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Chapter 15: Problem 16

Calculate the concentration of \(\mathrm{H}^{+}\) ions in a \(0.62 M\) NaOH solution.

Short Answer

Expert verified
The concentration of \(H^{+}\) ions in the 0.62 M NaOH solution is approximately \(1.61 \times 10^{-14} M\).

Step by step solution

01

Identify the Ionization Process

Sodium Hydroxide (NaOH) is a strong base that completely ionizes in water to sodium ions (Na+) and hydroxide ions (OH-). Hence, the concentration of OH- ions in the solution will be the same as that of NaOH which is \(0.62 M\).
02

Utilize the Relationship Between H+ and OH- Ions

The product of the concentrations of the H+ and OH- ions is equal to the ion product constant for water (\(K_w\)) which is \(1.0 \times 10^{-14}\) at 25 degrees Celsius. The relationship is given as \([H+][OH-] = K_w\). Hence, we can calculate the concentration of H+ ions using: \([H+] = \frac{K_w}{[OH-]}\)
03

Calculate the Concentration of H+ ions

Substituting the values into the equation, the concentration of H+ ions can be calculated as: \([H+] = \frac{1.0 \times 10^{-14}}{0.62}\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ionization Process
Ionization is the process by which molecules dissociate into their respective ions when dissolved in a solvent, typically water. This concept is crucial in solution chemistry, where we study how substances behave in solution. In the case of sodium hydroxide (NaOH), which is a strong base, it completely ionizes in water. This means that when NaOH is dissolved, it separates into sodium ions ( Na^+) and hydroxide ions ( OH^- ).
  • NaOH is well-known for its complete ionization; it doesn't partially ionize but rather dissociates fully, which is typical behavior for strong bases.
  • Since each molecule of NaOH provides one hydroxide ion, the concentration of OH^- ions becomes identical to the initial concentration of the NaOH solution.
This complete ionization process is vital as it allows us to predict the concentration of ions in the solution accurately.
Ion Product Constant
The ion product constant, often denoted as \( K_w \), is a fundamental property of water that describes the relationship between the concentrations of hydrogen ions (H^+) and hydroxide ions (OH^-) in aqueous solutions. At 25 degrees Celsius, \( K_w \) has a value of \( 1.0 \times 10^{-14} \). This constant is critical as it underpins the behavior of acids and bases in solution.
  • The equation \( [H^+][OH^-] = K_w \) expresses the ion product constant. This indicates that the product of these ion concentrations will always equal \( K_w \)in a neutral, acidic, or basic solution at 25 degrees Celsius.
  • If the concentration of OH^- is known (as in the NaOH solution), you can easily find the H^+ concentration using this relationship.
This balance maintained by \( K_w \) helps chemists understand solution pH and the nature of a solution鈥攚hether it is acidic, basic, or neutral.
Concentration Calculation
Calculating ion concentrations in a solution involves applying the relationship between the hydrogen and hydroxide ions, given by the ion product constant. In the exercise, knowing the concentration of OH^- ions in the NaOH solution (\(0.62 M\)), we aim to find the concentration of H^+ ions using the formula:
  • Start by writing the formula: \( [H^+] = \frac{K_w}{[OH^-]} \).
  • Substitute the known values into the formula. For \( K_w \), use \( 1.0 \times 10^{-14} \), and for \([OH^-]\), use the given \(0.62 M\).
  • Solve for [H^+] to get the concentration: \( [H^+] = \frac{1.0 \times 10^{-14}}{0.62} \).
This step-by-step substitution provides the precise concentration of hydrogen ions, showing how values are interdependent, based on ionization and fundamental constants like \( K_w \). This can clarify the acid-base balance in the solution and is essential for understanding solution chemistry.

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Most popular questions from this chapter

Predict whether the following solutions are acidic, basic, or nearly neutral: (a) \(\mathrm{NaBr}\), (b) \(\mathrm{K}_{2} \mathrm{SO}_{3}\) (c) \(\mathrm{NH}_{4} \mathrm{NO}_{2}\), (d) \(\operatorname{Cr}\left(\mathrm{NO}_{3}\right)_{3}\).

Compare the strengths of the following pairs of acids: (a) \(\mathrm{H}_{2} \mathrm{SO}_{4}\) and \(\mathrm{H}_{2} \mathrm{SeO}_{4}\), (b) \(\mathrm{H}_{3} \mathrm{PO}_{4}\) and \(\mathrm{H}_{3} \mathrm{AsO}_{4}\).

Compare the \(\mathrm{pH}\) of a \(0.040 \mathrm{M} \mathrm{HCl}\) solution with that of a \(0.040 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) solution. (Hint: \(\mathrm{H}_{2} \mathrm{SO}_{4}\) is a strong acid; \(K_{\mathrm{a}}\) for \(\left.\mathrm{HSO}_{4}^{-}=1.3 \times 10^{-2} .\right)\)

Like water, liquid ammonia undergoes autoionization: $$ \mathrm{NH}_{3}+\mathrm{NH}_{3} \rightleftharpoons \mathrm{NH}_{4}^{+}+\mathrm{NH}_{2}^{-} $$ (a) Identify the Br酶nsted acids and Br酶nsted bases in this reaction. (b) What species correspond to \(\mathrm{H}^{+}\) and \(\mathrm{OH}^{-}\) and what is the condition for a neutral solution?

Explain why small, highly charged metal ions are able to undergo hydrolysis.
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