Problem 23 - Chemistry: The Central Science in SI Units, Global Edition

Question

A buffer is prepared by adding \(15.0 \mathrm{~g}\) of sodium acetate \(\left(\mathrm{CH}_{3} \mathrm{COONa}\right)\) to \(500 \mathrm{~mL}\) of a \(0.100 \mathrm{M}\) acetic acid \(\left(\mathrm{CH}_{3} \mathrm{COOH}\right)\) solution. (a) Determine the pH of the buffer. (b) Write the complete ionic equation for the reaction that occurs when a few drops of nitric acid are added to the buffer. (c) Write the complete ionic equation for the reaction that occurs when a few drops of potassium hydroxide solution are added to the buffer.

Step-by-Step Solution

Answer

(a) The pH of the buffer is 6.10. (b) The net ionic equation for the reaction with nitric acid is: \(CH_3COO^-(aq) + H^+(aq) \rightarrow CH_3COOH(aq)\). (c) The net ionic equation for the reaction with potassium hydroxide is: \(CH_3COOH(aq) + OH^-(aq) \rightarrow CH_3COO^-(aq) + H_2O(l)\).

Step 1: (a) Calculate the amount of sodium acetate)

First, we need to find the amount of sodium acetate in moles. To do that, we can use the formula: moles = mass / molar mass. The molar mass of sodium acetate (CH3COONa) is: \(12.01g/mol(C) * 2 + 1.01g/mol(H) * 3 + 15.99g/mol(O) * 2 + 22.99g/mol(Na) = 82.03g/mol\) Given mass of sodium acetate = \(15.0 g\) So, moles of sodium acetate = \(15.0g / 82.03 g/mol = 0.183 mol\)

Step 2: (a) Calculate the concentration of sodium acetate)

Now, we calculate the concentration of sodium acetate by dividing the moles by the volume of the solution. Given volume of acetic acid solution = 500 mL Concentration of sodium acetate = \(\frac{0.183 mol}{0.500 L} = 0.366 M\)

Step 3: (a) Find the pKa of acetic acid)

We know that the \(K_a\) value for acetic acid is \(1.8 * 10 ^{-5}\). We can calculate the pKa value using the formula: pKa = -log \(K_a\). pKa = -log(1.8 * 10 ^{-5}) = 4.74

Step 4: (a) Determine the pH of the buffer)

Now we can calculate the pH of the buffer using the Henderson-Hasselbalch equation: pH = pKa + log \(\frac{[CH_3COO^-]}{[CH_3COOH]}\) pH = 4.74 + log \(\frac{0.366}{0.100}\) pH = 4.74 + 1.36 = \(6.10\) So, the pH of the buffer is 6.10.

Step 5: (b) Reaction with nitric acid)

When nitric acid (HNO3) is added to the buffer, it reacts with the acetate ions: \(CH_3COO^-(aq) + HNO_3(aq) \rightarrow CH_3COOH(aq) + NO_3^-(aq)\)

Step 6: (b) Complete ionic equation with nitric acid)

The complete ionic equation for the reaction above is: \(CH_3COO^-(aq) + H^+(aq) + NO_3^-(aq) \rightarrow CH_3COOH(aq) + NO_3^-(aq)\) NO3- ions are not involved in the reaction, so they can be considered as spectator ions. We can remove them to get the net ionic equation: \(CH_3COO^-(aq) + H^+(aq) \rightarrow CH_3COOH(aq)\)

Step 7: (c) Reaction with potassium hydroxide)

When potassium hydroxide (KOH) is added to the buffer, it reacts with the acetic acid: \(CH_3COOH(aq) + KOH(aq) \rightarrow CH_3COOK(aq) + H_2O(l)\)

Step 8: (c) Complete ionic equation with potassium hydroxide)

The complete ionic equation for the reaction above is: \(CH_3COOH(aq) + K^+(aq) + OH^-(aq) \rightarrow CH_3COO^-(aq) + K^+(aq) + H_2O(l)\) K+ ions are not involved in the reaction, so they can be considered as spectator ions. We can remove them to get the net ionic equation: \(CH_3COOH(aq) + OH^-(aq) \rightarrow CH_3COO^-(aq) + H_2O(l)\)

Key Concepts

Understanding the Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is pivotal in understanding buffer solutions. It is a mathematical formula that links pH, pKa, and the concentrations of an acid and its conjugate base. Here's the formula:

\( \text{pH} = \text{pKa} + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \)

Where:

• \([\text{A}^-]\) is the concentration of the conjugate base (e.g., acetate \(\text{CH}_3\text{COO}^-\)).
• \([\text{HA}]\) is the concentration of the weak acid (e.g., acetic acid \(\text{CH}_3\text{COOH}\)).
• pKa is the negative logarithm of the acid dissociation constant \(K_a\).

Using this equation, we can find the pH of the buffer solution. In our example, when you plug in the acetic acid and acetate concentrations along with the pKa of 4.74, you find a pH of 6.10. This shows that the solution can resist changes in pH, a crucial property of buffer solutions.

Ionic Equations in Buffer Reactions

Ionic equations help us understand what happens at the molecular level when acids or bases are added to a buffer. Take the buffer containing acetic acid and acetate ions:

• When a strong acid like nitric acid \(\text{HNO}_3\) is added, it donates protons \(\text{H}^+\). The protons react with acetate ions \(\text{CH}_3\text{COO}^-\) to form acetic acid.
• The ionic equation is: \(\text{CH}_3\text{COO}^-(\text{aq}) + \text{H}^+(\text{aq}) \rightarrow \text{CH}_3\text{COOH}(\text{aq})\).
• The nitrate ions \(\text{NO}_3^-\) are spectator ions and don't participate in the actual reaction.
• When a base like potassium hydroxide \(\text{KOH}\) is added, the hydroxide ions \(\text{OH}^-\) react with acetic acid to regenerate acetate ions and water.
• The ionic equation for this is: \(\text{CH}_3\text{COOH}(\text{aq}) + \text{OH}^-(\text{aq}) \rightarrow \text{CH}_3\text{COO}^-(\text{aq}) + \text{H}_2\text{O}(\text{l})\).
• Potassium ions \(\text{K}^+\) are spectator ions here.

By understanding these equations, we can see how buffers neutralize small amounts of added acids or bases.

Calculating pH in Buffer Solutions

Calculating the pH in buffer solutions involves a few key steps:

• First, determine the moles of each component in the buffer. For example, the mass of sodium acetate can be converted to moles using its molar mass.
• Next, find the concentrations by dividing the moles by the solution's volume in liters. This gives you the concentrations needed for the Henderson-Hasselbalch equation.
• Now, calculate the pH using the equation: \(\text{pH} = \text{pKa} + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \). Substitute the known values for the conjugate base and weak acid.
• In our exercise, using the concentration of acetate and acetic acid, and a pKa of 4.74, the calculated pH is 6.10.

This step-by-step process shows how buffers work to maintain stable pH levels, illustrating their importance in both chemistry and biological systems.