Question

question 4 (1 point)
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ph5. select all the correct statement(s) about a solution with a ph of 3.7.
$\ce{h_{3}o^{+}} < 1.0 \times 10^{-7}$ m
$\ce{h_{3}o^{+}} < \ce{oh^{-}}$
$\ce{oh^{-}} < 1.0 \times 10^{-7}$ m
$\ce{h_{3}o^{+}} > \ce{oh^{-}}$
it is acidic

Explanation:

Step1: Recall pH and ion concentrations

The pH of a solution is related to the hydronium ion concentration \([H_3O^+]\) by the formula \(pH = -\log_{10}[H_3O^+]\). For a neutral solution at 25°C, \(pH = 7\), \([H_3O^+] = [OH^-] = 1.0\times10^{-7}\ M\), and \(K_w = [H_3O^+][OH^-] = 1.0\times10^{-14}\) (at 25°C).

Step2: Calculate \([H_3O^+]\) for pH = 3.7

Using \(pH = -\log_{10}[H_3O^+]\), we can solve for \([H_3O^+]\):

$$ [H_3O^+] = 10^{-pH} = 10^{-3.7} \approx 2.0\times10^{-4}\ M $$

This is greater than \(1.0\times10^{-7}\ M\), so the first statement (\([H_3O^+] < 1.0\times10^{-7}\ M\)) is false.

Step3: Calculate \([OH^-]\)

Using \(K_w = [H_3O^+][OH^-]\), we can find \([OH^-]\):

$$ [OH^-] = \frac{K_w}{[H_3O^+]} = \frac{1.0\times10^{-14}}{2.0\times10^{-4}} = 5.0\times10^{-11}\ M $$

Step4: Compare \([H_3O^+]\) and \([OH^-]\)

We have \([H_3O^+] \approx 2.0\times10^{-4}\ M\) and \([OH^-] = 5.0\times10^{-11}\ M\). So \([H_3O^+] > [OH^-]\), which means the statement \([H_3O^+] < [OH^-]\) is false, and \([H_3O^+] > [OH^-]\) is true. Also, \([OH^-] = 5.0\times10^{-11}\ M < 1.0\times10^{-7}\ M\), so the statement \([OH^-] < 1.0\times10^{-7}\ M\) is true.

Step5: Determine if the solution is acidic

A solution is acidic if \(pH < 7\) (or \([H_3O^+] > [OH^-]\)). Since \(pH = 3.7 < 7\) and \([H_3O^+] > [OH^-]\), the solution is acidic, so the "It is acidic" statement is true.

Answer:

The correct statements are:

• \(\boldsymbol{[OH^-] < 1.0\times10^{-7}\ M}\)
• \(\boldsymbol{[H_3O^+] > [OH^-]}\)
• \(\boldsymbol{\text{It is acidic}}\)