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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°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

How many moles of \(\mathrm{Ba}(\mathrm{OH})_{2}\) does it take to neutralize \(0.50\) mole of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) ? Why does the resulting solution after the neutralization no longer conduct electricity? Why does the solution then conduct electricity after an excess of \(\mathrm{Ba}(\mathrm{OH})_{2}\) is added?

In water, the hydrogen phosphate ion, \(\mathrm{HPO}_{4}^{2-}\), can act both as a weak base and as a weak acid. (a) Write an equilibrium equation that shows hydrogen phosphate acting as a weak acid in water. (b) Write an equilibrium equation that shows bisulfate acting as a weak base in water. (c) What information would you need to help you determine whether a solution of hydrogen phosphate is going to be slightly acidic or slightly basic? How would you use that information to find out?

Hydride ion, \(\mathrm{H}\), is an exceptionally strong base, reacting with water to produce lots of hydroxide ion and \(\mathrm{H}_{2}\) gas. The \(K_{\mathrm{eq}}\) for this reaction is huge. (a) Write the balanced equation for the reaction between hydride and water. (b) Explain why \(\mathrm{H}_{2}\) gas forms. (Hint: Use the Bronsted-Lowry definition of base.)

Without using a calculator, what is the base- 10 logarithm of \(10^{0}\) ? Of 1?

True or false? In an aqueous solution at \(25^{\circ} \mathrm{C}\), you will always get the same number when you multiply the equilibrium \(\mathrm{H}_{3} \mathrm{O}^{+}\) concentration by the equilibrium OH \(^{-}\) concentration. Explain your answer.

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