# Buffers ## Buffer solutions - A **buffer solution** is a solution which resists changes in pH when small amounts of acids or alkalis are added - A buffer solution is used to keep the pH almost constant by maintaining an almost constant concentration of hydrogen and hydroxide ions in a solution

A buffer solution minimises pH changes on addition of small amounts of an acid or a base

### Demonstration of buffer action - To illustrate how effectively a buffer resists pH change, we can compare what happens when acid is added to a buffer solution versus when it's added to pure water:

- Notice how the **buffer solution** maintains a fairly constant pH (only drops from 4.76 to 4.57) - In contrast, the **water** experiences a dramatic pH drop (from 7.0 to 2.3) with the same amount of acid added - This demonstrates the effectiveness of buffer solutions in resisting pH changes ## Acidic buffers - Acidic buffers are made from a **weak acid** and a **soluble salt of the acid** - A common example is an aqueous mixture of ethanoic acid and sodium ethanoate - In solution, ethanoic acid **partially ionises**: $$\text{CH}_3\text{COOH (aq)} \rightleftharpoons \text{H}^+ \text{(aq)} + \text{CH}_3\text{COO}^- \text{(aq)}$$ ethanoic acid $\rightleftharpoons$ ethanoate high conc $\rightleftharpoons$ low conc - Sodium ethanoate is a salt which **fully dissociates** in solution: $$\text{CH}_3\text{COONa + aq} \rightarrow \text{Na}^+ \text{(aq)} + \text{CH}_3\text{COO}^- \text{(aq)}$$ sodium ethanoate $\rightarrow$ ethanoate ion low conc. $\rightarrow$ high conc. - This results in a buffer solution with high concentrations of $\text{CH}_3\text{COOH}$ and $\text{CH}_3\text{COO}^-$ - The establishes the **equilibrium** with hydrogen and ethanoate ions: $$\text{CH}_3\text{COOH (aq)} \rightleftharpoons \text{H}^+ \text{(aq)} + \text{CH}_3\text{COO}^- \text{(aq)}$$ high conc. high conc.

The buffer contains a reservoir of both the weak acid and its conjugate base, which can react with added $\text{H}^+$ or $\text{OH}^-$ to restore balance

## When $\text{H}^+$ ions (acid) are added: - The extra $\text{H}^+$ ions react with $\text{CH}_3\text{COO}^-$ to form $\text{CH}_3\text{COOH}$ - The equilibrium shifts **left** to reduce $[\text{H}^+]$ - This limits any drop in pH - Since there's a large supply of $\text{CH}_3\text{COO}^-$, its concentration doesn't change much - So, the pH remains almost constant **ETHANOATE IONS IN THE BUFFER SOLUTION REACT WITH THE ADDED $\text{H}^+$ IONS TO PREVENT THE pH FROM DECREASING** $$\text{CH}_3\text{COO}^- \text{(aq)} + \text{H}^+ \text{(aq)} \rightleftharpoons \text{CH}_3\text{COOH(aq)}$$ *Ethanoate ions react with added $\text{H}^+$ ions to prevent the pH from decreasing* ## When $\text{OH}^-$ ions (base) are added: - The $\text{OH}^-$ ions react with $\text{H}^+$ to form water: $$\text{OH}^- \text{(aq)} + \text{H}^+ \text{(aq)} \rightarrow \text{H}_2\text{O (l)}$$ - The $[\text{H}^+]$ decreases, shifting the equilibrium **right** $$\text{CH}_3\text{COOH (aq)} \rightarrow \text{H}^+ \text{(aq)} + \text{CH}_3\text{COO}^- \text{(aq)}$$ - More $\text{CH}_3\text{COOH}$ dissociates to replace $\text{H}^+$ ions - So, the pH stays nearly constant due to the buffer action **EQUILIBRIUM MOVES TO REPLACE THE REMOVED HYDROGEN IONS** *Hydroxide ions remove $\text{H}^+$ ions; equilibrium shifts to replace them* ## Calculating the pH of a buffer solution - The pH of a **buffer solution** can be calculated using: - The $K_a$ of the **weak acid** - The **equilibrium concentration** of the weak acid and its conjugate base (salt) - To determine the pH, the concentration of hydrogen ions is needed which can be found using the equilibrium expression: $$K_a = \frac{[\text{salt}][\text{H}^+]}{[\text{acid}]}$$ - This can be rearranged to give: $$[\text{H}^+] = K_a \times \frac{[\text{acid}]}{[\text{salt}]}$$ - To simplify the calculations, **logarithms** are used such that the expression becomes: $$-\log_{10}[\text{H}^+] = -\log_{10}K_a \times -\log_{10} \frac{[\text{acid}]}{[\text{salt}]}$$ - Since $-\log_{10}[\text{H}^+] = \text{pH}$, the expression can also be rewritten as: $$\text{pH} = \text{p}K_a + \log_{10} \frac{[\text{salt}]}{[\text{acid}]}$$ - This is known as the **Henderson-Hasselbalch equation**

Use $[\text{H}^+] = K_a \times \frac{[\text{acid}]}{[\text{salt}]}$ then calculate pH from $[\text{H}^+]$

### Worked Example Calculate the pH of a buffer solution containing 0.305 mol dm$^{-3}$ of ethanoic acid and 0.520 mol dm$^{-3}$ sodium ethanoate. The $K_a$ of ethanoic acid = 1.74 × 10$^{-5}$ mol dm$^{-3}$ **Answer:** Ethanoic acid is a weak acid that ionises as follows: $$\text{CH}_3\text{COOH (aq)} \rightleftharpoons \text{H}^+ \text{(aq)} + \text{CH}_3\text{COO}^- \text{(aq)}$$ **Step 1:** Write down the equilibrium expression to find $K_a$ $$K_a = \frac{[\text{CH}_3\text{COO}^-][\text{H}^+]}{[\text{CH}_3\text{COOH}]}$$ **Step 2:** Rearrange the equation to find $[\text{H}^+]$ $$[\text{H}^+] = K_a \times \frac{[\text{CH}_3\text{COOH}]}{[\text{CH}_3\text{COO}^-]}$$ **Step 3:** Substitute the values into the expression $$[\text{H}^+] = 1.74 \times \frac{[0.305]}{[0.520]}$$ $$[\text{H}^+] = 1.02 \times 10^{-5} \text{ mol dm}^{-3}$$ **Step 4:** Calculate the pH $$\text{pH} = -\log[\text{H}^+]$$ $$\text{pH} = -\log(1.02 \times 10^{-5})$$ $$\text{pH} = 4.99$$ ## Uses of buffers ### Controlling the pH of blood - In humans, $\text{HCO}_3^-$ ions act as a buffer to keep the blood pH between 7.35 and 7.45 - Body cells produce $\text{CO}_2$ during **aerobic respiration** - This $\text{CO}_2$ will combine with water in blood to form a solution containing $\text{H}^+$ ions $$\text{CO}_2 \text{(g)} + \text{H}_2\text{O (l)} \rightleftharpoons \text{H}^+ \text{(aq)} + \text{HCO}_3^- \text{(aq)}$$ - This equilibrium between $\text{CO}_2$ and $\text{HCO}_3^-$ is extremely important - If the concentration of $\text{H}^+$ ions is not regulated, the blood pH would drop and cause '**acidosis**' - **Acidosis** refers to a condition in which there is too much acid in the body fluids such as blood - This could cause body malfunctioning and eventually lead to coma

Blood pH must be maintained between 7.35 and 7.45 to prevent acidosis

**If there is an increase in $\text{H}^+$ ions** - The equilibrium position shifts to the **left** until equilibrium is restored $$\text{CO}_2 \text{(g)} + \text{H}_2\text{O (l)} \rightleftharpoons \text{H}^+ \text{(aq)} + \text{HCO}_3^- \text{(aq)}$$ - This reduces the concentration of $\text{H}^+$ and keeps the pH of the blood **constant** **If there is a decrease in $\text{H}^+$ ions** - The equilibrium position shifts to the **right** until equilibrium is restored $$\text{CO}_2 \text{(g)} + \text{H}_2\text{O (l)} \rightleftharpoons \text{H}^+ \text{(aq)} + \text{HCO}_3^- \text{(aq)}$$ - This increases the concentration of $\text{H}^+$ and keeps the pH of the blood **constant**