Question

Complete the following table: $$ \begin{array}{|l|l|l|l|} \hline\left[\mathrm{H}_{3} \mathrm{O}^{+}\right] & {\left[\mathrm{OH}^{-}\right]} & \mathrm{pH} & \begin{array}{l} \text { Acidic, Basic, } \\ \text { or Neutral? } \end{array} \\ \hline & 1 \times 10^{-6} \mathrm{M} & & \\ \hline & & 3.0 & \\ \hline 2 \times 10^{-5} \mathrm{M} & & & \\ \hline 1 \times 10^{-12} \mathrm{M} & & & \\ \hline & & 4.62 & \\ \hline \end{array} $$

Short answer

The solutions are: 1E-8 M, 8, Basic; 1E-3 M, 1E-11 M, Acidic; 5E-10 M, 4.69-4.70, Acidic; 1E-2 M, 12, Basic; 2.40E-10 M, Acidic.

Step-by-step solution

Compute missing \( [\text{H}_3\text{O}^+] \) or \( [\text{OH}^-] \)

For solutions, the product of \( [\text{H}_3\text{O}^+] \) and \( [\text{OH}^-] \) is always equal to the ion-product constant for water, \( K_w = 1.0 \times 10^{-14} \). Thus, \( [\text{H}_3\text{O}^+] = \frac{K_w}{[\text{OH}^-]} \) and vice-versa.

Calculate for the row with \( [\text{OH}^-] = 1 \times 10^{-6} \text{ M} \)

Use \( [\text{H}_3\text{O}^+] = \frac{1.0 \times 10^{-14}}{1 \times 10^{-6}} = 1 \times 10^{-8} \text{ M} \). Then, calculate the pH using \( \text{pH} = -\text{log} [\text{H}_3\text{O}^+] = 8 \). The solution is basic because \( \text{pH} > 7 \).

Calculate for the row with pH 3.0

Use \( [\text{H}_3\text{O}^+] = 10^{-\text{pH}} = 10^{-3} \text{ M} \), and \( [\text{OH}^-] = \frac{1.0 \times 10^{-14}}{10^{-3}} = 1 \times 10^{-11} \text{ M} \). The solution is acidic because \( \text{pH} < 7 \).

Calculate for the row with \( [\text{H}_3\text{O}^+] = 2 \times 10^{-5} \text{ M} \)

Use \( [\text{OH}^-] = \frac{1.0 \times 10^{-14}}{2 \times 10^{-5}} = 5 \times 10^{-10} \text{ M} \). Then, calculate the pH using \( \text{pH} = -\text{log}(2 \times 10^{-5}) \). Approximating, \( \text{pH} \) is between 4.69 and 4.70. The solution is acidic because \( \text{pH} < 7 \).

Calculate for the row with \( [\text{H}_3\text{O}^+] = 1 \times 10^{-12} \text{ M} \)

Use \( [\text{OH}^-] = \frac{1.0 \times 10^{-14}}{1 \times 10^{-12}} = 1 \times 10^{-2} \text{ M} \). Then, calculate the pH using \( \text{pH} = -\text{log}(1 \times 10^{-12}) = 12 \). The solution is basic because \( \text{pH} > 7 \).

Calculate for the row with pH 4.62

Use \( [\text{H}_3\text{O}^+] = 10^{-\text{pH}} = 10^{-4.62} \text{ M} \), and \( [\text{OH}^-] = \frac{1.0 \times 10^{-14}}{10^{-4.62}} = 2.40 \times 10^{-10} \text{ M} \). The solution is acidic because \( \text{pH} < 7 \).

Key concepts

pH Calculation

The pH of a solution is a measure of its acidity or basicity.
It is calculated using the concentration of hydronium ions, \( [\text{H}_3\text{O}^+] \).
The pH is given by the formula: \[ \text{pH} = -\log([\text{H}_3\text{O}^+]) \]
For example, if \( [\text{H}_3\text{O}^+] = 10^{-3} \text{ M} \), then: \[ \text{pH} = -\log(10^{-3}) = 3 \]
A lower pH value indicates a higher acidity, while a higher pH value indicates a higher basicity.

Ion-Product Constant

In water, the product of the concentrations of hydronium ions \( [\text{H}_3\text{O}^+] \) and hydroxide ions \( [\text{OH}^-] \) is a constant value.
This constant is known as the ion-product constant for water, \( K_w \), and its value is \( 1.0 \times 10^{-14} \) at 25°C.
This relationship is given by: \[ [\text{H}_3\text{O}^+] \times [\text{OH}^-] = K_w \]
This means that if you know the concentration of one ion, you can calculate the concentration of the other one.
For example, if \( [\text{OH}^-] = 1 \times 10^{-6} \text{ M} \), then: \[ [\text{H}_3\text{O}^+] = \frac{K_w}{[\text{OH}^-]} = \frac{1.0 \times 10^{-14}}{1 \times 10^{-6}} = 1 \times 10^{-8} \text{ M} \]

Hydronium Ion Concentration

Hydronium ions \( [\text{H}_3\text{O}^+] \) are central to the concept of acidity.
Their concentration can be directly related to the acidity of a solution.
We can calculate \( [\text{H}_3\text{O}^+] \) if we know the pH or the \( [\text{OH}^-] \).
For instance, if the pH is 4.62, then: \[ [\text{H}_3\text{O}^+] = 10^{-\text{pH}} = 10^{-4.62} \text{ M} \]
And if \( [\text{OH}^-] = 1 \times 10^{-12} \text{ M} \), then: \[ [\text{H}_3\text{O}^+] = \frac{K_w}{[\text{OH}^-]} = \frac{1.0 \times 10^{-14}}{1 \times 10^{-12}} = 1 \times 10^{-2} \text{ M} \]

Hydroxide Ion Concentration

Hydroxide ions \( [\text{OH}^-] \) determine the basicity of a solution.
Like hydronium ions, their concentration can be calculated using the ion-product constant, \( K_w \).
If the hydronium ion concentration is known, we can find \( [\text{OH}^-] \) using: \[ [\text{OH}^-] = \frac{K_w}{[\text{H}_3\text{O}^+]} \]
For instance, if \( [\text{H}_3\text{O}^+] = 2 \times 10^{-5} \text{ M} \), then: \[ [\text{OH}^-] = \frac{1.0 \times 10^{-14}}{2 \times 10^{-5}} = 5 \times 10^{-10} \text{ M} \]

Acidic and Basic Solutions

The pH scale ranges from 0 to 14 and categorizes solutions as acidic, neutral, or basic.

• Acidic: pH < 7
• Neutral: pH = 7
• Basic: pH > 7

The pH depends on the balance between hydronium and hydroxide ions.
For example, if the pH is 3.0, the solution is acidic.
For a pH of 12, the solution is basic.
Understanding these principles helps in predicting the behavior and reactivity of solutions in different chemical contexts.