Problem 33 - Chemistry The Science in Context

Question

Calculate the concentration of the following ions in the solution described: a. \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\) in \(8.4 \times 10^{-4} \mathrm{M} \mathrm{NaOH}\) b. \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\) in \(6.6 \times 10^{-5} M \mathrm{Ca}(\mathrm{OH})_{2}\) c. \(\left[\mathrm{OH}^{-}\right]\) in \(4.5 \times 10^{-3} \mathrm{MHCl}\) d. \(\left[\mathrm{OH}^{-}\right]\) in \(2.9 \times 10^{-5} M \mathrm{HCl}\)

Step-by-Step Solution

Answer

Question: Calculate the concentration of \(\mathrm{H}_{3}\mathrm{O}^{+}\) in \(8.4\times10^{-4}\,\mathrm{M}\,\mathrm{NaOH}\) and \(6.6\times10^{-5}\,\mathrm{M}\,\mathrm{Ca}(\mathrm{OH})_2\), and the concentration of \(\mathrm{OH}^{-}\) in \(4.5\times10^{-3}\,\mathrm{M\,HCl}\) and \(2.9\times10^{-5}\,\mathrm{M\,HCl}\). Answer: The concentration of \(\mathrm{H}_{3}\mathrm{O}^{+}\) in \(8.4\times10^{-4}\,\mathrm{M}\,\mathrm{NaOH}\) is \(1.19\times10^{-12}\) M and in \(6.6\times10^{-5}\,\mathrm{M}\,\mathrm{Ca}(\mathrm{OH})_2\) is \(7.58\times10^{-11}\) M. The concentration of \(\mathrm{OH}^{-}\) in \(4.5\times10^{-3}\,\mathrm{M\,HCl}\) is \(2.22\times10^{-12}\) M and in \(2.9\times10^{-5}\,\mathrm{M\,HCl}\) is \(3.45\times10^{-10}\) M.

Step 1: a. Calculate the concentration of \(\mathrm{H}_{3}\mathrm{O}^{+}\) in \(8.4\times10^{-4}\,\mathrm{M}\,\mathrm{NaOH}\).

In this solution, \(\mathrm{NaOH}\) is acting as a strong base and will completely dissociate: \(\mathrm{NaOH} \rightarrow \mathrm{Na}^{+} + \mathrm{OH}^{-}\) Since \(\mathrm{OH}^{-}\) is present in the solution, we can use the ion-product constant for water to relate it to the \(\mathrm{H}_{3}\mathrm{O}^{+}\) concentration: \(K_w = [\mathrm{H}_{3}\mathrm{O}^{+}][\mathrm{OH}^{-}]\) Rearrange the formula to find \([\mathrm{H}_{3}\mathrm{O}^{+}]\): \([\mathrm{H}_{3}\mathrm{O}^{+}] = \frac{K_w}{[\mathrm{OH}^{-}]}\) Now plug in the known values (\(K_w= 1.0\times10^{-14}\) and \([\mathrm{OH}^{-}] = 8.4\times10^{-4}\) M): \([\mathrm{H}_{3}\mathrm{O}^{+}] = \frac{1.0\times10^{-14}}{8.4\times10^{-4}}\) Calculate the result: \([\mathrm{H}_{3}\mathrm{O}^{+}] = 1.19\times10^{-12}\) M

Step 2: b. Calculate the concentration of \(\mathrm{H}_{3}\mathrm{O}^{+}\) in \(6.6\times10^{-5}\,\mathrm{M}\,\mathrm{Ca}(\mathrm{OH})_2\).

In this solution, \(\mathrm{Ca}(\mathrm{OH})_2\) is acting as a strong base and will completely dissociate: \(\mathrm{Ca}(\mathrm{OH})_2 \rightarrow \mathrm{Ca}^{2+} + 2\mathrm{OH}^{-}\) Since there is a 1:2 stoichiometry between \(\mathrm{Ca}(\mathrm{OH})_2\) and \(\mathrm{OH}^{-}\), we need to multiply the concentration of \(\mathrm{Ca}(\mathrm{OH})_2\) by 2: \([\mathrm{OH}^{-}] = 2 \times (6.6\times10^{-5}\,\mathrm{M}) = 1.32\times10^{-4}\) M Now we can use the ion-product constant for water as in part a: \([\mathrm{H}_{3}\mathrm{O}^{+}] = \frac{K_w}{[\mathrm{OH}^{-}]}\) Plug in the values (\(K_w= 1.0\times10^{-14}\) and \([\mathrm{OH}^{-}] = 1.32\times10^{-4}\) M): \([\mathrm{H}_{3}\mathrm{O}^{+}] = \frac{1.0\times10^{-14}}{1.32\times10^{-4}}\) Calculate the result: \([\mathrm{H}_{3}\mathrm{O}^{+}] = 7.58\times10^{-11}\) M

Step 3: c. Calculate the concentration of \(\mathrm{OH}^{-}\) in \(4.5\times10^{-3}\,\mathrm{M\,HCl}\).

In this solution, \(\mathrm{HCl}\) is acting as a strong acid and will completely dissociate: \(\mathrm{HCl} \rightarrow \mathrm{H}^{+} + \mathrm{Cl}^{-}\) The concentration of \(\mathrm{H}^{+}\) ions in the solution is equal to the concentration of \(\mathrm{HCl}\): \([\mathrm{H}^{+}] = 4.5\times10^{-3}\) M Since the concentration of \(\mathrm{OH}^{-}\) is related to the concentration of \(\mathrm{H}^{+}\) via the ion-product constant for water: \(K_w = [\mathrm{H}^{+}][\mathrm{OH}^{-}]\) Rearrange the formula to find \([\mathrm{OH}^{-}]\): \([\mathrm{OH}^{-}] = \frac{K_w}{[\mathrm{H}^{+}]}\) Plug in the values (\(K_w= 1.0\times10^{-14}\) and \([\mathrm{H}^{+}] = 4.5\times10^{-3}\) M): \([\mathrm{OH}^{-}] = \frac{1.0\times10^{-14}}{4.5\times10^{-3}}\) Calculate the result: \([\mathrm{OH}^{-}] = 2.22\times10^{-12}\) M

Step 4: d. Calculate the concentration of \(\mathrm{OH}^{-}\) in \(2.9\times10^{-5}\,\mathrm{M\,HCl}\).

As in part c, \(\mathrm{HCl}\) is acting as a strong acid. The concentration of \(\mathrm{H}^{+}\) ions in the solution is equal to the concentration of \(\mathrm{HCl}\): \([\mathrm{H}^{+}] = 2.9\times10^{-5}\) M Now we can calculate the concentration of \(\mathrm{OH}^{-}\) using the ion-product constant for water as before: \([\mathrm{OH}^{-}] = \frac{K_w}{[\mathrm{H}^{+}]}\) Plug in the values (\(K_w= 1.0\times10^{-14}\) and \([\mathrm{H}^{+}] = 2.9\times10^{-5}\) M): \([\mathrm{OH}^{-}] = \frac{1.0\times10^{-14}}{2.9\times10^{-5}}\) Calculate the result: \([\mathrm{OH}^{-}] = 3.45\times10^{-10}\) M

Key Concepts

Hydronium Ion

The hydronium ion, expressed as \( \mathrm{H}_{3}\mathrm{O}^{+} \), is a very important player in the chemistry of acids and bases. When an acid dissolves in water, it donates a proton (\( \mathrm{H}^{+} \)) to a water molecule, forming hydronium ion. This ion appears in countless chemical reactions and is pivotal in determining a solution's acidity.

The concentration of hydronium ions in a solution is directly tied to the solution's pH level, with higher concentrations leading to lower pH values, thereby indicating stronger acidity. Calculating the concentration of \( \mathrm{H}_{3} \mathrm{O}^{+} \) is crucial for understanding the properties of a solution. This is why the hydronium ion is frequently encountered in exercises involving acid-base equilibria and calculations.

For example, when a solution of a strong base like \( \mathrm{NaOH} \) is used, the hydroxide ion forms and the ion-product constant relationship allows for the determination of the corresponding \( \mathrm{H}_{3} \mathrm{O}^{+} \) concentration. It's a balancing act of sorts: when hydroxide levels are high, hydronium levels drop, and vice versa, maintaining the equilibrium dictated by water's ion-product constant.

Hydroxide Ion

A hydroxide ion (\( \mathrm{OH}^{-} \)), is commonly produced when a base dissolves in water. This ion is essential for basic (alkaline) solutions and various chemical processes. The presence of hydroxide ions can significantly shift a solution's pH toward higher values, indicating a more basic or alkaline nature. For instance, when you dissolve \( \mathrm{NaOH} \) (sodium hydroxide) in water, it disassociates completely into \( \mathrm{Na}^{+} \) and \( \mathrm{OH}^{-} \), increasing the hydroxide ion concentration.

Determining the concentration of \( \mathrm{OH}^{-} \) is crucial in many chemical contexts. It helps to assess the basicity of solutions like those involving strong bases.

• In the context of an exercise involving acids such as \( \mathrm{HCl} \), calculating the \( \mathrm{OH}^{-} \) ion concentration requires knowledge of how the ion-product constant for water links hydronium and hydroxide ion concentrations.
• For instance, when acid concentration overwhelms the basic constituent, you calculate \( \mathrm{OH}^{-} \) concentration using the ion-product constant, resolving how much the base has been neutralized.

This concept is applied when analyzing solutions involving acids and bases, ensuring students comprehend not only the numeric calculations but the underlying chemical principles at play.

Ion-Product Constant

The ion-product constant for water, denoted as \( K_w \), is fundamentally the equilibrium constant for the self-ionization of water. It's a cornerstone of acid-base chemistry, reflecting the balance between hydronium ions \( [\mathrm{H}_{3}\mathrm{O}^{+}] \) and hydroxide ions \( [\mathrm{OH}^{-}] \) in any aqueous solution.

Understanding \( K_w \) is essential because it interlinks these ion concentrations. With a constant value of \( 1.0 \times 10^{-14} \) at 25°C, \( K_w \) ensures that multiplying the concentrations of these ions always equals this constant. This relationship is critical across different types of aqueous solutions:

• In neutral solutions, both \( [\mathrm{H}_{3}\mathrm{O}^{+}] \) and \( [\mathrm{OH}^{-}] \) are approximately \( 1.0 \times 10^{-7} \) M.
• In acidic solutions, \( [\mathrm{H}_{3}\mathrm{O}^{+}] \) is greater than \( [\mathrm{OH}^{-}] \).
• In basic solutions, \( [\mathrm{OH}^{-}] \) exceeds \( [\mathrm{H}_{3}\mathrm{O}^{+}] \).

Therefore, the ion-product constant acts as a guiding star, helping chemists and students alike navigate the complex waters of acid and base reactions. This constant is integral in exercises requiring the calculation of ion concentrations, enabling a clear understanding of what happens at a molecular level.