Question

The \(\mid \mathrm{OH}\) I in \(100 \mathrm{~mL}\) of \(0.015 \mathrm{M} \mathrm{HCl}(\mathrm{aq})\) is (1) \(5 \times 10^{-12} \mathrm{M}\) (2) \(3 \times 10^{-10} \mathrm{M}\) (3) \(6.7 \times 10^{-13} \mathrm{M}\) (4) \(2.0 \times 10^{-9} \mathrm{M}\)

Short answer

The concentration of \[\text{OH}^- \] ions is \ 6.7 \times 10^{-13} \text{M} \.

Step-by-step solution

- Identify the Nature of HCl

Recognize that HCl is a strong acid. This means it fully dissociates in water. Therefore, the concentration of \[\text{H}^{+}\] ions will be equal to the concentration of HCl.

- Calculate the Concentration of \[ \text{H}^{+} \] Ions

Since the concentration of HCl is \[0.015 \text{M}\], the concentration of \[\text{H}^{+}\] ions is also \[0.015 \text{M}\].

- Use the Ion Product of Water

Recall that the product of the concentrations of \[\text{H}^{+}\] and \[\text{OH}^{-}\] ions in water is a constant: \[ K_w = 1 \times 10^{-14} \text{M}^2 \].

- Calculate the Concentration of \[ \text{OH}^{-} \] Ions

Use the formula \ [\text{OH}^-] = \frac{K_w}{ [\text{H}^+] } \ to calculate the concentration of \[\text{OH}^-\] ions. \ [\text{OH}^-] = \frac{1 \times 10^{-14}}{0.015} = 6.7 \times 10^{-13} \text{M} \.

Key concepts

headline of the respective core concept

To understand how to solve acid-base equilibrium problems, you first need to comprehend what happens during an acid-base reaction. In this problem, we deal with Hydrochloric Acid (HCl), which is a strong acid. Strong acids dissociate completely in water. This means they break down entirely into ions. For example, when HCl dissolves in water, it splits into \[\text{H}^{+}\] and \[\text{Cl}^{-}\].
Since HCl fully dissociates, the concentration of \[\text{H}^{+}\] ions in the solution is the same as the initial concentration of HCl. In our example, a 0.015 M HCl solution will have a 0.015 M concentration of \[\text{H}^{+}\] ions. Once you understand this, you're already ahead in solving such problems.
Knowing the concentration of \[\text{H}^{+}\] ions, you can use the relationship between \[\text{H}^{+}\] and \[\text{OH}^{-}\] ions, explained by the ion product of water.

headline of the respective core concept

The ion product of water (\[ K_w\]) is an essential concept in acid-base chemistry. It states that in any aqueous solution at 25°C, the product of the concentrations of \[\text{H}^{+}\] ions and \[\text{OH}^{-}\] ions is always equal to \[\text{1 \times 10^{-14}} \text{M}^2\]. This constant is crucial because, by knowing either the concentration of \[\text{H}^{+}\] ions or \[\text{OH}^{-}\] ions, you can always find the other.
In the textbook problem, we have a known concentration of \[\text{H}^{+}\] ions (0.015 M). Using the ion product constant, we can readily calculate the \[\text{OH}^{-}\] ion concentration. The formula is: \[ [\text{OH}^-] = \frac{K_w}{ [\text{H}^+] } \].
Substituting the known values: \[ [\text{OH}^-] = \frac{1 \times 10^{-14}}{0.015} = 6.7 \times 10^{-13} \text{M} \]. With this equation, you can convert between \[\text{H}^{+}\] and \[\text{OH}^{-}\] concentrations easily.

headline of the respective core concept

The dissociation of strong acids is a fundamental concept in understanding various acid-base reactions. A strong acid like HCl dissociates completely in water. Complete dissociation means every molecule of HCl will break down into \[\text{H}^{+}\] and \[\text{Cl}^{-}\].
Let's break this down with steps:

• First, recognize the substance (HCl in this case) is a strong acid.
• Know that complete dissociation means the molarity of the acid will equal the molarity of the resulting \[\text{H}^{+} \] ions.

In our example problem, a 0.015 M solution of HCl will entirely dissociate into 0.015 M \[\text{H}^{+}\] ions and 0.015 M \[\text{Cl}^{-}\] ions.
This concept is critical because it helps you understand the behavior of strong acids in different chemical reactions and how they influence the pH and equilibrium of a solution.
Understanding these simple yet vital pieces of information will allow you to solve many problems involving strong acid dissociation quickly and efficiently.