Hydroxide ion Concentration Calculator from pH and pOH

The Hydroxide Ion Concentration Calculator is a versatile tool designed to facilitate the calculation of the hydroxide ion concentration ([OH^–]) in a solution based on the given pH or pOH values.

This calculator is instrumental in understanding the basic principles of acid-base chemistry and provides insights into the alkalinity or basicity of a solution.

How to Use the Calculator?

Input Selection: Users have the flexibility to input either the pH or pOH value, choosing the appropriate selection from the dropdown menu.

Data Entry: Enter the pOH or pH value into the designated field.

Calculation: Upon clicking the “Calculate” button, the calculator swiftly processes the input and computes the corresponding [OH^–] concentration and displays it along with the associated pH or pOH value.

Measuring Hydroxide Ion Concentration from pOH and pH

Relationship between pH, pOH, and Hydroxide Ion Concentration: The hydroxide ion concentration ([OH^–]) in a solution can be calculated using the given pH or pOH values. Understanding the relationship between these parameters is fundamental:

pH and pOH Relationship: In an aqueous solution at 25°C, the sum of pH and pOH is always 14. This relationship remains constant due to the auto-ionization of water.

pH + pOH = 14

Calculating Hydroxide Ion Concentration [OH^–] from pOH

Given the pOH value, the hydroxide ion concentration can be determined using the formula:

From pOH Equation: If the pOH of a solution is known, the [OH^–] concentration can be calculated.

pOH = -log[OH^–]

Example: For a solution with a pOH of 4:

• Using the pOH equation:
  • pOH = 4
  • -log[OH^–] = 4
  • [OH^–] = 1.00 × 10^-4 M

Calculating Hydroxide Ion Concentration [OH⁻] from pH

Similarly, [OH⁻] can be calculated when the pH is given using the relationship between pH and pOH:

From pH Equation: If the pH of a solution is provided, the [OH⁻] concentration can be deduced.

• pH + pOH = 14
• pOH = 14 – pH

Example: For a solution with a pH of 11.6:

• Using the pH + pOH equation:
  • pH + pOH = 14
  • 11.6 + pOH = 14
  • pOH = 14 – 11.6
  • pOH = 2.4

Further Calculation:

From the pOH equation:

• pOH = 2.4
• -log[OH⁻] = 2.4
• [OH⁻] = 2.51 × 10^-12 M