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 $1M$ ($10^{0}M$).
  • At pH 14, the hydroxide ion concentration (${\mathrm{OH}}^{-}$) is $1M$, 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 $0.002M$ of $\mathrm{HCl}$.

1. Given values:
   • Concentration of $\mathrm{HCl}=0.002M$
   • 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 $2g$ pure ${\mathrm{H}}_3{\mathrm{PO}}_4$ per ${\mathrm{dm}}^3$ of solution? A1:

  1. Molar mass of ${\mathrm{H}}_3{\mathrm{PO}}_4$: $3(1.008)+30.97+4(16.00)=97.994gmo{l}^{-1}$
  2. Molarity of ${\mathrm{H}}_3{\mathrm{PO}}_4$ solution: Molarity = $\frac{\text{mass}}{\text{molar mass}\times \text{volume (L)}}=\frac{2g}{97.994gmo{l}^{-1}\times 1d{m}^3}=0.02041M$
  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 $[{\mathrm{H}}^{+}]\approx [{\mathrm{H}}_3{\mathrm{PO}}_4]$.
     • $[{\mathrm{H}}^{+}]\approx 0.02041M$
     • $\mathrm{pH}=-\log [{\mathrm{H}}^{+}]=-\log (0.02041)=1.69$ (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 $[{\mathrm{H}}^{+}]$: $[{\mathrm{H}}^{+}]=10^{-1.5}=0.0316M$