Calculate \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\) and \(\left[\mathrm{OH}^{-}\right]\) for each solution at \(25^{\circ} \mathrm{C} .\) a. \(\mathrm{pH}=8.55\) b. \(\mathrm{pH}=11.23\) c. \(\mathrm{pH}=2.87\)

Understanding the concentration of hydronium ions, denoted as \( [H_3O^+] \), is crucial for assessing the acidity of a solution. In an aqueous solution, hydronium ions are a measure of the number of free protons present, which directly influences the solution's pH level.

To calculate \( [H_3O^+] \) from a known pH, we can employ the formula \( [H_3O^+] = 10^{-pH} \). If the pH of a solution is given, just as in our exercise, applying this formula will immediately yield the concentration of hydronium ions. For a solution with \( pH = 8.55 \) as mentioned in our exercise, the concentration of hydronium ions is \( [H_3O^+] = 10^{-8.55} \).

This direct relationship between pH and \( [H_3O^+] \) allows for quick conversions and a fundamental understanding that as the pH decreases, indicating an increase in acidity, the concentration of hydronium ions rises.

The hydroxide ion concentration, \( [OH^-] \), is another key factor in characterizing the basicity of a solution. When \( [H_3O^+] \) is known, \( [OH^-] \) can be easily found using the ionic product of water, which is a constant at a given temperature and is \( 1.0 \times 10^{-14} \) at \( 25^\circ C \).

To find the \( [OH^-] \) we use the equation \( [OH^-] = \frac{K_w}{[H_3O^+]} \). As we determined earlier for a solution with \( pH = 8.55 \), the \( [H_3O^+] \) would be \( 10^{-8.55} \) and thus our \( [OH^-] \) is calculated as \( [OH^-] = \frac{1.0 \times 10^{-14}}{10^{-8.55}} \). It's important to recognize that as \( [H_3O^+] \) increases, indicating more acidic conditions, \( [OH^-] \) correspondingly decreases, signifying that the solution is less basic.

The ionic product of water \( (K_w) \) is a fundamental parameter in chemical equilibrium and is especially crucial in pH calculations. At a temperature of \( 25^\circ C \) \( K_w \) has a value of \( 1.0 \times 10^{-14} \). It represents the product of the molar concentrations of hydronium \( [H_3O^+] \) and hydroxide \( [OH^-] \) ions in pure water and provides a bridge between the two.

For every aqueous solution at \( 25^\circ C \) the product of \( [H_3O^+] \) and \( [OH^-] \) must be \( K_w \), meaning that an increase in the concentration of one will decrease the concentration of the other. This relationship holds true even if additional substances are dissolved in water, unless they affect the temperature or the ionic strength of the solution, which can lead to a different value of \( K_w \) at equilibrium.

Logarithmic equations are pivotal in understanding the relationship between pH, \( [H_3O^+] \) and \( [OH^-] \). pH is defined as the negative logarithm (to base 10) of the hydronium ion concentration: \( pH = -\log_{10}([H_3O^+]) \).

The use of logarithms simplifies the process of working with the very small numbers typical in these calculations. Applying the logarithm rules can help us switch between the pH and the corresponding hydronium or hydroxide ion concentrations with ease. For instance, to retrieve \( [H_3O^+] \) from pH, we rewrite the pH equation as \( [H_3O^+] = 10^{-pH} \). Solving logarithmic equations like these is an essential skill when dealing with acid-base chemistry and interpreting the resulting concentrations in a meaningful way.