pH of Strong Acids and Bases - Online Tutor, Practice Problems & Exam Prep

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.

Strong Acids and Bases

Recall that strong acids and strong bases completely ionized in water, and because of this fact, we can say here that the concentration of H⁺ ions and the concentration of OH^- ions are equal to the concentration of strong acids and bases respectively. So what exactly do I mean by this? Well here if we take a look we have 0.25 molar of hydrochloric acid, a strong acid. When it ionizes, it ionizes 100% into these two ions. That would mean that the concentration of the acid is the same concentration as my ions, so this would still be 0.25 molar and 0.25 molar.

So the concentration of HCl is equivalent to 0.25 molar for H⁺ for strong bases, they completely ionize as well. They would make 100% of calcium ion and two hydroxide ions here. So here we'd have 1.2 molar of calcium ions, but with the hydroxide ion we have to take into account that it's not just one hydroxide ion being produced, but two. What effect does this have on the concentration? Well, we'd say the real concentration of hydroxide ions will be 1.2 molar the original molarity times 2 because we make 2 OH^- ions. So the complete concentration of hydroxide ion in this case will be 2.4 molarity.

Now also recall that strong bases may contain the following ions. The four basic anions are hydroxide ions which we saw up above, but also hydride ion, amide ion and oxide ion. We are going to say when it comes to calculations dealing with pH or pOH, we are going to treat the concentration of OH^- ions as being the same as the concentration of hydride ions, amide ions and oxide ions. In addition to this, remember the formulas that are connected to pH and pOH. Remember that pH plus pOH equal 14.

If we know the H⁺ concentration, that allows us to find pH since it's the negative log of H⁺. If we know the pH, then we know what H⁺ concentration is because H⁺ = 10^-pH. pOH equals negative log of OH^- concentration. And if you know pOH itself, just remember OH^- concentration equals 10^-pOH.

pH of Strong Acids and Bases Example

If the concentration of barium hydride solution is 0.398 molar, calculate its pOH now barium hydride represents a strong base. Remember, strong base is completely ionized into ions. So here we form 1 barium barium ion +2 hydride ions.

The concentration of the of the strong base is 0.398 molar. Because we produce 2 hydride ions, it'd be 2 * 0.398 molar to get its real concentration. When we do that, we get 0.796 molar at the concentration of our hydride ion.

Now remember that the concentration of hydride ion is equal to the concentration of your hydroxide ion. This is important because now that we know the concentration of hydroxide ion, we can find the pOH itself equals negative log of OH- concentration.

So plug in the 0.796 and when we do that, we're going to get as our answer 0.0991 at the pOH of our barium hydride 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

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

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

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.