pH of Strong Acids and Bases: Videos & Practice Problems

Strong acids and bases are known for their complete ionization in water, leading to equal concentrations of their respective ions. For example, a 0.25 molar hydrochloric acid solution will fully dissociate into 0.25 molar hydrogen ions (H⁺) and chloride ions (Cl^-). Similarly, strong bases like calcium hydroxide not only dissociate into calcium ions (Ca²⁺) but also produce two hydroxide ions (OH^-) per molecule, doubling the concentration of OH^- ions in the solution. When calculating pH or pOH, it's essential to consider that the concentration of OH^- is equivalent to that of hydride (H^-), amide (NH₂^-), and oxide (O^2-) ions. The relationship between pH and pOH is captured by the equation:

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

and the concentrations of H⁺ and OH^- can be found using the equations:

$H^+={10}^{-\mathrm{pH}}$

and

$\mathrm{OH}^-={10}^{-\mathrm{pOH}}$

respectively. These equations are fundamental when analyzing the acidity or basicity of a solution.

pH of Strong Acids and Bases

Strong acids and strong bases are characterized by their complete ionization in water, which means that the concentration of hydrogen ions (H⁺) from strong acids and hydroxide ions (OH^-) from strong bases directly corresponds to their molarity. For instance, a 0.25 molar solution of hydrochloric acid (HCl), a strong acid, fully dissociates into H⁺ and Cl^- ions. Therefore, the concentration of H⁺ ions in this solution is also 0.25 molar.

Similarly, when considering strong bases, such as calcium hydroxide (Ca(OH)₂), which dissociates into one calcium ion (Ca²⁺) and two hydroxide ions (OH^-), the concentration of hydroxide ions must account for this stoichiometry. If the concentration of calcium ions is 1.2 molar, the resulting concentration of hydroxide ions will be calculated as follows: 1.2 molarity (Ca²⁺) × 2 = 2.4 molarity (OH^-).

In addition to hydroxide ions, strong bases can also produce other anions such as hydride ions (H^-), amide ions (NH₂^-), and oxide ions (O^2-). When performing calculations involving pH and pOH, the concentration of OH^- ions can be treated similarly to these other anions.

Understanding the relationship between pH and pOH is crucial. The formulas governing these relationships are:

• pH + pOH = 14
• pH = -log[H⁺]
• [H⁺] = 10^-pH
• pOH = -log[OH^-]
• [OH^-] = 10^-pOH

By utilizing these equations, one can easily convert between concentrations of H⁺ and OH^- ions, facilitating the calculation of pH and pOH in various chemical contexts.

pOH Calculation Example

To calculate the pOH of a barium hydride solution with a concentration of 0.398 molar, we first recognize that barium hydride (BaH₂) is a strong base that completely dissociates in solution. The dissociation can be represented as:

BaH₂ → Ba²⁺ + 2 H^-

From this reaction, we see that one mole of barium hydride produces two moles of hydride ions (H^-). Therefore, the concentration of hydride ions is:

[H^-] = 2 × 0.398 mol/L = 0.796 mol/L

Since hydride ions (H^-) can be considered equivalent to hydroxide ions (OH^-) in terms of their effect on pOH, we can use this concentration to find the pOH of the solution. The formula for calculating pOH is:

pOH = -log[OH^-]

Substituting the concentration of hydroxide ions into the equation gives us:

pOH = -log(0.796)

Calculating this yields a pOH of approximately 0.0991. Thus, the pOH of the barium hydride solution is 0.0991, indicating a highly basic solution.

An aqueous solution of HBrO₄ has a pH of 4.34. Find the molar concentration of HBrO₄ solution.

• A. 4.6 × 10⁻¹⁹ M
• B. 2.6 × 10⁻⁶ M
• C. 2.2 × 10⁻¹⁰ M
• D. 4.6 × 10⁻⁵ M

Answer: (D)

Calculate the pH of a 25 mL of 5.45 × 10⁻² M LiOH solution.

• A. 1.264
• B. 12.736
• C. 0.338
• D. 11.134

Answer: (B)

HI is a strong acid (K_a = 3.2 × 10⁹). Calculate [H⁺], [OH⁻], pH and pOH of a 7.1 × 10⁻² M HI solution.

• A. [H⁺] = 7.1 × 10⁻² M
  [OH⁻] = 1.4 × 10⁻¹³ M
  pH = 1.15
  pOH = 12.85
• B. [H⁺] = 7.1 × 10⁻² M
  [OH⁻] = 1.4 × 10⁻¹³ M
  pH = 8.50
  pOH = 5.50
• C. [H⁺] = 7.1 × 10⁻² M
  [OH⁻] = 7.1 × 10⁻¹⁶ M
  pH = 1.15
  pOH = 15.15
• D. [H⁺] = 7.1 × 10⁻² M
  [OH⁻] = 7.1 × 10⁻¹² M
  pH = 1.15
  pOH = 11.15

Answer: (A)

Here’s what students ask on this topic:

How do you calculate the pH of strong acids and bases?

To calculate the pH of strong acids and bases, you'll first need to know the concentration of the acid or base in moles per liter (Molarity, M). Strong acids and bases completely dissociate in water, so the concentration of the hydrogen ions (H⁺) for acids, or hydroxide ions (OH^-) for bases, will be equal to the concentration of the acid or base.

For strong acids:

1. Determine the concentration of the acid in the solution (e.g. 0.01 M HCl).
2. Since HCl is a strong acid, it dissociates completely, so [H⁺] = 0.01 M.
3. Calculate the pH using the formula: $\mathrm{pH}=-\log \left[H^{+}\right]$.
4. Plug in the concentration: $\mathrm{pH}=-\log \left(0.01\right)=2$.

For strong bases:

1. Determine the concentration of the base (e.g. 0.01 M NaOH).
2. Since NaOH is a strong base, it dissociates completely, so [OH^-] = 0.01 M.
3. Calculate the pOH using the formula: $\mathrm{pOH}=-\log \left[{\mathrm{OH}}^{-}\right]$.
4. Plug in the concentration: $\mathrm{pOH}=-\log \left(0.01\right)=2$.
5. To find the pH, use the relationship: $\mathrm{pH}=14-\mathrm{pOH}$.

What is the pH of a strong base?

The pH of a strong base is typically high, usually falling between 11 and 14. The pH scale measures how acidic or basic a solution is, with 7 being neutral. Values below 7 indicate acidity, while values above 7 indicate alkalinity (basicity). Strong bases, such as sodium hydroxide (NaOH) or potassium hydroxide (KOH), dissociate completely in water, releasing a large number of hydroxide ions (OH^-). This complete dissociation results in a high pH value, reflecting a high concentration of hydroxide ions, which makes the solution very basic. It's important to handle strong bases with care, as they can be corrosive and cause chemical burns upon contact with skin or other tissues.

How do you calculate the pH of a strong acid?

Calculating the pH of a strong acid is relatively straightforward because strong acids completely dissociate in water. Here's how you can do it:

1. Identify the concentration of the strong acid in the solution, usually given in molarity (M), which is moles per liter (mol/L).
2. Recognize that the pH is the negative logarithm (base 10) of the hydrogen ion concentration ([H⁺]) in the solution. The formula is $\mathrm{pH}=-\log \left[H^{+}\right]$.
3. Since strong acids fully dissociate, the concentration of hydrogen ions [H⁺] will be equal to the concentration of the acid.
4. Plug the concentration of the acid into the pH formula. For example, if you have a 0.01 M solution of hydrochloric acid (HCl), which is a strong acid, the [H⁺] is also 0.01 M.
5. Calculate the pH: $\mathrm{pH}=-\log \left(0.01\right)=2$.

So, the pH of a 0.01 M solution of a strong acid like HCl is 2. Remember that because the pH scale is logarithmic, a one-unit change in pH represents a tenfold change in hydrogen ion concentration.

How do you calculate the pH of a strong base?

To calculate the pH of a strong base, you need to know the concentration of the base in moles per liter (molarity). A strong base completely dissociates in water, yielding hydroxide ions (OH^-).

Here's how you do it:

1. Determine the concentration of the base (M). For example, if you have a 0.01 M solution of NaOH, the concentration of OH^- will also be 0.01 M because NaOH dissociates completely.
2. Calculate the pOH of the solution by taking the negative logarithm (base 10) of the hydroxide ion concentration. Using our example: $\mathrm{pOH}=-log\left[{\mathrm{OH}}^{-}\right]=-log\left[0.01\right]=2$.
3. To find the pH, you use the relationship between pH and pOH, which is $\mathrm{pH}+\mathrm{pOH}=14(25°C)$. So, $\mathrm{pH}=14-\mathrm{pOH}$.

Using the example: $\mathrm{pH}=14-2=12$.

So, the pH of a 0.01 M solution of NaOH is 12. Remember, the pH scale ranges from 0 to 14, with 7 being neutral. Values above 7 indicate a basic (alkaline) solution.