Question

(a) What are \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right],\left[\mathrm{OH}^{-}\right],\) and \(\mathrm{pOH}\) in a solution with a (18.32 \(\mathrm{pH}\) of \(9.85 ?\) (b) What are \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right],\left[\mathrm{OH}^{-}\right],\) and \(\mathrm{pH}\) in a solution with a pOH of \(9.43 ?\)

Short answer

(a) \( \text{[H}_3\text{O}^+] \) is \( 1.41 \times 10^{-10} \) M, \( \text{[OH}^-] \) is \( 7.08 \times 10^{-5} \) M, \( \text{pOH} \) is 4.15. (b) \( \text{[H}_3\text{O}^+] \) is \(2.69 \times 10^{-5} \) M, \( \text{[OH}^-] \) is \(3.71 \times 10^{-10} \) M, \( \text{pH} \) is 4.57.

Step-by-step solution

- Calculate \(\text{[H}_3\text{O}^+\)] from pH

Use the formula \( \text{[H}_3\text{O}^+] = 10^{-\text{pH}} \) to find the hydrogen ion concentration. Given the pH is 9.85, we have: \( \text{[H}_3\text{O}^+] = 10^{-9.85} \)

- Calculate \(\text{[OH}^- \)] from \(\text{[H}_3\text{O}^+]\)

Using the formula \( \text{[H}_3\text{O}^+ \times [OH}^- ] = 1 \times 10^{-14} \), rearrange it to find \( \text{[OH}^- ] \): \( \text{[OH}^- ] = \frac{1 \times 10^{-14}}{\text{[H}_3\text{O}^+}} \). Substitute \( \text{[H}_3\text{O}^+}\)'s value from Step 1.

- Calculate pOH from \(\text{[OH}^- \)]

Use the formula \( \text{pOH} = -\text{log}(\text{[OH}^- ]) \) to find the pOH.

- Calculate \(\text{[H}_3\text{O}^+\)] from pOH

For part (b), convert pOH to pH using \( \text{pH} = 14 - \text{pOH} \). Given pOH is 9.43, we have: \( \text{pH} = 14 - 9.43 = 4.57 \). Then find \( \text{[H}_3\text{O}^+] \) using \( \text{[H}_3\text{O}^+] = 10^{-4.57} \).

- Calculate \(\text{[OH}^- \)] from \(\text{[H}_3\text{O}^+]\)

Using the same approach as in Step 2, calculate \( \text{[OH}^- ] \) from the newly found \( \text{[H}_3\text{O}^+]\).

Key concepts

pH calculation

The pH is a measure of the acidity or basicity of a solution. It is defined as the negative logarithm of the hydronium ion concentration. This is given by the formula: \( \text{pH} = -\text{log}\text{[H}_3\text{O}^+]\text{ } \).
For instance, if you are given a pH of 9.85, you can find the hydronium ion concentration (\text{[H}_3\text{O}^+] ) by rearranging the formula to: \( \text{[H}_3\text{O}^+] = 10^{-\text{pH}} \).

This relationship helps you to quickly switch between knowing the pH of a solution and knowing the concentration of hydronium ions in that solution. An important note here is that as pH decreases, acidity increases, as there are more hydronium ions in the solution.

hydronium ion concentration

Hydronium ion concentration (\text{[H}_3\text{O}^+] ) is directly related to the pH of a solution. The equation to calculate this is \( \text{[H}_3\text{O}^+] = 10^{-\text{pH}} \).

For example, if the pH is 9.85, the concentration of \text{[H}_3\text{O}^+] ions will be: \( 10^{-9.85} \), which can be computed using a calculator.

Keep in mind that hydronium ions play a crucial role in determining the acidity of the solution. The higher the concentration of \text{[H}_3\text{O}^+] ions, the more acidic the solution. Additionally, hydronium ion concentration is a centerpiece in many calculations related to acids and bases in chemistry.

hydroxide ion concentration

The concentration of hydroxide ions (\text{[OH}^- ]) in a solution is tied to the concentration of hydronium ions. This relationship is given by the ion-product constant for water: \( [\text{H}_3\text{O}^+]\times [\text{OH}^- ] = 1 \times 10^{-14} \) at 25°C.

If you know the hydronium ion concentration, you can find the hydroxide ion concentration by rearranging the equation: \( \text{[OH}^- ] = \frac{1 \times 10^{-14}}{\text{[H}_3\text{O}^+}} \).

Let's say \text{[H}_3\text{O}^+} = \(10^{-9.85}\), plug this value into the equation to find \text{[OH}^- }. Thus, the higher the concentration of \text{OH}^- ions, the more basic (less acidic) the solution is. Understanding this balance is critical for grasping acid-base chemistry.

pOH calculation

pOH is a measure of hydroxide ion concentration (\text{[OH}^- }). It is found using the formula: \( \text{pOH} = -\text{log}(\text{[OH}^- }) \).
Just like pH, pOH makes it simpler to work with small numbers of concentrations. If you have calculated \text{[OH}^- } as in the previous step, for example, you can find the pOH of a solution easily by taking the negative logarithm of \text{[OH}^- }.

In addition, there's a relationship between pH and pOH: \( \text{pH} + \text{pOH} = 14 \) at 25°C. This can assist in switching between both values when only one is known, providing more versatility in analyzing a solution's properties.

acid-base equilibrium

Acid-base equilibrium involves the balance between acids and bases within a solution. This balance is typically referred to in context with the ion-product of water, and the equilibrium expressions like: \( [\text{H}_3\text{O}^+ ] \times [\text{OH}^-] = 1 \times 10^{-14} \).

Knowing the concentrations of \text{H}_3\text{O}^+ and \text{OH}^- ions allows for discussing the equilibrium state in the solution.
Any shift in this balance, such as adding more acid or base, will drive the reaction towards restoring equilibrium. For instance, increasing hydronium ion concentration drives the equilibrium towards reducing \text{OH}^- ions to maintain balance, and vice versa. This way, the concepts of \text{pH}, \text{pOH}, \text{[H}_3\text{O}^+], and \text{[OH}^-] are all interconnected under the umbrella of acid-base equilibrium.