Calculate the `[H^(o+)]` and `[overset(Theta)OH]` of `0.0315g` of `HNO_(3)` in `500 mL` of water. Calculate `pH` and `pOH` also.

Explanation

To solve the problem, we need to calculate the concentration of hydrogen ions \([H^+]\) and hydroxide ions \([OH^-]\) for a solution of \(0.0315 \, g\) of \(HNO_3\) in \(500 \, mL\) of water. We will also calculate the pH and pOH of the solution.

### Step 1: Calculate the Molar Mass of \(HNO_3\)

The molar mass of \(HNO_3\) can be calculated as follows:
- Hydrogen (H): \(1 \, g/mol\)
- Nitrogen (N): \(14 \, g/mol\)
- Oxygen (O): \(16 \, g/mol \times 3 = 48 \, g/mol\)

Adding these together:
\[ \text{Molar mass of } HNO_3 = 1 + 14 + 48 = 63 \, g/mol \]

### Step 2: Calculate the Number of Moles of \(HNO_3\)

Using the formula:
\[ \text{Moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]
Substituting the values:
\[ \text{Moles of } HNO_3 = \frac{0.0315 \, g}{63 \, g/mol} \approx 0.0005 \, mol \]

### Step 3: Calculate the Volume of the Solution in Liters

Convert \(500 \, mL\) to liters:
\[ 500 \, mL = \frac{500}{1000} = 0.5 \, L \]

### Step 4: Calculate the Molarity of the Solution

Using the formula for molarity:
\[ \text{Molarity} = \frac{\text{moles of solute}}{\text{volume of solution in L}} \]
Substituting the values:
\[ \text{Molarity} = \frac{0.0005 \, mol}{0.5 \, L} = 0.001 \, mol/L = 1 \times 10^{-3} \, mol/L \]

### Step 5: Calculate the Concentration of \([H^+]\)

Since \(HNO_3\) is a strong acid, it completely dissociates in solution:
\[ HNO_3 \rightarrow H^+ + NO_3^- \]
Thus, the concentration of \([H^+]\) is equal to the molarity of the solution:
\[ [H^+] = 1 \times 10^{-3} \, mol/L \]

### Step 6: Calculate the Concentration of \([OH^-]\)

Using the ion product of water at \(25^\circ C\):
\[ K_w = [H^+][OH^-] = 1 \times 10^{-14} \]
We can rearrange this to find \([OH^-]\):
\[ [OH^-] = \frac{K_w}{[H^+]} = \frac{1 \times 10^{-14}}{1 \times 10^{-3}} = 1 \times 10^{-11} \, mol/L \]

### Step 7: Calculate the pH of the Solution

Using the formula for pH:
\[ \text{pH} = -\log[H^+] \]
Substituting the value:
\[ \text{pH} = -\log(1 \times 10^{-3}) = 3 \]

### Step 8: Calculate the pOH of the Solution

Using the relationship between pH and pOH:
\[ \text{pH} + \text{pOH} = 14 \]
Thus,
\[ \text{pOH} = 14 - \text{pH} = 14 - 3 = 11 \]

### Summary of Results

- \([H^+] = 1 \times 10^{-3} \, mol/L\)
- \([OH^-] = 1 \times 10^{-11} \, mol/L\)
- \(\text{pH} = 3\)
- \(\text{pOH} = 11\)