Question

Calculate the \(\left[\mathrm{OH}^{-}\right]\) of each aqueous solution with the following \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\) : a. \(\mathrm{NaOH}\) solution, \(1.0 \times 10^{-12} \mathrm{M}\) b. aspirin, \(6.0 \times 10^{-4} \mathrm{M}\) c. milk of magnesia, \(1.0 \times 10^{-9} \mathrm{M}\) d. stomach acid, \(5.2 \times 10^{-2} \mathrm{M}\)

Short answer

a) \(\text{[OH}^{-}\text{]} = 1.0 \times 10^{-2} \) M, b) \(\text{[OH}^{-}\text{]} = 1.7 \times 10^{-11} \) M, c) \(\text{[OH}^{-}\text{]} = 1.0 \times 10^{-5} \) M, d) \(\text{[OH}^{-}\text{]} = 1.9 \times 10^{-13} \) M

Step-by-step solution

Understanding the relation

Use the relationship between \(\text{[H}_{3}\text{O}^{+}\text{]}\) and \(\text{[OH}^{-}\text{]}\) in water: \[ \text{[H}_{3}\text{O}^{+}\text{]} \times \text{[OH}^{-}\text{]} = 1.0 \times 10^{-14} \]

Calculate \(\text{[OH}^{-}\text{]}\) for \(\text{NaOH}\) solution

Given \[ \text{[H}_{3}\text{O}^{+}\text{]} = 1.0 \times 10^{-12} \], solve for \[ \text{[OH}^{-}\text{]} \] using the relationship: \[ \text{[OH}^{-}\text{]} = \frac{1.0 \times 10^{-14}}{\text{[H}_{3}\text{O}^{+}\text{]}} = \frac{1.0 \times 10^{-14}}{1.0 \times 10^{-12}} = 1.0 \times 10^{-2} \text{M} \]

Calculate \(\text{[OH}^{-}\text{]}\) for aspirin solution

Given \[ \text{[H}_{3}\text{O}^{+}\text{]} = 6.0 \times 10^{-4} \], solve for \[ \text{[OH}^{-}\text{]} \] using the relationship: \[ \text{[OH}^{-}\text{]} = \frac{1.0 \times 10^{-14}}{\text{[H}_{3}\text{O}^{+}\text{]}} = \frac{1.0 \times 10^{-14}}{6.0 \times 10^{-4}} \], which simplifies to \[ \text{[OH}^{-}\text{]} = 1.7 \times 10^{-11} \text{M} \]

Calculate \(\text{[OH}^{-}\text{]}\) for milk of magnesia solution

Given \[ \text{[H}_{3}\text{O}^{+}\text{]} = 1.0 \times 10^{-9} \], solve for \[ \text{[OH}^{-}\text{]} \] using the relationship: \[ \text{[OH}^{-}\text{]} = \frac{1.0 \times 10^{-14}}{\text{[H}_{3}\text{O}^{+}\text{]}} = \frac{1.0 \times 10^{-14}}{1.0 \times 10^{-9}} = 1.0 \times 10^{-5} \text{M} \]

Calculate \(\text{[OH}^{-}\text{]}\) for stomach acid solution

Given \[ \text{[H}_{3}\text{O}^{+}\text{]} = 5.2 \times 10^{-2} \], solve for \[ \text{[OH}^{-}\text{]} \] using the relationship: \[ \text{[OH}^{-}\text{]} = \frac{1.0 \times 10^{-14}}{\text{[H}_{3}\text{O}^{+}\text{]}} = \frac{1.0 \times 10^{-14}}{5.2 \times 10^{-2}} = 1.9 \times 10^{-13} \text{M} \]

Key concepts

Acid-Base Chemistry

Acid-base chemistry is all about understanding the balance between acids and bases in a solution. An acid is something that can donate a proton (H\textsuperscript{+}), and a base can accept one. All aqueous solutions have a relationship between their hydrogen ion concentration (\text{[H}_{3}\text{O}^{+}\text{]}) and hydroxide ion concentration (\text{[OH}^{-}\text{]}). This relationship is defined by the ion-product constant for water: \[ \text{[H}_{3}\text{O}^{+}\text{]} \times \text{[OH}^{-}\text{]} = 1.0 \times 10^{-14} \] This equilibrium helps predict behaviors of acids and bases and guides pH and pOH calculations. By knowing one of these concentrations, you can easily calculate the other.

Hydronium Ion Concentration

Hydronium ion concentration (\text{[H}_{3}\text{O}^{+}\text{]}) represents the number of \text{H}^{+} ions available in a solution. The hydronium ion is formed when a proton (\text{H}^{+}) is accepted by a water molecule, resulting in \text{H}_{3}\text{O}^{+}\text{].
\[ \text{H}_{2}\text{O} + \text{H}^{+} \rightarrow \text{H}_{3}\text{O}^{+} \] High \text{[H}_{3}\text{O}^{+}\text{]} means the solution is more acidic, whereas low levels indicate it’s more basic. When you know the hydronium ion concentration, you can use the relationship \[ \text{[H}_{3}\text{O}^{+}\text{]} \times \text{[OH}^{-}\text{]} = 1.0 \times 10^{-14} \] to find the hydroxide ion concentration (\text{[OH}^{-}\text{]}) .

Hydroxide Ion Concentration

Hydroxide ion concentration (\text{[OH}^{-}\text{]}) tells us how many hydroxide ions are present in a solution. It's closely linked to the acidity or basicity of the solution. The more \text{[OH}^{-}\text{]} ions, the more basic the solution. Conversely, a higher hydronium ion concentration means fewer hydroxide ions, indicating an acidic solution.
Let’s see a specific example to put things into perspective. Given \text{[H}_{3}\text{O}^{+}\text{]} = 1.0 \times 10^{-12} \text{M} for a \text{NaOH} solution: \[ \text{[OH}^{-}\text{]} = \frac{1.0 \times 10^{-14}}{1.0 \times 10^{-12}} = 1.0 \times 10^{-2} \text{M} \] Use this reciprocal relationship to solve for \text{[OH}^{-}\text{]} when given \text{[H}_{3}\text{O}^{+}\text{]}.

Neutralization Reactions

Neutralization reactions occur when an acid and a base react to form water and a salt. This reaction is crucial in balancing the pH of various solutions and involves a transfer of protons. For instance, when you mix hydrochloric acid (HCl) with sodium hydroxide (NaOH), they neutralize each other: \[ \text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_{2}\text{O} \] The products (NaCl and H\textsubscript{2}O) are neutral compared to reactants. This process reduces the acidity or basicity of the original solutions.
Neutralization is essential in many biological and chemical processes, ensuring that environments remain at optimal pH levels.