# 4.2 The pH Scale The pH scale is a fundamental concept in chemistry used to express the acidity or alkalinity of an aqueous solution. It provides a numerical measure that indicates the concentration of hydrogen ions ($\mathrm{H}^+$) in a solution. ## Understanding the pH Scale * **Range:** The pH scale typically ranges from 0 to 14. * **Acidity:** * Substances with pH values less than 7 are considered **acidic**. * The lower the pH value (closer to 0), the stronger the acid and the more acidic the solution. Extremely acidic substances have pH values below 1. * **Alkalinity (Basicity):** * Substances with pH values greater than 7 are considered **alkaline** (or basic). * The higher the pH value (closer to 14), the stronger the base and the more alkaline the solution. * **Neutrality:** * A pH value of exactly 7 indicates a **neutral** solution (e.g., pure water at $25^\circ\mathrm{C}$). ## pH and Hydrogen Ion Concentration The pH value is an inverse logarithmic indication of the concentration of hydrogen ions (or hydronium ions, $\mathrm{H}_3\mathrm{O}^+$) in a solution. * **Relationship:** * At pH 0, the hydronium ion concentration is $1\,M$ ($10^0\,M$). * At pH 14, the hydroxide ion concentration ($\mathrm{OH}^-$) is $1\,M$, which implies a very low $\mathrm{H}^+$ concentration ($10^{-14}\,M$) in water. * **Formula:** The pH is calculated using the formula: $$ \mathrm{pH} = -\log[\mathrm{H}^+] $$ where $[\mathrm{H}^+]$ is the molar concentration of hydrogen ions in moles per liter ($M$). * **Inverse Relationship:** * A change of one pH unit represents a tenfold change in the $[\mathrm{H}^+]$ concentration. For example, a solution with pH 3 has ten times the $[\mathrm{H}^+]$ concentration of a solution with pH 4. ## Worked Examples ### Example 4.1: Finding pH from $\mathrm{H}^+$ concentration Find the pH of a solution of $0.002\,M$ of $\mathrm{HCl}$. 1. **Given values:** * Concentration of $\mathrm{HCl} = 0.002\,M$ * Since $\mathrm{HCl}$ is a strong acid, it dissociates completely, so $[\mathrm{H}^+] = [\mathrm{HCl}]$. * Therefore, $[\mathrm{H}^+] = 2.0 \times 10^{-3}\,M$. 2. **Apply the formula:** $$ \mathrm{pH} = -\log[\mathrm{H}^+] $$ 3. **Show calculation:** $$ \mathrm{pH} = -\log[2.0 \times 10^{-3}] = 2.70 $$ ### Example 4.2: Finding $\mathrm{H}^+$ concentration from pH If moist soil has a pH of 7.84, what is the $\mathrm{H}^+$ concentration of the soil solution? 1. **Given values:** * $\mathrm{pH} = 7.84$ 2. **Apply the formula (rearranged):** From $\mathrm{pH} = -\log[\mathrm{H}^+]$, we can derive $[\mathrm{H}^+] = 10^{-\mathrm{pH}}$. 3. **Show calculation:** $$ [\mathrm{H}^+] = 10^{-7.84} $$ $$ [\mathrm{H}^+] = 1.45 \times 10^{-8}\,M $$ * **Hint:** To calculate this, place -7.84 in your calculator and take the antilog (often inverse $\log$ or $10^x$). ## Possible Questions/Answers Here are some practice questions based on the concepts discussed: * **Q1:** What is the pH of a solution of $2\,g$ pure $\mathrm{H}_3\mathrm{PO}_4$ per $\mathrm{dm}^3$ of solution? **A1:** 1. **Molar mass of $\mathrm{H}_3\mathrm{PO}_4$**: $3(1.008) + 30.97 + 4(16.00) = 97.994\,g\,mol^{-1}$ 2. **Molarity of $\mathrm{H}_3\mathrm{PO}_4$ solution**: Molarity = $\frac{\text{mass}}{\text{molar mass} \times \text{volume (L)}} = \frac{2\,g}{97.994\,g\,mol^{-1} \times 1\,dm^3} = 0.02041\,M$ 3. **Calculate $\mathrm{pH}$**: * *Note:* Phosphoric acid ($\mathrm{H}_3\mathrm{PO}_4$) is a **weak acid**. For a precise pH calculation, its acid dissociation constants ($K_a$ values) would be needed. However, in many introductory contexts without explicit $K_a$ values, a simplified approach might be expected. If we assume it behaves as a strong *monoprotic* acid for estimation purposes (i.e., only the first proton dissociates completely), then $[\mathrm{H}^+] \approx [\mathrm{H}_3\mathrm{PO}_4]$. * $[\mathrm{H}^+] \approx 0.02041\,M$ * $\mathrm{pH} = -\log[\mathrm{H}^+] = -\log(0.02041) = 1.69$ *(This is an approximation. A more rigorous calculation for a weak polyprotic acid is more complex.)* * **Q2:** Calculate the concentration of hydrogen ion ($\mathrm{H}^+$) in a solution of sulphuric acid having pH of 1.5. **A2:** 1. **Given $\mathrm{pH}$**: $\mathrm{pH} = 1.5$ 2. **Apply formula**: $[\mathrm{H}^+] = 10^{-\mathrm{pH}}$ 3. **Calculate $[\mathrm{H}^+]$**: $[\mathrm{H}^+] = 10^{-1.5} = 0.0316\,M$