Question

What is the \(\mathrm{pH}\) of a solution made by mixing \(0.40 \mathrm{~mol}\) \(\mathrm{NaOH}, 0.25 \mathrm{~mol} \mathrm{Na}_{2} \mathrm{HPO}_{4}\), and \(0.30 \mathrm{~mol} \mathrm{H}_{3} \mathrm{PO}_{4}\) with water and diluting to \(2.00 \mathrm{~L} ?\)

Short answer

The pH of the solution made by mixing \(0.40 \mathrm{~mol}\) NaOH, \(0.25 \mathrm{~mol}\) Na₂HPO₄, and \(0.30 \mathrm{~mol}\) H₃PO₄ with water and diluting to \(2.00 \mathrm{~L}\) is approximately \(6.92\).

Step-by-step solution

Write the chemical reactions

First, identify the chemical reactions that will occur when NaOH reacts with H₃PO₄ and Na₂HPO₄. These reactions will be: \[ \begin{cases} \mathrm{H_{3}PO_{4} + NaOH\ \rightarrow H_{2}PO_{4}^{-} + Na^{+} + H_{2}O} \\ \\ \mathrm{Na_{2}HPO_{4} + NaOH\ \rightarrow Na_{3}PO_{4} + H_{2}O} \end{cases} \]

Determine the limiting reactant

In order to verify which reaction (if not both) will occur, we need to find the limiting reactant. Calculate the moles of NaOH needed to react with H₃PO₄ and Na₂HPO₄. Moles of NaOH required to react with \(0.30 \mathrm{~mol}\) \(H_{3}PO_{4}\): \(0.30 \cdot {1}/{1} = 0.30 \mathrm{~mol}\), Moles of NaOH required to react with \(0.25 \mathrm{~mol}\) \(Na_{2}HPO_{4}\): \(0.25 \cdot {1}/{1} = 0.25 \mathrm{~mol}\), Total moles of NaOH required: \(0.30 + 0.25 = 0.55 \mathrm{~mol}\) Since we have \( 0.40 \mathrm{~mol}\) of NaOH initially, it is the limiting reactant.

Determine molar concentrations of reactants

Calculate the molar concentrations of each reactant after the reaction has occurred, using the initial moles of reactants and the limiting reactant (NaOH). Moles of NaOH left: \(0 \mathrm{~mol}\), Moles of H₃PO₄ left: \(0.30 - 0.40 = -0.10 \mathrm{~mol} \to 0 \mathrm{~mol}\) (since it reacts completely), Moles of Na₂HPO₄ left: \(0.25 - (0.40 - 0.30) = 0.15 \mathrm{~mol}\), Moles of \(H_{2}PO_{4}^{-}\) formed: \(0.40 - (0.40 - 0.30) = 0.30 \mathrm{~mol}\), Now, divide the moles by the total volume of the solution to get molar concentrations: \[[H_{3}PO_{4}] = \frac{0 \mathrm{~mol}}{2.00 \mathrm{~L}} = 0 \mathrm{M},\] \[[Na_{2}HPO_{4}] = \frac{0.15 \mathrm{~mol}}{2.00 \mathrm{~L}} = 0.075 \mathrm{M},\] \[[H_{2}PO_{4}^{-}] = \frac{0.30 \mathrm{~mol}}{2.00 \mathrm{~L}} = 0.15 \mathrm{M}.\]

Calculate the pH of the solution

Since we have a mixture of \(H_{2}PO_{4}^{-}\) and \(Na_{2}HPO_{4}\), we can consider the mixture as a buffer solution. Use the Henderson-Hasselbalch equation to find the pH of the solution: \[pH = pK_{a2} + \log{\frac{[A^{-}]}{[HA]}}\] Where \(pK_{a2}\) is the second ionization constant of phosphoric acid (\(H_{3}PO_{4}\)). From a table of ionization constants, we find that \(pK_{a2} \approx 7.21\). Now, plug in the values for the concentrations of \(H_{2}PO_{4}^{-}\) and \(Na_{2}HPO_{4}\): \[pH = 7.21 + \log{\frac{0.075}{0.15}}\] \[pH \approx 6.92\] So, the pH of the solution made by mixing the given reagents and diluting to \(2.00 \mathrm{~L}\) is approximately \(6.92\).

Key concepts

Buffer Solution

A buffer solution is a special type of solution that resists drastic changes in its pH, even when small amounts of acids or bases are added. This stability is due to the presence of a weak acid and its conjugate base or a weak base and its conjugate acid. These pairs can neutralize small additions of other acids or bases.
In the given scenario, the mixture of \(H_2PO_4^-\) and \(Na_2HPO_4\) acts as a buffer solution. The \(H_2PO_4^-\) acts as the weak acid, and \(Na_2HPO_4\) serves as the conjugate base. This blend is effective in maintaining a constant pH, especially around the pKa of the weak acid involved in the buffer system.

• Buffers are crucial for biological systems where enzymes function optimally within narrow pH ranges.
• A well-known application is blood, which contains buffer systems to maintain pH between 7.35 and 7.45.

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is a simple formula that allows us to calculate the pH of a buffer solution. It relates the pH of the solution to the pKa of the acid component and the ratio of the concentrations of the conjugate base and the acid.
The equation is given by:
\[pH = pK_a + \log{\frac{[A^-]}{[HA]}}\]
Where:

• \(pK_a\) is the negative logarithm of the acid dissociation constant (Ka) for the acid.
• \([A^-]\) is the concentration of the conjugate base.
• \([HA]\) is the concentration of the weak acid.

This equation is particularly useful for buffer solutions because it allows us to understand and predict how the pH will change with varying ratios of \(A^-\) and \(HA\). The exercise involves using this equation to calculate the pH of the buffer solution created from mixing \(H_2PO_4^-\) and \(Na_2HPO_4\). The small change in this ratio affects the buffer's effectiveness in maintaining pH stability.

Ionization Constant

An ionization constant or acid dissociation constant (Ka) measures the strength of an acid in solution. It represents the extent to which an acid ionizes in water, producing hydrogen ions.

The pKa is derived from Ka and is defined as:\[pK_a = -\log(Ka)\]
Smaller values of Ka (or larger values of pKa) indicate weaker acids. This is critical for buffer solutions, as the stability of the pH in such systems hinges on the equilibrium position governed by the ionization constants of the involved acids.

• In the exercise, the second ionization constant of phosphoric acid \(pK_{a2}\) is used, which is around 7.21.
• This value was chosen because the buffer solution is primarily the reaction of \(H_2PO_4^-\) and \(Na_2HPO_4\).

Understanding ionization constants is fundamental for predicting and calculating the equilibrium conditions in chemical solutions.

Limiting Reactant

The concept of the limiting reactant is key in chemical reactions. This is the substance that is completely consumed first and limits the amount of product formed as well as the extent of the reaction.

In the exercise, comparing the available moles of NaOH with phosphoric acid \(H_3PO_4\) and sodium hydrogen phosphate \(Na_2HPO_4\) helped determine that NaOH is the limiting reactant. This conclusion arises because the total moles of NaOH required to fully react with both acids exceed the initial moles of NaOH present.
This concept is essential in stoichiometry for:

• Calculating theoretical yields of products.
• Determining the concentrations of reactants and products after completion of the reaction.

Understanding limiting reactants helps prevent errors in predicting the outcomes of chemical reactions, which is crucial for efficient laboratory and industrial processes.