Question

a. Calculate the molarity of \(\mathrm{H}_{3} \mathrm{O}^{+}\) in a solution that has a ph of \(8.90 .\) b. Calculate the concentration of \(\mathrm{OH}^{-}\) for this solution.

Short answer

[\textrm{H}_3\textrm{O}^+] = 1.26 \times 10^{-9} \text{ M}, [\textrm{OH}^-] = 7.94 \times 10^{-6} \text{ M}

Step-by-step solution

- Understand pH to \(\textrm{H}_3\textrm{O}^+\) Relationship

Recognize that pH is related to the concentration of \(\textrm{H}_3\textrm{O}^+\) ions by the equation \[ \textrm{pH} = -\textrm{log}[\textrm{H}_3\textrm{O}^+] \].

- Rearrange pH Equation

Rearrange the equation to solve for \[\textrm{H}_3\textrm{O}^+\] concentration: \[ [\textrm{H}_3\textrm{O}^+] = 10^{-\textrm{pH}} \].

- Substitute Given pH Value

Substitute the known pH of 8.90 into the equation: \[ [\textrm{H}_3\textrm{O}^+] = 10^{-8.90} \].

- Calculate \(\textrm{H}_3\textrm{O}^+\) Molarity

Use a calculator to find the value: \[ [\textrm{H}_3\textrm{O}^+] = 1.26 \times 10^{-9} \text{ M} \].

- Understand \(\textrm{H}_3\textrm{O}^+\) to \(\textrm{OH}^-\) Relationship

Recall the relationship between \(\textrm{H}_3\textrm{O}^+\) and \(\textrm{OH}^-\) ions: \[ [\textrm{H}_3\textrm{O}^+][\textrm{OH}^-] = 1.0 \times 10^{-14} \].

- Substitute \(\textrm{H}_3\textrm{O}^+\) Value to Find \(\textrm{OH}^-\) Concentration

Substitute \([\textrm{H}_3\textrm{O}^+] = 1.26 \times 10^{-9} \text{ M}\) into the expression: \[ [\textrm{OH}^-] = \frac{1.0 \times 10^{-14}}{1.26 \times 10^{-9}} \].

- Calculate \(\textrm{OH}^-\) Concentration

Use a calculator to find the value: \[ [\textrm{OH}^-] = 7.94 \times 10^{-6} \text{ M} \].

Key concepts

pH calculation

The pH of a solution is a measure of how acidic or basic it is. It is defined by the equation: \(\textrm{pH} = -\textrm{log}[\textrm{H}_3\textrm{O}^+ ]\). This equation allows us to understand the concentration of hydronium ions (\textrm{H}_3\textrm{O}^+) in the solution. When calculating the pH, remember:

• A low pH (below 7) indicates an acidic solution.
• A high pH (above 7) indicates a basic solution.
• A pH of 7 means the solution is neutral.

In this specific problem, the given pH is 8.90. You'll need to use the inverse of the pH equation to find the concentration of hydronium ions.

molarity of hydronium ion

To find the molarity of hydronium ions (\textrm{H}_3\textrm{O}^+), you rearrange the pH equation: \[ [\textrm{H}_3\textrm{O}^+] = 10^{-\textrm{pH}} \]. This formula helps convert pH back to the hydronium ion concentration. By substituting the given pH value of 8.90, you'll end up with: \[ [\textrm{H}_3\textrm{O}^+] = 10^{-8.90} \]. Using a calculator for precision, it yields: \[ [\textrm{H}_3\textrm{O}^+] = 1.26 \times 10^{-9} \text{ M} \]. This indicates the molarity, or concentration, of hydronium ions in the solution.

relationship between \textrm{H}_3\textrm{O}^+ and \textrm{OH}^- ions

In solution chemistry, \textrm{H}_3\textrm{O}^+ and \textrm{OH}^- ions have a unique relationship. They must satisfy the water dissociation constant (\textrm{K}_w), which is \[ [\textrm{H}_3\textrm{O}^+][\textrm{OH}^-] = 1.0 \times 10^{-14} \]. Given the \textrm{H}_3\textrm{O}^+ concentration, you can find \textrm{OH}^- by rearranging the equation: \[ [\textrm{OH}^-] = \frac{1.0 \times 10^{-14}}{[\textrm{H}_3\textrm{O}^+} \]. Substituting \[ [\textrm{H}_3\textrm{O}^+] = 1.26 \times 10^{-9} \], we calculate: \[ [\textrm{OH}^-] = \frac{1.0 \times 10^{-14}}{1.26 \times 10^{-9}} \]. This results in \[ [\textrm{OH}^-] = 7.94 \times 10^{-6} \text{ M} \]. Therefore, the balance of hydronium and hydroxide ions maintains the solution's overall pH and properties.