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Calculate \(\left[\mathrm{H}^{+}\right]\)for each of the following solutions, and indicate whether the solution is acidic, basic, or neutral: (a) \([\mathrm{OH}]=0.00045 \mathrm{M} ;\) (b) \([\mathrm{OH}]=8.8 \times 10^{-9} \mathrm{M}\); (c) a solution in which \(\left[\mathrm{OH}^{-}\right]\)is 100 times greater than \(\left[\mathrm{H}^{+}\right]\).

Short Answer

Expert verified
For the given solutions: (a) \([\mathrm{H}^{+}] = 2.22 \times 10^{-11} \mathrm{M}\) and the solution is basic, (b) \([\mathrm{H}^{+}] = 1.14 \times 10^{-6} \mathrm{M}\) and the solution is acidic, (c) \([\mathrm{H}^{+}] = 1 \times 10^{-8} \mathrm{M}\) and the solution is basic.

Step by step solution

01

Calculate \([\mathrm{H}^{+}]\) for solution (a) using the ion product constant of water#

To find the concentration of hydrogen ions in solution (a), we can use the ion product constant for water, which states that: \(K_w = [\mathrm{H}^{+}] [\mathrm{OH}^{-}]\). We know that \(K_w = 1 \times 10^{-14}\) at 25°C and that \([\mathrm{OH}^{-}] = 0.00045 \mathrm{M}\). We can rearrange the equation to solve for \([\mathrm{H}^{+}]\): \[[\mathrm{H}^{+}] = \frac{K_w}{[\mathrm{OH}^{-}]} = \frac{1 \times 10^{-14}}{0.00045}\]
02

Evaluate \([\mathrm{H}^{+}]\) for solution (a) and determine its acidity#

Using a calculator, we find that: \[[\mathrm{H}^{+}] = \frac{1 \times 10^{-14}}{0.00045} = 2.22 \times 10^{-11} \mathrm{M}\] Since \([\mathrm{H}^{+}] < [\mathrm{OH}^{-}]\), solution (a) is basic.
03

Calculate \([\mathrm{H}^{+}]\) for solution (b) using the ion product constant of water#

Similar to solution (a), we can use the ion product constant for water to find the concentration of hydrogen ions in solution (b). We know \([\mathrm{OH}^{-}] = 8.8 \times 10^{-9}\mathrm{M}\), so: \[[\mathrm{H}^{+}] = \frac{K_w}{[\mathrm{OH}^{-}]} = \frac{1 \times 10^{-14}}{8.8 \times 10^{-9}}\]
04

Evaluate \([\mathrm{H}^{+}]\) for solution (b) and determine its acidity#

Using a calculator, we find that: \[[\mathrm{H}^{+}] = \frac{1 \times 10^{-14}}{8.8 \times 10^{-9}} = 1.14 \times 10^{-6} \mathrm{M}\] Since \([\mathrm{H}^{+}] > [\mathrm{OH}^{-}]\), solution (b) is acidic.
05

Calculate \([\mathrm{H}^{+}]\) for solution (c) using the given relationship between \([\mathrm{H}^{+}]\) and \([\mathrm{OH}^{-}]\)#

We're given that \(\left[\mathrm{OH}^{-}\right]\) in solution (c) is 100 times greater than \(\left[\mathrm{H}^{+}\right]\), so we can write: \[[\mathrm{OH}^{-}] = 100 [\mathrm{H}^{+}]\] Now, we can use the ion product constant for water to find the concentration of hydrogen ions: \[[\mathrm{H}^{+}] [\mathrm{OH}^{-}] = K_w\] Substitute the relationship between the concentrations: \[[\mathrm{H}^{+}] (100 [\mathrm{H}^{+}]) = 1 \times 10^{-14}\]
06

Solve for \([\mathrm{H}^{+}]\) in solution (c) and determine its acidity#

Solve the equation for \([\mathrm{H}^{+}]\): \[[\mathrm{H}^{+}] = \sqrt{\frac{1 \times 10^{-14}}{100}} = \sqrt{1 \times 10^{-16}} = 1 \times 10^{-8} \mathrm{M}\] Since \(\left[\mathrm{OH}^{-}\right] = 100 [\mathrm{H}^{+}] = 1 \times 10^{-6} \mathrm{M}\) and \(\left[\mathrm{H}^{+}\right] = 1 \times 10^{-8} \mathrm{M}\), we can conclude that \([\mathrm{H}^{+}] < [\mathrm{OH}^{-}]\), making solution (c) basic.

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Key Concepts

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

pH Calculations
Understanding pH calculations is crucial in acid-base chemistry. pH is a measure of how acidic or basic a solution is. The formula for calculating pH is:
\[\text{pH} = -\log_{10} [\mathrm{H}^+]\]
This formula tells us the pH based on the concentration of hydrogen ions \(\mathrm{H}^+\) in a solution. A lower pH value (below 7) indicates acidity, while a higher pH value (above 7) indicates basicity.
For neutral water at 25°C, the pH is 7. By computing the pH from the \(\mathrm{H}^+\) concentration, we can determine the nature of the solution:
  • pH < 7: Acidic
  • pH = 7: Neutral
  • pH > 7: Basic
Ion Product Constant of Water
The ion product constant of water, denoted as \(K_w\), is a fundamental concept in acid-base chemistry.
At 25°C, \(K_w\) is valued at \(1 \times 10^{-14}\), signifying the product of the concentrations of hydrogen ions \(\mathrm{H}^+\) and hydroxide ions \(\mathrm{OH}^-\) in pure water:
\[K_w = [\mathrm{H}^+][\mathrm{OH}^-] = 1 \times 10^{-14}\]
This constant helps in identifying the concentration of either ion if the other is known. For instance, if \(\mathrm{OH}^-\) is known, you can find \(\mathrm{H}^+\) by rearranging the equation:
\[[\mathrm{H}^+] = \frac{K_w}{[\mathrm{OH}^-]}\]
This relationship is essential for calculations related to solution acidity and basicity.
Solution Acidity Determination
Determining solution acidity involves analyzing \(\mathrm{H}^+\) and \(\mathrm{OH}^-\) concentrations. By comparing these concentrations, we can classify a solution as acidic, basic, or neutral.
Here's how you can determine acidity based on concentrations:
  • If \(\mathrm{H}^+ > \mathrm{OH}^-\): The solution is acidic.
  • If \(\mathrm{H}^+ < \mathrm{OH}^-\): The solution is basic.
  • If \(\mathrm{H}^+ = \mathrm{OH}^-\): The solution is neutral.
This analysis provides insight into the relative strength of acids and bases. It underscores the central role of hydrogen and hydroxide ions in determining a solution's character. Understanding and applying these principles allow you to predict how a solution will behave chemically.

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Most popular questions from this chapter

Calculate the pH of each of the following strong acid solutions: (a) \(8.5 \times 10^{-3} \mathrm{M} \mathrm{HBr}\), (b) \(1.52 \mathrm{~g}\) of \(\mathrm{HNO}_{3}\) in \(575 \mathrm{~mL}\) of solution, (c) \(5.00 \mathrm{~mL}\) of \(0.250 \mathrm{M} \mathrm{HClO}_{4}\) diluted to \(50.0 \mathrm{~mL}\), (d) a solution formed by mixing \(10.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{HBr}\) with \(20.0 \mathrm{~mL}\) of \(0.200 \mathrm{M} \mathrm{HCl}\).

The iodate ion is reduced by sulfite according to the following reaction: $$ \mathrm{IO}_{3}^{-}(a q)+3 \mathrm{SO}_{3}^{2-}(a q) \longrightarrow \Gamma(a q)+3 \mathrm{SO}_{4}^{2-}(a q) $$ The rate of this reaction is found to be first order in \(\mathrm{IO}_{3}^{-}\), first order in \(\mathrm{SO}_{3}^{2-}\), and first order in \(\mathrm{H}^{+}\). (a) Write the rate law for the reaction. (b) By what factor will the rate of the reaction change if the \(\mathrm{pH}\) is lowered from \(5.00\) to \(3.50\) ? Does the reaction proceed more quickly or more slowly at the lower \(\mathrm{pH}\) ? (c) By using the concepts discussed in Section 14.6, explain how the reaction can be \(\mathrm{pH}\)-dependent even though \(\mathrm{H}^{+}\)does not appear in the overall reaction.

Write the chemical equation and the \(K_{b}\) expression for the reaction of each of the following bases with water: (a) dimethylamine, \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH}_{\text {; }}\) (b) carbonate ion, \(\mathrm{CO}_{3}^{2-} ;\) (c) formate ion, \(\mathrm{CHO}_{2}^{-}\).

Is each of the following statements true or false? (a) All strong acids contain one or more \(\mathrm{H}\) atoms. (b) A strong acid is a strong electrolyte. (c) A \(1.0-M\) solution of a strong acid will have \(\mathrm{pH}=1.0\).

Pyridinium bromide \(\left(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NHBr}\right)\) is a strong electrolyte that dissociates completely into $\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+}\( and \)\mathrm{Br}^{-}$. An aqueous solution of pyridinium bromide has a pH of \(2.95 .\) (a) Write out the reaction that leads to this acidic pH. (b) Using Appendix D, calculate the \(K_{a}\) for pyridinium bromide. (c) A solution of pyridinium bromide has a pH of 2.95 . What is the concentration of the pyridinium cation at equilibrium, in units of molarity?

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