic species, the pOH of a solution can be found directly by measuring the concentration of hydroxide ions and taking the negative logarithm. In our specific exercise, because the presence of disodium hydrogen phosphate (Na_2HPO_4) and trisodium phosphate (Na_3PO_4) creates a basic environment, we can determine the pOH by first calculating the pH from the hydrolysis of these salts and then using the relation that pH + pOH = 14 at standard conditions (298 K). This is a direct application of the principles of hydrolysis and chemical equilibrium, tailored for solutions with basic characteristics.

Chemical Equilibrium

Chemical equilibrium is the state reached when the rates of the forward and reverse reactions are equal, resulting in no net change in the concentration of reactants and products over time. It's fundamental to the study of reactions in solution, such as the hydrolysis of salts or ionization of weak acids or bases.

Using this concept, we can describe the behavior of phosphates in aqueous solutions. The phosphate ion (HPO_4^{2-}) can lose a proton to become (PO_4^{3-}), or it can gain a proton to revert to its previous state. At equilibrium, the rate at which (HPO_4^{2-}) loses a proton is equal to the rate at which (PO_4^{3-}) gains a proton. By applying this to our exercise, students can explore how to mathematically express this equilibrium using the ionization constant (K_3) for phosphoric acid and determine the concentrations of ions involved at equilibrium.

Hydrolysis of Salts

Hydrolysis of salts is a chemical reaction in which a salt reacts with water to produce an acid and a base. It plays a crucial role in understanding the pH of solutions, especially in the context of salts resulting from the reaction of strong bases and weak acids, or vice versa. In this process, the anions of the weak acid or the cations of the weak base can react with water molecules to produce hydroxide (OH^-) or hydronium (H_3O^+) ions, respectively.

In the context of our textbook exercise, disodium hydrogen phosphate (Na_2HPO_4) undergoes hydrolysis to form sodium ions (Na^+) and hydrogen phosphate ions (HPO_4^{2-}). The hydrogen phosphate ions can further lose a proton, resulting in an increase in the hydroxide ion concentration in the solution, contributing to its basic nature. Understanding the hydrolysis of these salts allows us to calculate the concentration of hydroxide ions and ultimately the pH and pOH to describe the solution's acidity or basicity.

Phosphoric Acid Ionization Constants

Phosphoric acid (H_3PO_4) is a triprotic acid, meaning it can donate three protons (H^+), each with its own ionization constant (K_a). These constants are essential in predicting the extent of ionization at each stage and, hence, the concentration of ions in solution. For phosphoric acid, the constants (K_1, K_2, K_3) decrease with each successive proton donation, indicating that the first proton is the easiest to lose while the third one is the hardest.

The third ionization constant (K_3) is particularly relevant in our problem because it pertains to the equilibrium involving the hydrogen phosphate ion (HPO_4^{2-}) and the phosphate ion (PO_4^{3-}). By understanding and applying this constant, we can predict the behavior of various phosphate species in solution, allowing the student to determine the resultant pH and pOH of phosphate-containing solutions. Phosphoric acid's ionization constants thus serve as a beacon in navigating the complex seas of acid-base equilibria and solution chemistry.