The value of the ion product constant for water, `(K_(w))` at `60^(@)`C is `9.6xx10^(-14)` `M^(2)` what is the `[H_(3)O^(+)]` of a neutral aqueous solutoin at `60^(@)`C and an aqueous solution with a pH=7.0 at `60 ^(@)`C are respectively?

Options

A. `3.1xx10^(-8)` acidic
B. `3.1xx10^(-7)`, neutral
C. `3.1xx10^(-8)`, basic
D. `3.1xx10^(-7)`, basic

Answer

To solve the problem, we need to find the concentration of \([H_3O^+]\) in two scenarios: a neutral aqueous solution at \(60^\circ C\) and an aqueous solution with a pH of 7.0 at \(60^\circ C\).

### Step-by-Step Solution:

1. **Understanding the Ion Product Constant for Water (\(K_w\))**:
The value of \(K_w\) at \(60^\circ C\) is given as \(9.6 \times 10^{-14} \, M^2\). This constant is defined as:
\[ K_w = [H_3O^+][OH^-] \]

2. **Finding \([H_3O^+]\) in a Neutral Solution**:
In a neutral solution, the concentrations of \([H_3O^+]\) and \([OH^-]\) are equal. Let’s denote this concentration as \(x\):
\[ K_w = x \cdot x = x^2 \]
Therefore, we can write:
\[ x^2 = 9.6 \times 10^{-14} \]
To find \(x\), we take the square root:
\[ x = \sqrt{9.6 \times 10^{-14}} \approx 3.1 \times 10^{-7} \, M \]
Thus, the concentration of \([H_3O^+]\) in a neutral solution at \(60^\circ C\) is approximately \(3.1 \times 10^{-7} \, M\).

3. **Finding \([H_3O^+]\) in a Solution with pH = 7.0**:
The pH of a solution is related to the concentration of \([H_3O^+]\) by the formula:
\[ \text{pH} = -\log[H_3O^+] \]
For a solution with pH = 7.0:
\[ 7.0 = -\log[H_3O^+] \]
To find \([H_3O^+]\), we can rearrange this equation:
\[ [H_3O^+] = 10^{-7} \, M \]
Thus, the concentration of \([H_3O^+]\) in a solution with pH = 7.0 at \(60^\circ C\) is \(1.0 \times 10^{-7} \, M\).

4. **Conclusion**:
- The concentration of \([H_3O^+]\) in a neutral aqueous solution at \(60^\circ C\) is \(3.1 \times 10^{-7} \, M\).
- The concentration of \([H_3O^+]\) in an aqueous solution with pH = 7.0 at \(60^\circ C\) is \(1.0 \times 10^{-7} \, M\).

### Summary of Results:
- \([H_3O^+]\) in neutral solution: \(3.1 \times 10^{-7} \, M\)
- \([H_3O^+]\) in solution with pH = 7.0: \(1.0 \times 10^{-7} \, M\)