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Chapter 8: Problem 71

What are the \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\) and \(\left[\mathrm{OH}^{-}\right]\) for a solution with the following \(\mathrm{pH}\) values? a. \(3.0\) b. \(6.00\) c. \(8.0\) d. \(11.0\) e. \(9.20\)

Short Answer

Expert verified
a.\text{[H}_{3}\text{O}^{+}\text{]}=1.0\times10^{-3}\text{M}, \text{[OH}^{-}\text{]}=1.0\times10^{-11}\text{M}; b.\text{[H}_{3}\text{O}^{+}\text{]}=1.0\times10^{-6}\text{M}, \text{[OH}^{-}\text{]}=1.0\times10^{-8}\text{M}; c.\text{[H}_{3}\text{O}^{+}\text{]}=1.0\times10^{-8}\text{M}, \text{[OH}^{-}\text{]}=1.0\times10^{-6}\text{M}; d.\text{[H}_{3}\text{O}^{+}\text{]}=1.0\times10^{-11}\text{M}, \text{[OH}^{-}\text{]}=1.0\times10^{-3}\text{M}; e.\text{[H}_{3}\text{O}^{+}\text{]}=6.3\times10^{-10}\text{M}, \text{[OH}^{-}\text{]}=1.6\times10^{-5}\text{M}

Step by step solution

01

- Understanding pH and concentration relationship

The pH of a solution is related to the concentration of hydrogen ions \(\text{[H}_{3}\text{O}^{+}\text{]}\) by the formula: \(\text{pH} = -\text{log} \text{[H}_{3}\text{O}^{+}\text{]}\). For each given pH value, use this equation to find \(\text{[H}_{3}\text{O}^{+}\text{]}\) by rearranging the formula: \(\text{[H}_{3}\text{O}^{+}\text{]} = 10^{-\text{pH}}\).
02

- Calculating \[ \text{[H}_{3}\text{O}^{+}\text{]} \] for each pH

For each pH given: \[ a. \text{pH} = 3.0 \rightarrow \text{[H}_{3}\text{O}^{+}\text{]} = 10^{-3} = 1 \times 10^{-3} \rightarrow \text{[H}_{3}\text{O}^{+}\text{]} = 1.0 \times 10^{-3}\text{ M} \] \[ b. \text{pH} = 6.00 \rightarrow \text{[H}_{3}\text{O}^{+}\text{]} = 10^{-6} = 1 \times 10^{-6} \rightarrow \text{[H}_{3}\text{O}^{+}\text{]} = 1.0 \times 10^{-6}\text{ M} \] \[ c. \text{pH} = 8.0 \rightarrow \text{[H}_{3}\text{O}^{+}\text{]} = 10^{-8} = 1 \times 10^{-8} \rightarrow \text{[H}_{3}\text{O}^{+}\text{]} = 1.0 \times 10^{-8}\text{ M} \] \[ d. \text{pH} = 11.0 \rightarrow \text{[H}_{3}\text{O}^{+}\text{]} = 10^{-11} = 1 \times 10^{-11} \rightarrow \text{[H}_{3}\text{O}^{+}\text{]} = 1.0 \times 10^{-11}\text{ M} \] \[ e. \text{pH} = 9.20 \rightarrow \text{[H}_{3}\text{O}^{+}\text{]} = 10^{-9.20} \rightarrow \text{[H}_{3}\text{O}^{+}\text{]} = 6.3 \times 10^{-10}\text{ M} \]
03

- Calculating \[ \text{[OH}^{-}\text{]} \] using the ion product of water

The relationship between \[ \text{[H}_{3}\text{O}^{+}\text{]} \] and \[ \text{[OH}^{-}\text{]} \] is given by the ion product of water: \[ \text{[H}_{3}\text{O}^{+}\text{]} \times \text{[OH}^{-}\text{]} = 1.0 \times 10^{-14} \]. To find \[ \text{[OH}^{-}\text{]} \]: \[ \text{[OH}^{-}\text{]} = \frac{1.0 \times 10^{-14}}{\text{[H}_{3}\text{O}^{+}\text{]}} \] For each \[ \text{[H}_{3}\text{O}^{+}\text{]} \] calculated before: a. \[ \text{[OH}^{-}\text{]} = \frac{1.0 \times 10^{-14}}{1.0 \times 10^{-3}} = 1.0 \times 10^{-11} \text{ M} \] b. \[ \text{[OH}^{-}\text{]} = \frac{1.0 \times 10^{-14}}{1.0 \times 10^{-6}} = 1.0 \times 10^{-8} \text{ M} \] c. \[ \text{[OH}^{-}\text{]} = \frac{1.0 \times 10^{-14}}{1.0 \times 10^{-8}} = 1.0 \times 10^{-6} \text{ M} \] d. \[ \text{[OH}^{-}\text{]} = \frac{1.0 \times 10^{-14}}{1.0 \times 10^{-11}} = 1.0 \times 10^{-3} \text{ M} \] e. \[ \text{[OH}^{-}\text{]} = \frac{1.0 \times 10^{-14}}{6.3 \times 10^{-10}} = 1.6 \times 10^{-5} \text{ M} \]

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Key Concepts

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

Hydronium Ion Concentration
The hydronium ion concentration, denoted as \([\text{H}_{3}\text{O}^{+}]\), is a critical concept in understanding pH. The pH of a solution gives us information about its acidity or basicity. It is calculated using the formula: \(\text{pH} = -\text{log} [\text{H}_{3}\text{O}^{+}]\). To find \([\text{H}_{3}\text{O}^{+}]\) from pH, we rearrange this formula: \([\text{H}_{3}\text{O}^{+}] = 10^{-\text{pH}}\). This means if the pH is known, the hydronium concentration can be determined easily.
For example:
  • For a pH of 3.0, \([\text{H}_{3}\text{O}^{+}] = 10^{-3} = 1 \times 10^{-3} \text{ M}\).
  • For a pH of 6.00, \([\text{H}_{3}\text{O}^{+}] = 10^{-6} = 1 \times 10^{-6} \text{ M}\).
  • For a pH of 8.0, \([\text{H}_{3}\text{O}^{+}] = 10^{-8} = 1 \times 10^{-8} \text{ M}\).
  • For a pH of 11.0, \([\text{H}_{3}\text{O}^{+}] = 10^{-11} = 1 \times 10^{-11} \text{ M}\).
  • For a pH of 9.20, \([\text{H}_{3}\text{O}^{+}] = 10^{-9.20} \approx 6.3 \times 10^{-10} \text{ M}\).
Hydroxide Ion Concentration
After determining the hydronium ion concentration, it is essential to find the hydroxide ion concentration, denoted as \[ \text{[OH}^{-}\]. This concentration is linked to \([\text{H}_{3}\text{O}^{+}]\) through the ion product of water. To calculate \([\text{OH}^{-}]\), you can use the relationship:
\[ \text{[H}_{3}\text{O}^{+}] \times \text{[OH}^{-}] = 1.0 \times 10^{-14} \].
From this, we arrive at:
\[ \text{[OH}^{-}] = \frac{1.0 \times 10^{-14}}{\text{[H}_{3}\text{O}^{+}} \].
For instance, using the previously found \([\text{H}_{3}\text{O}^{+}]\) values:
  • At pH 3.0, \([\text{H}_{3}\text{O}^{+}] = 1.0 \times 10^{-3}\), so \[ \text{[OH}^{-}] = \frac{1.0 \times 10^{-14}}{1.0 \times 10^{-3}} = 1.0 \times 10^{-11} \text{ M} \].
  • At pH 6.00, \([\text{H}_{3}\text{O}^{+}] = 1.0 \times 10^{-6}\), so \[ \text{[OH}^{-}] = 1.0 \times 10^{-8} \text{ M} \].
  • At pH 8.0, \([\text{H}_{3}\text{O}^{+}] = 1.0 \times 10^{-8}\), so \[ \text{[OH}^{-}] = 1.0 \times 10^{-6} \text{ M} \].
  • At pH 11.0, \([\text{H}_{3}\text{O}^{+}] = 1.0 \times 10^{-11}\), so \[ \text{[OH}^{-}] = 1.0 \times 10^{-3} \text{ M} \].
  • At pH 9.20, \([\text{H}_{3}\text{O}^{+}] = 6.3 \times 10^{-10}\), so \[ \text{[OH}^{-}] = \frac{1.0 \times 10^{-14}}{6.3 \times 10^{-10}} \approx 1.6 \times 10^{-5} \text{ M} \].
Ion Product of Water
The ion product of water, represented as \K_w\, is essential for understanding the relationship between hydronium and hydroxide ions. Pure water autoionizes slightly to produce \([\text{H}_{3} \text{O}^{+}]\) and \[ \text{[OH}^{-}]\] ions. This product is always equal to \1.0 \times 10^{-14} \text{ at 25掳C}\. This means:
\[\text{[H}_{3}\text{O}^{+}] \times \text{[OH}^{-}] = 1.0 \times 10^{-14}\].
Using this relationship, if one ion concentration is known, the other can be computed easily. This is particularly useful when working with acidic or basic solutions.
For instance, if \[\text{[H}_{3}\text{O}^{+}]\] is high (acidic solution), \[ \text{[OH}^{-}]\] will be low and vice versa. This balance maintains the \K_w\ value constant in water.
The ion product of water is fundamental to understanding pH and pOH calculations, ensuring precise measurement of acidity and basicity in solutions.
pH and pOH Relationship
pH and pOH are related measures of acidity and basicity. The pH measures the concentration of hydronium ions, while pOH measures the concentration of hydroxide ions. Their relationship is given by the equation:
\[ \text{pH} + \text{pOH} = 14 \text{ at 25掳C}\].
This tells us that if one is known, the other can be easily calculated. For instance:
  • If you have a solution with a pH of 3.0, the pOH = 14 - 3.0 = 11.0.
  • If the solution has a pH of 8.0, the pOH = 14 - 8.0 = 6.0.

This relationship helps in simplifying the calculation process, especially when switching between acidity and basicity measurements. Understanding this relationship is crucial for balancing chemical reactions and predicting the behavior of solutions in various scenarios.
Logarithmic Calculations
Logarithmic calculations are integral to understanding and calculating pH and pOH. The logarithmic function transforms a wide range of concentrations into a more manageable scale. Specifically, the pH is calculated using:
\[ \text{pH} = -\text{log} \text{[H}_{3}\text{O}^{+}] \].
This implies:
\[ [\text{H}_{3}\text{O}^{+}] = 10^{-\text{pH}} \].
Finding \[ \text{pOH} \] works similarly:
\[ \text{pOH} = -\text{log} \text{[OH}^{-}] \].
The usefulness of logarithms lies in simplifying calculations involving large range differences. For example, \[ [\text{H}_{3}\text{O}^{+}] \] can vary from \[ 1 M \] to \[ 1 \times 10^{-14} M \]. Using logarithms makes it easier to handle these numbers, simplifying our work with solution concentrations. These calculations are fundamental when working with acids, bases, and their reactions.

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Most popular questions from this chapter

Write the formula for each of the following acids and bases: a. barium hydroxide b. hydroiodic acid c. nitric acid d. strontium hydroxide e. sodium hydroxide f. chloric acid

In the blood plasma, \(\mathrm{pH}\) is maintained by the carbonic acid- bicarbonate buffer system. a. How is pH maintained when acid is added to the buffer system? b. How is pH maintained when base is added to the buffer system?

If \(29.7 \mathrm{~mL}\) of a \(0.205 \mathrm{M} \mathrm{KOH}\) solution is required to neutralize \(25.0 \mathrm{~mL}\) of \(\mathrm{a} \mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\) solution, what is the molarity of the acetic acid solution? \(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}(a q)+\mathrm{KOH}(a q) \longrightarrow \mathrm{KC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\)

Indicate whether each of the following is acidic, basic, or neutral: a. \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=6.0 \times 10^{-12} \mathrm{M}\) b. \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=1.4 \times 10^{-4} \mathrm{M}\) c. \(\left[\mathrm{OH}^{-}\right]=5.0 \times 10^{-12} \mathrm{M}\) d. \(\left[\mathrm{OH}^{-}\right]=4.5 \times 10^{-2} \mathrm{M}\)

Determine the \(\mathrm{pH}\) for the following solutions: a. \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=1.0 \times 10^{-8} \mathrm{M}\) b. \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=5.0 \times 10^{-2} \mathrm{M}\) c. \(\left[\mathrm{OH}^{-}\right]=3.5 \times 10^{-4} \mathrm{M}\) d. \(\left[\mathrm{OH}^{-}\right]=0.005 \mathrm{M}\)
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