pH of Strong Acids & Bases: Videos & Practice Problems

Understanding the relationship between pH and pOH is crucial when studying strong acids and strong bases, which are classified as strong electrolytes. Strong electrolytes are substances that completely ionize in solution, meaning they dissociate 100% into their constituent ions. For instance, hydrochloric acid (HCl) is a strong binary acid, while sodium hydroxide (NaOH) is a strong base, characterized by the presence of a group 1A ion (Na⁺) paired with hydroxide (OH^-).

When a strong acid or base is dissolved in water, it fully dissociates, leading to a reaction that can be represented with a single arrow pointing forward, indicating that the products are highly favored. This complete ionization simplifies the calculation of pH and pOH. For strong acids, the pH can be determined using the formula:

pH = -\log[H^+]

Similarly, for strong bases, pOH is calculated as:

pOH = -\log[OH^-]

These calculations eliminate the need for an ICE (Initial, Change, Equilibrium) chart, streamlining the process of determining the acidity or basicity of a solution. In the following example, we will apply these principles to calculate pH and pOH for specific strong acid and base scenarios.

To calculate the pH of a 0.0782 molar solution of calcium hydride (CaH₂), it is essential to recognize that calcium hydride is a strong base. This is due to the presence of calcium ions (Ca²⁺) and hydride ions (H^-). In the case of calcium hydride, each formula unit releases two hydride ions, which means that the effective concentration of hydroxide ions (OH^-) must be considered when determining the pH.

For strong bases, the concentration of hydroxide ions is derived from the number of hydroxide or hydride ions produced. In this example, since there are two hydride ions per formula unit, the actual concentration of hydroxide ions is calculated by multiplying the initial concentration by 2. Thus, the effective concentration of OH^- becomes:

$\left(0.0782\right)2=0.1564$

Next, to find the pOH, we use the formula:

$\mathrm{pOH}=-log\left(0.1564\right)$

This calculation yields a pOH of approximately 0.81. However, since the question specifically asks for the pH, we can use the relationship between pH and pOH, given by the equation:

$\mathrm{pH}+\mathrm{pOH}=14$

Substituting the calculated pOH into this equation allows us to solve for pH:

$\mathrm{pH}=14-0.81=13.19$

Thus, the pH of the 0.0782 molar solution of calcium hydride is 13.19. It is important to note that this method of multiplying the concentration by the number of hydroxide ions applies specifically to strong bases. In contrast, for strong acids, such as sulfuric acid (H₂SO₄), the first hydrogen ion is released strongly, while the second is weakly dissociated, so we do not multiply the concentration in that case.

To calculate the pH of a strong acid solution, such as hydrochloric acid (HBr), we can directly use the concentration of hydrogen ions, as strong acids fully dissociate in solution. For a 0.000550 M HBr solution, the pH can be determined using the formula:

pH = -\log[H^+]

Substituting the concentration into the equation gives:

pH = -\log(0.000550) ≈ 3.25964

However, it is essential to express the pH to the correct number of significant figures. The concentration of HBr has three significant figures, as determined by counting from the first non-zero digit (5) in the number 0.000550. Therefore, the pH must also be reported with three decimal places. Rounding 3.25964 to three decimal places results in a pH of 3.260.

For another example, consider calculating the pH of a 50 mL solution of sulfuric acid (H₂SO₄) with a concentration of 4.3 x 10^-7 M. Since H₂SO₄ is also a strong acid, we can use the same approach without needing to adjust for the number of protons contributed by the acid. The concentration remains as is, and we can apply the pH formula:

pH = -\log(4.3 \times 10^{-7})

Calculating this will yield the pH value, which should also be reported with the appropriate number of significant figures based on the concentration provided.