Given that `K_(w)` for water is `10^(-13)` `M^(2)` at `62^(@)`C, compute the sum of pOH and pH for a neutral aqueous solution at `62^(@)`C:
(a)`7.0`
(b)`13.30`
(c)`14.0`
(d)`13.0`

Options

A. `7.0`
B. `13.30`
C. `14.0`
D. `13.0`

Answer

To solve the problem, we need to compute the sum of pH and pOH for a neutral aqueous solution at 62°C, given that the ion product of water (Kw) is \(10^{-13} \, M^2\).

### Step-by-step Solution:

1. **Understand the relationship between pH, pOH, and pKw**:
The relationship is given by the equation:
\[ pH + pOH = pK_w \]
where \(pK_w\) is the negative logarithm of the ion product of water (\(K_w\)).

2. **Calculate pKw**:
We are given that:
\[ K_w = 10^{-13} \, M^2 \]
To find \(pK_w\), we use the formula:
\[ pK_w = -\log(K_w) \]
Substituting the value of \(K_w\):
\[ pK_w = -\log(10^{-13}) = 13 \]

3. **Sum of pH and pOH**:
Now, substituting \(pK_w\) back into the equation:
\[ pH + pOH = 13 \]

4. **Conclusion**:
Therefore, the sum of pH and pOH for a neutral aqueous solution at 62°C is:
\[ \boxed{13.0} \]

### Answer:
The correct option is (d) 13.0.