Question

A buffer contains 0.10 mol of acetic acid and 0.13 mol of sodium acetate in 1.00 $\mathrm{L}$ (a) What is the pH of this buffer? (b) What is the pH of the buffer after the addition of 0.02 mol of $\mathrm{KOH}$ ? (c) What is the $\mathrm{pH}$ of the buffer after the addition of 0.02 $\mathrm{mol}$ of ${\mathrm{HNO}}_3?$

Short answer

The pH of the initial buffer solution is 4.92. After adding 0.02 mol of KOH, the pH increases to 5.24. When 0.02 mol of HNO3 is added, the pH decreases to 4.67.

Step-by-step solution

Understand the components of the buffer solution

The buffer solution contains acetic acid and sodium acetate. Acetic acid, $C{H}_{3}COOH$, is a weak acid and will create an equilibrium with its conjugate base, acetate ($C{H}_{3}CO{O}^{-}$). Sodium acetate is the sodium salt of the acetate ion, which will increase the amount of $C{H}_{3}CO{O}^{-}$ in the solution. This forms the buffer solution, which maintains its pH despite the addition of small amounts of acidic or basic substances.

Calculate the pH of the buffer using the Henderson-Hasselbalch equation

The pH of the buffer solution can be calculated using the Henderson-Hasselbalch equation: $$pH=p{K}_a+{\log }_{10}\frac{[A^{-}]}{[HA]}$$ In this case, $HA$ represents acetic acid ($C{H}_{3}COOH$) and $A^{-}$ represents acetate ($C{H}_{3}CO{O}^{-}$). The $p{K}_a$ of acetic acid is 4.74. Using the given amounts of acetic acid and sodium acetate, we find the concentrations: $$[C{H}_{3}COOH]=\frac{0.10mol}{1.00L}=0.10M$$ $$[C{H}_{3}CO{O}^{-}]=\frac{0.13mol}{1.00L}=0.13M$$ Now, we can plug these values into the Henderson-Hasselbalch equation to calculate the pH of the buffer: $$pH=4.74+{\log }_{10}\frac{0.13}{0.10}=4.92$$

Calculate the pH after adding 0.02 mol of KOH

KOH is a strong base, which will react completely with the acetic acid, forming water and more acetate: $$C{H}_{3}COOH+O{H}^{-}⟶C{H}_{3}CO{O}^{-}+H_{2}O$$ In this case, 0.02 mol of KOH was added, meaning 0.02 mol of acetic acid will react, forming 0.02 mol of acetate. After adding KOH: $$[C{H}_{3}COOH]=0.10mol-0.02mol=0.08mol$$ $$[C{H}_{3}CO{O}^{-}]=0.13mol+0.02mol=0.15mol$$ Now, plug these values into the Henderson-Hasselbalch equation: $$pH=4.74+{\log }_{10}\frac{0.15}{0.08}=5.24$$

Calculate the pH after adding 0.02 mol of HNO3

HNO3 is a strong acid, which will react completely with the acetate, forming water and acetic acid: $$C{H}_{3}CO{O}^{-}+H^{+}⟶C{H}_{3}COOH$$ In this case, 0.02 mol of HNO3 was added, meaning 0.02 mol of acetate will react, forming 0.02 mol of acetic acid. After adding HNO3: $$[C{H}_{3}COOH]=0.10mol+0.02mol=0.12mol$$ $$[C{H}_{3}CO{O}^{-}]=0.13mol-0.02mol=0.11mol$$ Now, plug these values into the Henderson-Hasselbalch equation: $$pH=4.74+{\log }_{10}\frac{0.11}{0.12}=4.67$$ In conclusion, the pH of the initial buffer is 4.92, the pH after adding 0.02 mol of KOH is 5.24, and the pH after adding 0.02 mol of HNO3 is 4.67.

Key concepts

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is a crucial tool for calculating the pH of buffer solutions. It helps you determine how the solution resists changes in pH with the addition of acids or bases. The equation is presented as: $$pH=p{K}_a+{\log }_{10}\frac{[A^{-}]}{[HA]}$$ Here,

• $[A^{-}]$ represents the concentration of the conjugate base,
• $[HA]$ is the concentration of the weak acid, and
• $p{K}_a$ is the negative logarithm of the acid dissociation constant, $K_a$.

This equation explains the balance within a buffer solution where both a weak acid and its conjugate base are present. It essentially describes how a buffer can absorb added acids or bases while maintaining a relatively stable pH. Understanding this concept is essential for solving problems related to buffer solutions.

pH Calculation

Calculating pH is an exercise in understanding the balance of hydrogen ions in a solution. The pH scale ranges from 0 to 14 and measures how acidic or basic a solution is. Each unit change signifies a tenfold change in hydrogen ion concentration. In a buffer solution, such as one consisting of acetic acid and sodium acetate, the calculation of pH involves applying the Henderson-Hasselbalch equation. For instance, given the concentrations of acetic acid and acetate, you first identify the ratio $\frac{[A^{-}]}{[HA]}$. By computing $${\log }_{10}\left(\frac{[A^{-}]}{[HA]}\right)$$ and adding it to the $p{K}_a$ of the acid (4.74 for acetic acid), you can pinpoint the specific pH of the buffer under those conditions. This meticulous calculation shows how buffers keep the pH steady, even when small amounts of stronger acids or bases are added.

Acid-Base Equilibrium

Acid-base equilibrium is a concept focused on the balance between acids and bases in a chemical system. It plays an integral role in buffer solutions, directly influencing their ability to stabilize pH. In this scenario, the equilibrium involves a weak acid, like acetic acid, and its conjugate base, acetate.When you introduce a base, such as $O{H}^{-}$ from KOH, the equilibrium shifts as follows: $$C{H}_{3}COOH+O{H}^{-}⟶C{H}_{3}CO{O}^{-}+H_{2}O$$ Because the equilibrium shifts to counteract the added $O{H}^{-}$, more acetate forms, consequently increasing the solution's pH. Conversely, when an acid like HNO3 is added, it reacts with the acetate ions: $$C{H}_{3}CO{O}^{-}+H^{+}⟶C{H}_{3}COOH$$ In this case, the equilibrium shifts to produce more acetic acid, which reduces the solution's pH. Understanding these reactions explains how buffers adjust to and maintain pH, emphasizing the dynamic yet stable nature of acid-base equilibrium.