Question

What is the pH of a solution made by mixing \(0.30 \mathrm{~mol}\) \(\mathrm{NaOH}, 0.25 \mathrm{~mol} \mathrm{Na}_{2} \mathrm{HPO}_{4}\), and \(0.20 \mathrm{~mol} \mathrm{H}_{3} \mathrm{PO}_{4}\) with water and diluting to \(1.00 \mathrm{~L}\) ?

Short answer

The pH of the solution made by mixing 0.30 mol NaOH, 0.25 mol Na₂HPO₄, and 0.20 mol H₃PO₄ with water and diluting to 1.00 L is approximately 13.37.

Step-by-step solution

Calculate initial concentrations of the species

First, let's calculate the initial concentrations of the species in the solution. For NaOH: \[Initial \mathrm{~OH^{-}}~concentration = \frac{0.30 \mathrm{~mol}~OH^{-}}{1.00~L} = 0.30~M\] For Na₂HPO₄: \[Initial\mathrm{~HPO^{2-}}~concentration = \frac{0.25 \mathrm{~mol}~HPO^{2-}}{1.00 L} = 0.25~M\] For H₃PO₄: \[Initial\mathrm{~H_{3}PO_{4}}~concentration = \frac{0.20 \mathrm{~mol}~H_{3}PO_{4}}{1.00 L} = 0.20~M\]

Write the equilibrium constant expressions

We can write the possible reactions and their corresponding equilibrium constant expressions for the acid-base reaction between the species. Since H₃PO₄ is a triprotic acid, we must consider all possible reactions involving its ionization. Reaction 1: \(OH^{-} + H_{3}PO_{4} \rightleftharpoons H_{2}PO^{2-} +H_{2}O\) \[K_{a1} = \frac{[H_{2}PO^{2-}]}{[OH^{-}][H_{3}PO_{4}]}\] Reaction 2: \(OH^{-} + H_{2}PO^{2-} \rightleftharpoons HPO^{3-}+ H_{2}O\) \[K_{a2} = \frac{[HPO^{3-}]}{[OH^{-}][H_{2}PO^{2-}]}\] Reaction 3: \(OH^{-} + HPO^{3-} \rightleftharpoons PO^{4-}+ H_{2}O\) \[K_{a3} = \frac{[PO^{4-}]}{[OH^{-}][HPO^{3-}]}\]

Calculate final concentrations of the species

In this step, let's use the initial concentrations calculated in Step 1, along with the equilibrium constant expressions, to calculate the final concentrations of the species. For simplification purposes, we can ignore reaction 3 since it involves the third ionization of H₃PO₄, which does not contribute much to pH. Using Reaction 1: \[K_{a1} = \frac{[H_{2}PO^{2-}]}{[OH^{-}][H_{3}PO_{4}]} = 7.5×10^{-3}\] Plugging in the concentrations from Step 1: \[7.5×10^{-3} = \frac{(0.25-x)}{(0.30-x)(0.20)}\] \[x ≈ 0.067\] For Reaction 2: \[K_{a2} = \frac{[HPO^{3-}]}{[OH^{-}][H_{2}PO^{2-}]} = 6.2×10^{-8}\] Using the concentrations from Reaction 1: \[6.2×10^{-8} = \frac{y}{(0.30-x)(0.25-x)}\] \[y ≈ 4.35×10^{-7}\]

Calculate the pH of the solution

Now, we can calculate the pH using the final concentrations of the species and the equation for calculating pH. \[pOH = -\log([OH^{-}])\] First, the pOH has to be calculated. \[pOH = -\log\left(0.30 - 0.067\right) ≈ 0.63\] Next, we find the pH by considering that: \[pH + pOH = 14\] \[pH = 14 - 0.63 ≈ 13.37\] Therefore, the pH of the solution made by mixing 0.30 mol NaOH, 0.25 mol Na₂HPO₄, and 0.20 mol H₃PO₄ with water and diluting to 1.00 L is approximately 13.37.

Key concepts

Understanding Acid-Base Equilibrium

Acid-base equilibrium refers to the state of balance between an acid and its conjugate base (or a base and its conjugate acid) in a solution. This delicate balance is governed by the principle known as the equilibrium constant for the dissociation reaction of the acid/base in water. When an acid dissolves in water, it donates protons (H+) to form its conjugate base, while a base accepts protons to form its conjugate acid.

For example, when we look at the third ionization of H₃PO₄ in the textbook solution, it's clear why it was deemed insignificant: the acid is weak, and the amount of ionization is too low to affect the pH substantially. Focusing on the first and second ionizations allowed for a more practical approach to finding the solution's pH without unnecessary complications. By understanding how acid-base equilibrium works, students can predict the behavior of a solution and calculate its pH with greater accuracy.

Concentration of Solution and its Role in pH

The concentration of a solution is a measure of the amount of solute that is dissolved in a given amount of solvent. It's vital in pH calculations because the pH scale is logarithmic, and even small changes in concentration can lead to significant pH shifts.

In the exercise, we start by finding the initial molar concentrations of hydroxide (OH-), dihydrogen phosphate (HPO2-), and phosphoric acid (H3PO4). It's these concentrations that set the starting point for understanding the acid-base reactions that will occur. Knowing the initial amounts allows us to use equilibrium constants to find the final concentrations after reactions have taken place, which are necessary to calculate the pH. This illustrates why precision in measuring concentration is so crucial in chemistry.

Equilibrium Constant Expressions and pH Calculations

Equilibrium constant expressions, represented by the symbol K, describe the ratio of the concentrations of products to reactants of a chemical reaction at equilibrium. The higher the value of K, the more the reaction favors the formation of products. For acids and bases, we have specific types of K values, such as the dissociation constants (Ka) for acids and (Kb) for bases.

The textbook solution shows how to use these constants to find the final concentrations of species in a solution. We set up expressions for each reaction's equilibrium, then solve for unknowns, representing how much each species has reacted. It's crucial not to confuse these with initial concentrations or assume that all reactants convert entirely into products. In weak acid-base reactions, only a fraction of the molecules dissociate, and this nuanced understanding is essential for accurately determining pH, which is critical in fields ranging from biochemistry to environmental science.