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Chapter 15: Problem 154

The \(\mathrm{pH}\) of a solution is 4 . (a) What is the \(\mathrm{H}_{3} \mathrm{O}^{+}\) concentration? (b) What is the \(\mathrm{OH}^{-}\) concentration? (c) Is this solution acidic or basic?

Short Answer

Expert verified
(a) The \(H_3O^+\) concentration in the solution is \(10^{-4} M\). (b) The \(OH^-\) concentration in the solution is \(10^{-10} M\). (c) Since the \(H_3O^+\) concentration is higher than the \(OH^-\) concentration, the solution is acidic.

Step by step solution

01

(a) Calculating \(H_3O^+\) concentration

To calculate the \(H_3O^+\) concentration, we need to use the formula relating pH and \(H_3O^+\) concentration, which is: \[pH = -\log_{10}(H_3O^+)\] Given that the pH of the solution is 4, we have: \[4 = -\log_{10}(H_3O^+)\] We need to solve this equation for \(H_3O^+\). First, we will take the power of 10 on both sides: \[10^4 = 10^{-\log_{10}(H_3O^+)}\] Now, since \(10^{-\log_{10}(x)}=x\), we have: \[H_3O^+ = 10^{-4}\]
02

(a) Result

Thus, the \(H_3O^+\) concentration in the solution is \(10^{-4} M\).
03

(b) Calculating \(OH^-\) concentration

To calculate the \(OH^-\) concentration, we need to use the ion-product constant of water, which is given by: \[K_w = [H_3O^+] [OH^-]\] The ion-product constant of water, \(K_w\), is equal to \(1.0 \times 10^{-14} M^2\). Given that we know the \(H_3O^+\) concentration, we can find the \(OH^-\) concentration by rearranging the equation and solving for the \(OH^-\) concentration: \[[OH^-] = \frac{K_w}{[H_3O^+]}\] Now, substitute the given values: \[[OH^-] = \frac{1.0 \times 10^{-14} M^2}{10^{-4} M}\]
04

(b) Result

Thus, the \(OH^-\) concentration in the solution is \(10^{-10} M\).
05

(c) Determining if the solution is acidic or basic

To determine if the solution is acidic or basic, we can compare the concentrations of \(H_3O^+\) and \(OH^-\). If the concentration of \(H_3O^+\) is higher than the concentration of \(OH^-\), the solution is acidic. If the concentration of \(OH^-\) is higher than the concentration of \(H_3O^+\), the solution is basic. In this case, we have \( H_3O^+ = 10^{-4} M \) and \( OH^- = 10^{-10} M \). Since the \(H_3O^+\) concentration is higher than the \(OH^-\) concentration, the solution is acidic.

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Key Concepts

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

H3O+ concentration
The concentration of hydronium ions (\(H_3O^+\)) in a solution can be easily determined using the solution's pH. The pH scale is a measure of how acidic or basic a solution is, ranging from 0 to 14, where a pH of 7 is neutral. Solutions with a pH less than 7 are acidic, and those with a pH greater than 7 are basic.
To find the \(H_3O^+\) concentration, you use the following relationship:\[pH = -\log_{10}(H_3O^+)\]
To isolate \(H_3O^+\), we essentially reverse the logarithm by exponentiating, resulting in:\[H_3O^+ = 10^{-pH}\]
For instance, if a solution has a pH of 4, the \(H_3O^+\) concentration would be:\[H_3O^+ = 10^{-4} \text{ M}\]
This indicates a relatively high presence of hydronium ions, characteristic of an acidic solution.
OH- concentration
The concentration of hydroxide ions (\(OH^-\)) in a solution is related to the \(H_3O^+\) concentration through the ion-product constant of water (\(K_w\)). This constant at 25掳C is always:\[K_w = [H_3O^+][OH^-] = 1.0 \times 10^{-14} \text{ M}^2\]
When you know one ion's concentration, you can easily calculate the other. First, rearrange the equation:\[[OH^-] = \frac{K_w}{[H_3O^+]}\]
For a solution with a known \(H_3O^+\) concentration of \(10^{-4} \text{ M}\), the \(OH^-\) concentration is found by:\[OH^- = \frac{1.0 \times 10^{-14} \text{ M}^2}{10^{-4} \text{ M}} = 10^{-10} \text{ M}\]
This lower concentration of \(OH^-\) ions supports the acidic nature of the solution.
acidic or basic solution determination
Determining whether a solution is acidic or basic is an essential step in understanding its chemical behavior. This is done by comparing the concentrations of \(H_3O^+\) and \(OH^-\).
  • If \([H_3O^+] > [OH^-] \), the solution is acidic.
  • If \([H_3O^+] < [OH^-] \), the solution is basic.
  • \([H_3O^+] = [OH^-] \) implies a neutral solution.
In our example, the solution has a \(H_3O^+\) concentration of \(10^{-4} \text{ M}\) and an \(OH^-\) concentration of \(10^{-10} \text{ M}\).
Since \(H_3O^+\) is greater than \(OH^-\), the solution is definitely acidic, reinforcing its pH level of 4 as an indicator of acidity.

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Most popular questions from this chapter

(a) List all the weak bases phosphoric acid, \(\mathrm{H}_{3} \mathrm{PO}_{4}\), can produce via successive losses of its protons. (b) Of the bases you listed, which has no ability to serve as a weak acid? (c) Phosphorous acid, \(\mathrm{H}_{3} \mathrm{PO}_{3}\), is a diprotic acid, even though there are three \(\mathrm{H}\) atoms in the formula. Draw a Lewis dot diagram for phosphorous acid. (Hint: Only H bound directly to \(\mathrm{O}\) is acidic).

Consider the autoionization of water. (a) What do we mean by autoionization of water (b) Write a balanced equation to go along with your explanation. (c) An alternate name for autoionization of water is autodissociation. Exactly how is water dissociating?

Ethanol, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\), and hydride ion, \(\mathrm{H}^{-}\), react to produce \(\mathrm{H}_{2}\) gas and the \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{O}^{-}\) anion. Arrhenius could not tell you what is going on, but Bronsted and Lowry would have no trouble. How would they explain this reaction?

Without calculating \(\mathrm{pH}\) values, list these solutions in order of acidity, from lowest \(\mathrm{pH}\) to highest \(\mathrm{pH}\) : (a) \(0.10 \mathrm{M} \mathrm{LiNO}_{3}\) (b) \(0.10 \mathrm{M} \mathrm{NaF}\) (c) \(0.10 \mathrm{M} \mathrm{KOH}\) (d) \(0.10 \mathrm{M} \mathrm{HCN}\) (e) \(0.10 \mathrm{M} \mathrm{HNO}_{3}\) (f) \(0.20 \mathrm{M} \mathrm{HCl}\)

Pyridine, \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{~N}\), is a weak base. (a) Write the chemical equation for the reaction between pyridine and water. (b) List all species present in an aqueous solution of pyridine in order of concentration, highest to lowest. (c) In a 0.100 M pyridine solution, \(3.2 \%\) of the pyridine has reacted with water to form products. Calculate the concentration of all species present (except water) in \(1 \mathrm{~L}\) of a \(0.100\) M pyridine solution. (d) What is the pH of this solution?
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