Which of the following solution will have `pH = 13`?

Options

A. `2g NaOH "in" 500 mL` solution.
B. `100 mL` solution fo `0.05M Ca(OH)_(2)`.
C. `100mL` solution of `1.0N Ca(OH)_(2)`.
D. `4gNaOH` in `500 mL` solution.

Explanation

To determine which of the following solutions will have a pH of 13, we need to analyze the concentration of hydroxide ions (OH⁻) in each solution, as pH and pOH are related through the equation:

\[ \text{pH} + \text{pOH} = 14 \]

Given that a pH of 13 corresponds to a pOH of 1, we can find the concentration of OH⁻ ions using the formula:

\[ \text{pOH} = -\log[\text{OH}^-] \]

From this, we can deduce that:

\[ [\text{OH}^-] = 10^{-\text{pOH}} = 10^{-1} = 0.1 \, \text{M} \]

Thus, we are looking for solutions that provide a hydroxide ion concentration of 0.1 M.

Let's evaluate each option step by step:

### Step 1: Evaluate Option 1
**2 grams of NaOH in 500 mL of solution.**

1. Calculate the molarity:
- Molar mass of NaOH = 40 g/mol.
- Moles of NaOH = \(\frac{2 \, \text{g}}{40 \, \text{g/mol}} = 0.05 \, \text{mol}\).
- Volume of solution in liters = \(\frac{500 \, \text{mL}}{1000} = 0.5 \, \text{L}\).
- Molarity (M) = \(\frac{0.05 \, \text{mol}}{0.5 \, \text{L}} = 0.1 \, \text{M}\).

2. Since NaOH is a strong base, it completely dissociates:
- \[ \text{NaOH} \rightarrow \text{Na}^+ + \text{OH}^- \]
- Therefore, \([\text{OH}^-] = 0.1 \, \text{M}\).

**Conclusion:** This option is correct.

### Step 2: Evaluate Option 2
**100 mL of 0.05 M calcium hydroxide.**

1. Calculate the concentration of OH⁻:
- Calcium hydroxide dissociates as follows:
- \[ \text{Ca(OH)}_2 \rightarrow \text{Ca}^{2+} + 2\text{OH}^- \]
- Therefore, 0.05 M of Ca(OH)₂ will produce \(0.05 \times 2 = 0.1 \, \text{M}\) of OH⁻.

**Conclusion:** This option is also correct.

### Step 3: Evaluate Option 3
**100 mL of 1 N calcium hydroxide.**

1. Understand normality:
- 1 normal (N) means 1 equivalent of OH⁻ per liter.
- Since calcium hydroxide provides 2 OH⁻ ions, the molarity (M) is:
- \[ \text{Molarity} = \frac{1 \, \text{N}}{2} = 0.5 \, \text{M} \]
- Thus, the concentration of OH⁻ ions is \(0.5 \, \text{M} \).

**Conclusion:** This option is incorrect because we need 0.1 M.

### Step 4: Evaluate Option 4
**4 grams of NaOH in 500 mL of solution.**

1. Calculate the molarity:
- Moles of NaOH = \(\frac{4 \, \text{g}}{40 \, \text{g/mol}} = 0.1 \, \text{mol}\).
- Volume of solution in liters = 0.5 L.
- Molarity (M) = \(\frac{0.1 \, \text{mol}}{0.5 \, \text{L}} = 0.2 \, \text{M}\).

2. Since NaOH is a strong base:
- \([\text{OH}^-] = 0.2 \, \text{M}\).

**Conclusion:** This option is incorrect because we need 0.1 M.

### Final Conclusion:
The solutions with pH = 13 are:
- Option 1: 2 grams of NaOH in 500 mL.
- Option 2: 100 mL of 0.05 M calcium hydroxide.