# 9.8 Buffer Solutions and Their Applications A **buffer solution** is a solution that resists significant changes in pH upon the addition of a small amount of acid or base. Such solutions maintain a relatively constant pH, even over long periods. ## 9.8.1 Types of Buffer Solutions There are two primary types of buffer solutions: 1. **Acidic Buffers:** These are formed by mixing a weak acid with a salt of that acid and a strong base. These buffers have a pH value of less than 7. * **Example:** A solution of acetic acid ($\mathrm{CH_3COOH}$) and sodium acetate ($\mathrm{CH_3COONa}$). 2. **Basic Buffers:** These are formed by mixing a weak base with a salt of that base and a strong acid. These buffers have a pH value greater than 7. * **Example:** A solution of ammonium hydroxide ($\mathrm{NH_4OH}$) and ammonium chloride ($\mathrm{NH_4Cl}$). ## 9.8.2 Calculation of pH of Buffer Solutions The pH of a buffer solution can be calculated using the **Henderson-Hasselbalch equation**. This equation relates the pH (or pOH) of a solution to the acid dissociation constant ($pK_a$) or base dissociation constant ($pK_b$) and the ratio of the concentrations of the conjugate acid-base pair. The concentration of the conjugate base (for acidic buffers) or conjugate acid (for basic buffers) is primarily supplied by the salt, which is a strong electrolyte. Therefore, we can approximate the concentration of the conjugate species to be equal to the concentration of the salt. ### For Acidic Buffers The Henderson-Hasselbalch equation is: $$ \text{pH} = \text{pK}_{\text{a}} + \log \frac{[\text{salt}]}{[\text{acid}]} $$ ### For Basic Buffers The Henderson-Hasselbalch equation is: $$ \text{pOH} = \text{pK}_{\text{b}} + \log \frac{[\text{salt}]}{[\text{base}]} $$ ## Worked Examples ### Example 9.6 What is the pH of a buffer if the concentration of $\mathrm{CH_3COOH}$ is 0.1 M and $\mathrm{CH_3COONa}$ is 1.0 M? The $pK_a$ for $\mathrm{CH_3COOH}$ is 4.76. #### Solution 1. **Write the given values:** * Concentration of acid, $[\mathrm{CH_3COOH}] = 0.1 \, \text{M}$ * Concentration of salt, $[\mathrm{CH_3COONa}] = 1.0 \, \text{M}$ * $pK_a = 4.76$ 2. **Apply the formula (Henderson-Hasselbalch equation for an acidic buffer):** $$ \text{pH} = \text{pK}_{\text{a}} + \log \frac{[\text{salt}]}{[\text{acid}]} $$ 3. **Show the calculation:** $$ \text{pH} = 4.76 + \log \frac{[1.0]}{[0.1]} $$ $$ \text{pH} = 4.76 + \log(10) $$ Since $\log(10) = 1$: $$ \text{pH} = 4.76 + 1.00 $$ $$ \text{pH} = 5.76 $$ ## Possible Questions/Answers - **Q:** Calculate the pH of a buffer solution in which 0.11 M $\mathrm{CH_3COONa}$ and 0.09 M $\mathrm{CH_3COOH}$ solutions are present. The $K_a$ for $\mathrm{CH_3COOH}$ is $1.8 \times 10^{-5}$. - **A:** 1. **Given:** * $[\mathrm{CH_3COOH}] = 0.09 \, \text{M}$ * $[\mathrm{CH_3COONa}] = 0.11 \, \text{M}$ * $K_a = 1.8 \times 10^{-5}$ 2. **First, calculate the $pK_a$ from the given $K_a$:** $$ pK_a = -\log(K_a) $$ $$ pK_a = -\log(1.8 \times 10^{-5}) $$ $$ pK_a = 4.74 $$ 3. **Apply the Henderson-Hasselbalch equation:** $$ \text{pH} = \text{pK}_{\text{a}} + \log \frac{[\text{salt}]}{[\text{acid}]} $$ $$ \text{pH} = 4.74 + \log \frac{[0.11]}{[0.09]} $$ $$ \text{pH} = 4.74 + \log(1.222) $$ $$ \text{pH} = 4.74 + 0.087 $$ $$ \text{pH} \approx 4.83 $$ ## Buffers in Biological Systems Buffers are crucial for maintaining stable pH levels in biological systems, which is essential for the proper functioning of enzymes and other biochemical processes. | Biological System | pH Range | | :---------------- | :----------- | | Human blood | 7.35 to 7.45 | | Tears | 7.40 | | Stomach | 1.65 to 1.75 | | Milk | 6.7 to 6.8 | | Egg white (protein) | 8.0 to 8.1 |

9.8 Buffer Solutions And Their Applications | Acids Bases Chemistry | Chemistry 11