pH Scale and Calculations

Introduction

Key Concepts

Understanding the pH Scale

The pH scale is a logarithmic scale used to specify the acidity or basicity of an aqueous solution. It ranges from 0 to 14, with 7 being neutral. Values below 7 indicate acidic solutions, while those above 7 signify basic (alkaline) solutions.

Definition of pH

pH is defined as the negative logarithm of the hydrogen ion concentration: $$\text{pH} = -\log[H^+]$$ where $[H^+]$ represents the concentration of hydrogen ions in moles per liter (M).

Calculating pH

To calculate the pH of a solution, determine the hydrogen ion concentration and apply the pH formula. For example, if $[H^+] = 1.0 \times 10^{-3} \text{ M}$, then: $$\text{pH} = -\log(1.0 \times 10^{-3}) = 3$$

Relationship Between pH and pOH

pH and pOH are related through the equation: $$\text{pH} + \text{pOH} = 14$$ This relationship is derived from the ion product of water ($K_w$) at 25°C: $$K_w = [H^+][OH^-] = 1.0 \times 10^{-14} \text{ M}^2$$

Calculating pOH

Similar to pH, pOH is calculated using the hydroxide ion concentration: $$\text{pOH} = -\log[OH^-]$$ For instance, if $[OH^-] = 1.0 \times 10^{-5} \text{ M}$, then: $$\text{pOH} = -\log(1.0 \times 10^{-5}) = 5$$

Strong Acids and Bases

Strong acids and bases dissociate completely in water. For a strong acid like hydrochloric acid ($HCl$), the concentration of $H^+$ ions equals the concentration of the acid: $$[H^+] = [HCl]$$ Similarly, for a strong base like sodium hydroxide ($NaOH$): $$[OH^-] = [NaOH]$$

Weak Acids and Bases

Unlike strong acids and bases, weak acids and bases only partially dissociate in water. The degree of dissociation is characterized by the acid dissociation constant ($K_a$) or the base dissociation constant ($K_b$).

Buffer Solutions

Buffer solutions resist changes in pH upon the addition of small amounts of acid or base. They are typically composed of a weak acid and its conjugate base or a weak base and its conjugate acid. The Henderson-Hasselbalch equation describes the pH of buffer solutions: $$\text{pH} = \text{p}K_a + \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right)$$ where $[\text{A}^-]$ is the concentration of the conjugate base and $[\text{HA}]$ is the concentration of the weak acid.

Applications of pH in Chemical Reactions

pH plays a pivotal role in various chemical reactions, including enzyme activity in biological systems, solubility of compounds, and corrosion processes. Understanding pH allows chemists to manipulate reaction conditions to favor desired products.

pH Indicators

pH indicators are substances that change color in response to pH changes. They are often weak acids or bases themselves and can provide a visual representation of the pH level of a solution. Common indicators include litmus, phenolphthalein, and bromothymol blue.

Titration and pH Calculations

Titration is a technique used to determine the concentration of an unknown acid or base by reacting it with a standard solution of base or acid, respectively. The equivalence point, where moles of acid equal moles of base, can be identified using pH indicators or pH meters.

Calculating pH in Mixed Solutions

In solutions containing both acids and bases, the net $[H^+]$ can be determined by considering the strengths and concentrations of each component. For example, in a mixture of $HCl$ and $NaOH$, the excess $H^+$ or $OH^-$ ions dictate the resulting pH.

Common Mistakes in pH Calculations

Students often make errors in pH calculations by:

• Incorrectly applying the logarithmic function.
• Misidentifying strong and weak acids/bases.
• Forgetting to consider the autoionization of water in certain calculations.

Practice Problems

To reinforce understanding, consider the following problems:

1. Calculate the pH of a 0.025 M $HCl$ solution.
2. Determine the pOH of a solution with $[OH^-] = 2.5 \times 10^{-4} \text{ M}$.
3. Find the pH of a buffer solution containing 0.1 M acetic acid and 0.1 M sodium acetate.

Solutions to Practice Problems

Problem 1:

For $HCl$, a strong acid: $$[H^+] = 0.025 \text{ M}$$ $$\text{pH} = -\log(0.025) \approx 1.60$$

Problem 2:

$$\text{pOH} = -\log(2.5 \times 10^{-4}) \approx 3.60$$ $$\text{pH} = 14 - 3.60 = 10.40$$

Problem 3:

Using the Henderson-Hasselbalch equation: $$\text{pH} = \text{p}K_a + \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right)$$ Given $[\text{A}^-] = 0.1 \text{ M}$ and $[\text{HA}] = 0.1 \text{ M}$: $$\text{pH} = \text{p}K_a + \log(1) = \text{p}K_a$$ If $K_a$ for acetic acid is $1.8 \times 10^{-5}$: $$\text{pH} \approx 4.74$$

Comparison Table

Summary and Key Takeaways