Problem 16 - Chemistry: Principles and Reactions

Question

Calculate \(\left[\mathrm{H}^{+}\right]\) and \(\left[\mathrm{OH}^{-}\right]\) in solutions with the following \(\mathrm{pH}\). (a) \(9.0\) (b) \(3.20\) (c) \(-1.05\) (d) \(7.46\)

Step-by-Step Solution

Answer

Given the pH values (a) 9.0, (b) 3.20, (c) -1.05, and (d) 7.46, calculate the concentrations of \([\mathrm{H}^+]\) and \([\mathrm{OH}^{-}]\) ions in the solutions. (a) pH = 9.0: \([\mathrm{H}^{+}] = 1 \times 10^{-9}\mathrm{M}\) and \([\mathrm{OH}^{-}] = 1 \times 10^{-5}\mathrm{M}\) (b) pH = 3.20: \([\mathrm{H}^{+}] = 6.31 \times 10^{-4}\mathrm{M}\) and \([\mathrm{OH}^{-}] = 1.58 \times 10^{-11}\mathrm{M}\) (c) pH = -1.05: \([\mathrm{H}^{+}] = 1.12 \times 10^{1}\mathrm{M}\) and \([\mathrm{OH}^{-}] = 8.91 \times 10^{-16}\mathrm{M}\) (d) pH = 7.46: \([\mathrm{H}^{+}] = 3.46\times 10^{-8}\mathrm{M}\) and \([\mathrm{OH}^{-}] = 2.89 \times 10^{-7}\mathrm{M}\)

Step 1: Calculate the concentration of \([\mathrm{H}^{+}]\) using the pH formula

The formula for pH is given by \(pH = -\log{[\mathrm{H}^{+}]}\). To find the concentration of \([\mathrm{H}^{+}]\), we need to rearrange this formula to \([\mathrm{H}^{+}] = 10^{-pH}\). Calculate the concentration of \([\mathrm{H}^{+}]\) for each given pH value. (a) For pH = 9.0, \([\mathrm{H}^{+}] = 10^{-9.0} = 1 \times 10^{-9}\mathrm{M}\) (b) For pH = 3.20, \([\mathrm{H}^{+}] = 10^{-3.20} = 6.31 \times 10^{-4}\mathrm{M}\) (c) For pH = -1.05, \([\mathrm{H}^{+}] = 10^{-(-1.05)} = 1.12 \times 10^{1}\mathrm{M}\) (d) For pH = 7.46, \([\mathrm{H}^{+}] = 10^{-7.46} = 3.46 \times 10^{-8}\mathrm{M}\)

Step 2: Calculate the pOH for each solution

The relationship between pH and pOH is given by \(pH + pOH = 14\). We can calculate the pOH using this relationship for each given pH value. (a) For pH = 9.0, \(pOH = 14 - 9.0 = 5.0\) (b) For pH = 3.20, \(pOH = 14 - 3.20 = 10.8\) (c) For pH = -1.05, \(pOH = 14 - (-1.05) = 15.05\) (d) For pH = 7.46, \(pOH = 14 - 7.46 = 6.54\)

Step 3: Calculate the concentration of \([\mathrm{OH}^{-}]\) using the pOH formula

The formula for pOH is given by \(pOH = -\log{[\mathrm{OH}^{-}]}\). To find the concentration of \([\mathrm{OH}^{-}]\), we need to rearrange this formula to \([\mathrm{OH}^{-}] = 10^{-pOH}\). Calculate the concentration of \([\mathrm{OH}^{-}]\) for each given pOH value. (a) For pOH = 5.0, \([\mathrm{OH}^{-}] = 10^{-5.0} = 1 \times 10^{-5}\mathrm{M}\) (b) For pOH = 10.8, \([\mathrm{OH}^{-}] = 10^{-10.8} = 1.58 \times 10^{-11}\mathrm{M}\) (c) For pOH = 15.05, \([\mathrm{OH}^{-}] = 10^{-15.05} = 8.91 \times 10^{-16}\mathrm{M}\) (d) For pOH = 6.54, \([\mathrm{OH}^{-}] = 10^{-6.54} = 2.89 \times 10^{-7}\mathrm{M}\) In conclusion, for each given pH value, the concentrations of \([\mathrm{H}^{+}]\) and \([\mathrm{OH}^{-}]\) ions are: (a) pH = 9.0: \([\mathrm{H}^{+}] = 1 \times 10^{-9}\mathrm{M}\) and \([\mathrm{OH}^{-}] = 1 \times 10^{-5}\mathrm{M}\) (b) pH = 3.20: \([\mathrm{H}^{+}] = 6.31 \times 10^{-4}\mathrm{M}\) and \([\mathrm{OH}^{-}] = 1.58 \times 10^{-11}\mathrm{M}\) (c) pH = -1.05: \([\mathrm{H}^{+}] = 1.12 \times 10^{1}\mathrm{M}\) and \([\mathrm{OH}^{-}] = 8.91 \times 10^{-16}\mathrm{M}\) (d) pH = 7.46: \([\mathrm{H}^{+}] = 3.46\times 10^{-8}\mathrm{M}\) and \([\mathrm{OH}^{-}] = 2.89 \times 10^{-7}\mathrm{M}\)

Key Concepts

Hydrogen Ion Concentration

Understanding hydrogen ion concentration is essential for evaluating the acidity of a solution. The concentration of \([H^{+}]\) ions indicates how acidic or basic a solution is, expressed through the pH scale. The formula to determine hydrogen ion concentration from pH is \([H^{+}] = 10^{-pH}\). This uses the fact that pH is the negative logarithm of the hydrogen ion concentration.
To grasp this concept completely:

• A lower pH means a higher concentration of hydrogen ions, which makes the solution more acidic.
• A higher pH implies fewer hydrogen ions, indicating a basic solution.

Utilizing the logarithmic function allows us to deal with the wide-ranging \([H^{+}]\) values in a manageable scale. For instance, a solution with pH 3.2 has \(6.31 \times 10^{-4}\) M of hydrogen ions, demonstrating its acidic nature.

Hydroxide Ion Concentration

Hydroxide ion concentration \([OH^{-}]\) is a key factor in determining a solution’s basicity. Like hydrogen ion concentration, it leverages a logarithmic scale termed pOH. Calculating hydroxide ion concentration from pOH is expressed as \([OH^{-}] = 10^{-pOH}\). Here’s why it matters:

• Solutions with a high pOH will have a high concentration of hydroxide ions, making them more basic.
• Conversely, a lower pOH signifies fewer hydroxide ions, indicative of an acidic environment.

For accurate calculations, remember the relationship between pH and pOH: \(pH + pOH = 14\). This lets you frequently derive one value to find the other. Take, for example, a solution with a pH of 9.0; this computes to a pOH of 5.0, leading to a hydroxide concentration of \(1 \times 10^{-5}\) M.

pOH Calculation

The calculation of pOH is a fundamental aspect of understanding solution chemistry. It complements the concept of pH, revealing the levels of hydroxide ions present. The equation \(pOH = 14 - pH\) allows the direct calculation of pOH from a known pH.

• Calculating the pOH helps you understand a solution's basicity or acidity better, especially when transitioning between pH and pOH values.
• The calculation highlights the inverse relationship between hydrogen and hydroxide ions in a solution.

For instance, if a solution has a pH of 7.46, its pOH calculates to 6.54, illustrating a more neutral-leaning environment. This dual relationship, where high pH equals low pOH and vice versa, is crucial in assessing the chemical nature of solutions.

Logarithmic Functions in Chemistry

Logarithmic functions are pivotal in the field of chemistry for simplifying measurement of very large or small numbers. In the context of pH and pOH calculations, logarithms represent exponential scales between hydrogen or hydroxide ion concentrations.
The use of \(log\) helps in several ways:

• It transforms multiplicative relationships into additive ones, making it easier to handle biological and chemical processes like acid-base equilibria.
• It simplifies the interpretation and comparison of concentrations that vary by orders of magnitude, which is particularly common in acid-base chemistry.

Through operations like converting between \([H^{+}]\) or \([OH^{-}]\) concentrations to pH and pOH, logarithms enable straightforward calculations and deepen understanding of underlying chemical models.