What Is The PH Of A Solution That Results From Mixing 25.0 ML Of 0.200 M HA With 12.5 ML Of 0.400 M NaOH?

What Is The PH Of A Solution That Results From Mixing 25.0 ML Of 0.200 M HA With 12.5 ML Of 0.400 M NaOH?

Understanding the pH of a solution resulting from mixing a weak acid with a strong base is a fundamental concept in chemistry, particularly in acid-base titrations and buffer solutions. In this scenario, we are dealing with the weak acid hydroxyapatite (HA) and sodium hydroxide (NaOH), a strong base. Calculating the final pH involves determining the initial amounts of each reactant, understanding their reaction, and applying principles of stoichiometry and equilibrium chemistry. This article explores these steps in detail to find the pH of the resulting solution.

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Background Concepts

Understanding pH and pOH

• pH is a measure of the acidity or alkalinity of a solution, defined as:

pH = -log[H⁺]

• pOH relates to the hydroxide ion concentration:

pOH = -log[OH⁻]

• The relationship between pH and pOH at 25°C is:

pH + pOH = 14

Weak Acid and Strong Base Reactions

• Hydroxyapatite (HA) is a weak acid, which means it does not fully dissociate in water.
• Sodium hydroxide (NaOH) is a strong base and dissociates completely.
• When mixed, NaOH will react with HA, neutralizing some of it and possibly leading to a solution that is either acidic, basic, or neutral depending on the amounts.

Stoichiometry in Acid-Base Reactions

• Key to calculating the final pH is understanding how much of each reactant is present and how they react.
• The reaction between HA and NaOH can be represented (assuming HA is a monoprotic acid for simplicity):

HA + OH⁻ → A⁻ + H₂O

• The amount of HA neutralized depends on the molar amounts of NaOH added.

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Step 1: Calculate Initial Moles of Reactants

Calculating Moles of HA

• Volume of HA solution = 25.0 mL = 0.025 L
• Concentration of HA = 0.200 M

Number of moles of HA: \[ \text{moles HA} = \text{volume} \times \text{concentration} = 0.025\,L \times 0.200\, \text{mol/L} = 0.005\, \text{mol} \]

Calculating Moles of NaOH

• Volume of NaOH solution = 12.5 mL = 0.0125 L
• Concentration of NaOH = 0.400 M

Number of moles of NaOH: \[ \text{moles NaOH} = 0.0125\,L \times 0.400\, \text{mol/L} = 0.005\, \text{mol} \]

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Step 2: Analyze the Reaction

Reaction Stoichiometry

• Since both reactants have 0.005 mol, and assuming a 1:1 reaction ratio:

\[ \text{HA} + \text{OH}^- \rightarrow \text{A}^- + \text{H}_2\text{O} \]

• The entire amount of HA will be neutralized by the NaOH present.

Extent of Reaction

• Moles of HA initially: 0.005 mol
• Moles of NaOH initially: 0.005 mol

Since they are equal, all HA will react with NaOH, producing:

• 0.005 mol of A⁻ (the conjugate base)
• Water

Remaining reactants:

• None, as all HA is neutralized.

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Step 3: Determine the Composition of the Final Solution

Final Volume of the Mixture

• Total volume = 25.0 mL + 12.5 mL = 37.5 mL = 0.0375 L

Remaining Species in Solution

• Since all HA has reacted, the solution contains:
• 0 mol of HA
• 0 mol of free NaOH (fully reacted)
• 0.005 mol of A⁻ (the conjugate base formed)
• The solution also contains water and any additional ions from the initial solutions, but these do not affect pH significantly.

Nature of the Final Solution

• The solution is now a mixture of the conjugate base (A⁻) in water.
• Because the strong base fully reacted, the pH depends on the hydrolysis of A⁻.

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Step 4: Determine the pH Based on Hydrolysis of A⁻

Hydrolysis of the Conjugate Base

• A⁻ can undergo hydrolysis:

\[ \text{A}^- + \text{H}_2\text{O} \rightleftharpoons \text{HA} + \text{OH}^- \]

• The equilibrium constant for this reaction is the base dissociation constant \(Kb\), related to the acid dissociation constant \(Ka\) of HA:

\[ Kb = \frac{Kw}{K_a} \]

• \(K_a\) of HA must be known or estimated to proceed.

Estimating \(K_a\) for HA

• The problem does not provide \(Ka\), but typical weak acids have \(Ka\) values ranging from \(10^{-3}\) to \(10^{-7}\).
• For hydroxyapatite, \(K_a\) is approximately \(10^{-3}\) to \(10^{-4}\). Let's assume:

\[ K_a = 1 \times 10^{-4} \]

• Therefore:

\[ Kb = \frac{Kw}{K_a} = \frac{1 \times 10^{-14}}{1 \times 10^{-4}} = 1 \times 10^{-10} \]

Calculating Hydroxide Ion Concentration

• Moles of A⁻ = 0.005 mol
• Final volume = 0.0375 L

Concentration of A⁻: \[ [A^-] = \frac{0.005\, \text{mol}}{0.0375\, \text{L}} \approx 0.133\, \text{M} \]

• Using the hydrolysis equilibrium:

\[ K_b = \frac{[OH^-]^2}{[A^-]} \]

• Solving for \([OH^-]\):

\[ [OH^-] = \sqrt{K_b \times [A^-]} = \sqrt{1 \times 10^{-10} \times 0.133} \]

\[ [OH^-] \approx \sqrt{1.33 \times 10^{-11}} \approx 3.65 \times 10^{-6}\, \text{M} \]

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Step 5: Calculate the Final pH

Determine pOH

\[ pOH = -\log [OH^-] = -\log (3.65 \times 10^{-6}) \approx 5.44 \]

Determine pH

\[ pH = 14 - pOH = 14 - 5.44 = 8.56 \]

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Conclusion

The final pH of the solution, after mixing 25.0 mL of 0.200 M HA with 12.5 mL of 0.400 M NaOH, is approximately 8.56. This indicates a slightly basic solution, which makes sense given the complete neutralization of the weak acid and the hydrolysis of the conjugate base formed. The pH is above 7, confirming the basic nature of the resulting solution due to the presence of hydrolyzed A⁻ ions.

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Additional Considerations

• Assumptions Made: The calculation assumes complete reaction and uses an estimated \(K_a\) for HA. Actual values may vary based on the specific compound.
• Effect of Ionic Strength: In real solutions, ionic strength can influence activity coefficients, slightly affecting the pH.
• Buffer Capacity: Since a weak acid and its conjugate base are present, the solution can act as a buffer, resisting drastic changes in pH upon small additions of acids or bases.

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Summary

• Initial moles of HA: 0.005 mol
• Moles of NaOH: 0.005 mol
• Reaction: Complete neutralization to form conjugate base A⁻
• Final species: A⁻ in water undergoing hydrolysis
• Estimated pH: approximately 8.56

This detailed approach demonstrates how stoichiometry, acid-base equilibria, and hydrolysis reactions