Problem 11 - Chemical Principles

Question

A buffer solution of volume \(100.0 \mathrm{~mL}\) is \(0.100 \mathrm{M}\) \(\mathrm{CH}_{3} \mathrm{COOH}(\mathrm{aq})\) and \(0.100 \mathrm{M} \mathrm{NaCH} \mathrm{CO}_{2}(\mathrm{aq})\). (a) What are the \(\mathrm{pH}\) and the pH change resulting from the addition of \(10.0 \mathrm{~mL}\) of \(0.950 \mathrm{M} \mathrm{NaOH}(\mathrm{aq})\) to the buffer solution? (b) What are the \(\mathrm{pH}\) and the \(\mathrm{pH}\) change resulting from the addition of \(20.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{HNO}_{3}(\mathrm{aq})\) to the initial buffer solution?

Step-by-Step Solution

Answer

The pH after adding NaOH is 6.351 with a change of 1.591. After adding HNO3, the pH is 4.584 with a change of -0.176.

Step 1: Calculate the initial amounts of acetic acid and acetate

Since both CH3COOH and NaCH3CO2 are 0.100 M in a 100 mL solution, to find the number of moles we use the formula: moles = concentration × volume (in liters). For CH3COOH: moles_acid = 0.100 M × 0.100 L = 0.010 moles. For NaCH3CO2: moles_base = 0.100 M × 0.100 L = 0.010 moles.

Step 2: Calculate the moles of NaOH added

Use the formula, moles = concentration × volume (in liters), to find the moles of NaOH. moles_NaOH = 0.950 M × 0.010 L = 0.0095 moles.

Step 3: Determine the new concentrations after the addition of NaOH

The added NaOH will react with CH3COOH to form NaCH3CO2 and water. moles_acid_remaining = 0.010 moles_acid - 0.0095 moles_NaOH = 0.0005 moles. moles_base_formed = 0.010 moles_base + 0.0095 moles_NaOH = 0.0195 moles. The total volume after adding NaOH is 100 mL + 10 mL = 110 mL or 0.110 L.

Step 4: Calculate the new concentrations of acetic acid and acetate

Concentration_acid = moles_acid_remaining / total_volume = 0.0005 moles / 0.110 L = 0.004545 M. Concentration_base = moles_base_formed / total_volume = 0.0195 moles / 0.110 L = 0.177272 M.

Step 5: Use the Henderson-Hasselbalch equation to calculate the new pH

The Henderson-Hasselbalch equation is pH = pKa + log([base]/[acid]). For acetic acid, pKa is about 4.76. pH_after_NaOH = 4.76 + log(0.177272/0.004545) = 4.76 + log(38.98) = 4.76 + 1.591 = 6.351.

Step 6: Calculate the initial pH of the buffer solution

Using the Henderson-Hasselbalch equation with initial concentrations: initial_pH = 4.76 + log(0.100/0.100) = 4.76 + log(1) = 4.76.

Step 7: Calculate the change in pH after addition of NaOH

Change_in_pH = pH_after_NaOH - initial_pH = 6.351 - 4.76 = 1.591.

Step 8: Calculate the moles of HNO3 added

Using the formula, moles = concentration × volume (in liters), for HNO3: moles_HNO3 = 0.100 M × 0.020 L = 0.002 moles.

Step 9: Determine the new concentrations after the addition of HNO3

The added HNO3 will react with NaCH3CO2 to form CH3COOH and water. moles_base_remaining = 0.010 moles_base - 0.002 moles_HNO3 = 0.008 moles. moles_acid_formed = 0.010 moles_acid + 0.002 moles_HNO3 = 0.012 moles. The total volume after adding HNO3 is 100 mL + 20 mL = 120 mL or 0.120 L.

Step 10: Calculate the new concentrations of acetic acid and acetate after HNO3 addition

Concentration_acid = moles_acid_formed / total_volume = 0.012 moles / 0.120 L = 0.100 M. Concentration_base = moles_base_remaining / total_volume = 0.008 moles / 0.120 L = 0.066667 M.

Step 11: Use the Henderson-Hasselbalch equation to calculate the new pH after HNO3 addition

pH_after_HNO3 = 4.76 + log(0.066667/0.100) = 4.76 + log(0.66667) = 4.76 - 0.176 = 4.584.

Step 12: Calculate the change in pH after addition of HNO3

Change_in_pH = pH_after_HNO3 - initial_pH = 4.584 - 4.76 = -0.176.

Key Concepts

pH Calculation

Understanding the pH calculation is crucial for students studying chemistry, especially when dealing with solutions that resist changes in pH, known as buffers. pH is a measure of the acidity or basicity of a solution and is calculated as the negative logarithm of the hydrogen ion concentration.

For a buffer solution, consisting of a weak acid and its conjugate base, the pH is not easily changed by the addition of small amounts of strong acid or base. This property is essential for many biological and chemical systems that require a stable pH. When you add NaOH, a strong base, to the buffer solution, it reacts with the weak acid, in this case, acetic acid, neutralizing some of it and forming more of the conjugate base, acetate.

After the reaction, we must recalculate the moles of acid and conjugate base to determine the new concentrations. From there, we use these concentrations to find the new pH value. In the exercise, the addition of NaOH led to the pH increasing from 4.76 to 6.351, indicating that the solution became less acidic (more basic).

Conversely, adding a strong acid like HNO3 would decrease the concentration of the conjugate base and increase the concentration of the weak acid, thereby lowering the pH, as reflected in the change from 4.76 to 4.584. These calculations are essential for predicting how a buffer will behave when subjected to changes and are fundamental in various applications, from industrial processes to physiological systems.

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is a mathematical formula that relates the pH of a buffer solution to the pKa of the weak acid and the ratio of the concentrations of the conjugate base and the weak acid. Derived from the equilibrium constant expression for an acid-base reaction, this equation is given by:

\[\begin{equation} pH = pKa + \log \left( \frac{[\text{base}]}{[\text{acid}]} \right) \end{equation}\] Where:

• pKa is the acid dissociation constant, a measure of the strength of the weak acid.
• [base] is the molarity of the conjugate base.
• [acid] is the molarity of the weak acid.

In our example, the pKa value for acetic acid is 4.76, reflecting its relatively weak acidic nature. By inserting the molarities of acetate and acetic acid into the Henderson-Hasselbalch equation, we can calculate the pH of the buffer before and after adding other chemicals.

This equation is particularly powerful because it allows us to anticipate how the pH will change with the introduction of additional acid or base, making it a vital tool for anyone working with buffer solutions in a laboratory or industrial setting.

Acid-Base Reaction

Acid-base reactions are at the heart of understanding buffer solutions and their behavior. These reactions are exchanges where an acid donates a proton (H+ ion) to a base. In the context of buffers, we often work with a weak acid and its conjugate base.

A buffer effectively minimizes pH changes upon the addition of an acid or base due to the presence of species that can 'soak up' added H+ or OH- ions. When NaOH, a strong base, is added to the buffer, it reacts with the acetic acid to form sodium acetate and water in a neutralization reaction.

\[\begin{equation} CH_3COOH + NaOH \rightarrow NaCH_3CO_2 + H_2O \end{equation}\] Similarly, when we add HNO3, a strong acid, to the buffer, it reacts with the sodium acetate, forming more acetic acid and water. This is an example of an acid-base reaction where the conjugate base is converted back to the weak acid:

\[\begin{equation} NaCH_3CO_2 + HNO_3 \rightarrow CH_3COOH + NaNO_3 \end{equation}\] These reactions demonstrate the buffer's ability to adapt and maintain its pH within a narrow range, even in the presence of strong acids or bases. Understanding these concepts is vital for students, as they apply not only in chemistry but also in biological systems where maintaining a stable pH is crucial for proper function.