Question

Calculate the hydronium ion concentration in \(\mathrm{mol} / \mathrm{L}\) for each of the following solutions: (a) a solution whose \(\mathrm{pH}\) is \(5.20,(\mathrm{~b})\) a solution whose \(\mathrm{pH}\) is \(16.00,(\mathrm{c})\) a solution whose hydroxide concentration is \(3.7 \times 10^{-9} \mathrm{M}\).

Short answer

(a) \(6.31 \times 10^{-6} \text{ M}\), (b) \(1 \times 10^{-16} \text{ M}\), (c) \(2.70 \times 10^{-6} \text{ M}\).

Step-by-step solution

Understanding pH and Hydronium Ion Concentration

The pH of a solution is the negative logarithm (base 10) of the hydronium ion concentration \([ ext{H}_3 ext{O}^+]\). This is expressed as \( ext{pH} = - ext{log}([ ext{H}_3 ext{O}^+]) \). To find \([ ext{H}_3 ext{O}^+]\), we use the formula \([ ext{H}_3 ext{O}^+] = 10^{- ext{pH}}\). Let's apply this to find the solutions for parts (a) and (b).

Calculating Hydronium Ion Concentration for Solution (a)

For a solution with a pH of 5.20, use the formula \([ ext{H}_3 ext{O}^+] = 10^{- ext{pH}}\). This gives \([ ext{H}_3 ext{O}^+] = 10^{-5.20}\). Use a calculator to compute this value.

Calculate the Result for Solution (a)

Upon calculation, \(10^{-5.20} \approx 6.31 \times 10^{-6}\). Therefore, the hydronium ion concentration for part (a) is \(6.31 \times 10^{-6} \text{ M}\).

Calculating Hydronium Ion Concentration for Solution (b)

For a solution with a pH of 16.00, use the same formula: \([ ext{H}_3 ext{O}^+] = 10^{-16.00}\). Calculate this value using a calculator.

Calculate the Result for Solution (b)

Upon calculation, \(10^{-16.00} = 1 \times 10^{-16}\). Thus, the hydronium ion concentration for part (b) is \(1 \times 10^{-16} \text{ M}\).

Using OH- Concentration to Find Hydronium Ion Concentration in Solution (c)

For a solution where the hydroxide concentration \([ ext{OH}^-]\) is given, use the water dissociation constant: \( [ ext{H}_3 ext{O}^+][ ext{OH}^-] = 1.0 \times 10^{-14} \). Rearranging gives \([ ext{H}_3 ext{O}^+] = \frac{1.0 \times 10^{-14}}{[ ext{OH}^-]}\). Substitute the given \([ ext{OH}^-] = 3.7 \times 10^{-9} \text{ M}\) to find \([ ext{H}_3 ext{O}^+]\).

Calculate the Result for Solution (c)

Substitute into the equation: \([ ext{H}_3 ext{O}^+] = \frac{1.0 \times 10^{-14}}{3.7 \times 10^{-9}} \approx 2.70 \times 10^{-6} \text{ M}\). Therefore, for part (c), the hydronium ion concentration is approximately \(2.70 \times 10^{-6} \text{ M}\).

Key concepts

Hydronium Ion Concentration

The hydronium ion concentration, represented as \([\text{H}_3\text{O}^+]\), is a key factor in determining the acidity of a solution. When we talk about \(\text{pH}\), we're referring to a specific way of measuring how acidic or basic a solution is. The relationship is established by the equation:

• \( \text{pH} = -\log([\text{H}_3\text{O}^+]) \)

To find the hydronium ion concentration from the pH, we rearrange this equation:

• \([\text{H}_3\text{O}^+] = 10^{-\text{pH}}\)

For example, if a solution has a pH of 5.20, plug this into the formula, and you get \([\text{H}_3\text{O}^+] \approx 6.31 \times 10^{-6}\) M.
This conversion is essential for chemists and students alike, as it allows them to understand the level of activity of hydrogen ions in solution.

Logarithm

Logarithms are a mathematical concept used to solve equations involving exponential functions. In pH calculations, the logarithm helps link the pH scale to hydronium ion concentrations. Here's how it works:

• The pH is the negative logarithm (base 10) of the hydronium ion concentration.
• This means \( \text{pH} = -\log([\text{H}_3\text{O}^+]) \).

This relationship simplifies converting pH values back to hydronium ion concentrations using inverse operations. If you're comfortable with using logs, you can quickly go from a pH like 16.00 to a concentration of \(1 \times 10^{-16} \text{ M}\).
Understanding logarithms in this context helps decipher why changes in pH reflect logarithmic changes in acidity.

Water Dissociation Constant

The water dissociation constant, often denoted as \(K_w\), is crucial for understanding the balance between hydronium and hydroxide ions in water:

• For pure water at 25°C, \(K_w = [\text{H}_3\text{O}^+][\text{OH}^-] = 1.0 \times 10^{-14}\).

This constant tells us that, at equilibrium, the product of the concentrations of these ions is always \(1.0 \times 10^{-14}\). If one is known, the other can be calculated.
For instance, if the hydroxide ion concentration \([\text{OH}^-]\) is \(3.7 \times 10^{-9} \text{ M}\), the hydronium ion concentration is found by:

• \([\text{H}_3\text{O}^+] = \frac{1.0 \times 10^{-14}}{[\text{OH}^-]}\)

Using this relationship ensures you can confidently switch between these measurements and understand how they affect water's acidity or basicity.

Hydroxide Concentration

Hydroxide ion concentration \([\text{OH}^-]\) plays a vital role in determining a solution's basicity:

• A higher \([\text{OH}^-]\) means a more basic solution.

This measurement links directly with hydronium ion concentration through the water dissociation constant \(K_w\). Knowing \([\text{OH}^-]\) allows for the calculation of \([\text{H}_3\text{O}^+]\) using:

• \([\text{H}_3\text{O}^+] = \frac{1.0 \times 10^{-14}}{[\text{OH}^-]}\)

This relationship is particularly useful when working with solutions where the hydroxide ion is more accessible to measure. For example, if \([\text{OH}^-] = 3.7 \times 10^{-9} \text{ M}\), the \([\text{H}_3\text{O}^+]\) calculates to approximately \(2.70 \times 10^{-6} \text{ M}\).
Understanding both ions helps clarify the full picture of a solution's properties, whether acidic or basic.