Question

A solution is prepared that is \(0.150 \mathrm{M} \mathrm{CH}_{3} \mathrm{COOH}\) and \(0.250 \mathrm{M} \mathrm{NaHCOO}\) (a) Show that this is a buffer solution. (b) Calculate the pH of this buffer solution. (c) What is the final pH if 1.00 L of 0.100 M HCl is added to \(1.00 \mathrm{L}\) of this buffer solution?

Short answer

The solution is indeed a buffer solution. The original pH of the solution is calculated by using the Henderson-Hasselbalch equation. After the addition of HCl, the final pH is calculated by finding the new concentrations of the base and the acid, and substituting them into the Henderson-Hasselbalch equation.

Step-by-step solution

Prove it's a buffer solution

The components of the solution are CH3COOH (acetic acid, a weak acid) and NaHCOO (sodium acetate, the conjugate base of acetic acid). Since we have a weak acid and its conjugate base, we can conclude that this is indeed a buffer solution.

Calculate original pH

The Henderson-Hasselbalch equation is given by: pH = pKa + log([base]/[acid]) Acetic acid (CH3COOH) has a known pKa of 4.76. The concentrations of the base (NaHCOO) and the acid (CH3COOH) are given as 0.250 M and 0.150 M respectively. Substituting these values, we have the original pH = 4.76 + log(0.250/0.150).

Calculate final pH after addition of HCl

Upon the addition of HCl, it will react with the base NaHCOO present in the buffer solution. For every moles of HCl added, equal moles of NaHCOO get used up and equal moles of CH3COOH get formed. Since 1.00 L of 0.100 M HCl is added, 0.100 mol of NaHCOO will be used up and 0.1 mol of CH3COOH will be formed. We can calculate the new concentrations of NaHCOO and CH3COOH and substitute them into the Henderson-Hasselbalch equation to find the final pH.

Key concepts

Henderson-Hasselbalch Equation

Understanding the Henderson-Hasselbalch equation is key to grasping pH calculations in buffer solutions. This equation is a simplified method to calculate the pH of a buffer solution, which is a mixture of a weak acid and its conjugate base or a weak base and its conjugate acid. The Henderson-Hasselbalch equation is written as:
\[ \text{pH} = \text{pKa} + \text{log}\frac{\text{[conjugate base]}}{\text{[weak acid]}} \]

In this formula, pKa is the acid dissociation constant, which is a measure of the strength of the acid. The ratio of the concentrations of the conjugate base and the weak acid provides insight into the balance between the acidic and basic components of the buffer. The logarithm in the equation suggests that even small changes in the ratios of these components can result in significant pH shifts. For buffer solutions, the closer the pH is to the pKa value of the weak acid, the more resistant the solution is to changes in pH upon addition of strong acids or bases.

Acetic Acid and Sodium Acetate Buffer

An acetic acid and sodium acetate buffer system is a classical example of a buffer solution utilized in laboratories. Acetic acid \( (CH_3COOH) \) is a weak acid, meaning it only partially dissociates into ions in water. Its conjugate base is acetate \( (CH_3COO^-) \), which can be provided by a salt like sodium acetate \( (NaCH_3COO) \). These two components in solution form a buffer because the acetate ions can react with any added strong acid \( (H^+) \), transforming into acetic acid, thereby maintaining the acidity of the solution within a narrow range. Likewise, any added strong base \( (OH^-) \) would be neutralized by the acetic acid in a similar manner. The effectiveness of the buffer relies on the presence of significant amounts of both the weak acid and its conjugate base.

pH Calculation of a Buffer Solution

To calculate the pH of a buffer solution, you would commonly use the Henderson-Hasselbalch equation. The calculation involves knowing the molar concentrations of the weak acid and its conjugate base as well as the pKa of the weak acid. With the provided acetic acid (with a pKa of 4.76) and sodium acetate buffer, it works as follows:
First, identify the concentrations of acetic acid and sodium acetate. Then, you apply the Henderson-Hasselbalch equation:
\[ \text{pH} = 4.76 + \text{log}\frac{0.250}{0.150} \]

This application yields the pH of the buffer before any additional substances are added. The successful application of this equation also demonstrates why maintaining proper proportions between the components of the buffer is crucial for its stability.

Adding HCl to a Buffer

When a strong acid like hydrochloric acid \( (HCl) \) is added to a buffer solution, the acidic component (HCl) will react primarily with the conjugate base (acetate ions) present in the buffer. The chemical reaction between \( HCl \) and \( CH_3COO^- \) yields \( CH_3COOH \) and \( NaCl \). The beautiful aspect of a buffer solution is its ability to absorb such shocks to its pH by neutralizing the added acid.
For the given problem, when 1.00 L of 0.100 M HCl is added, the buffer will use 0.100 moles of sodium acetate to form 0.100 moles of acetic acid. The Henderson-Hasselbalch equation is then used once again to calculate the new pH, based on the updated concentrations of sodium acetate and acetic acid. Proper calculation after such an addition shows how the pH of the buffer changes, but only slightly, demonstrating the strength of the buffer's resistance to pH variations.