A `0.025 m` solution of monobasic acids has a freezing point of `-0.060^(@)C`. What are `K_(a)` and `pK_(a)` of the acid? `(K_(f)=1.86^(@)C)`

Explanation

To solve the problem, we need to determine the acid dissociation constant (\(K_a\)) and the negative logarithm of the dissociation constant (\(pK_a\)) for a monobasic acid given its molality and freezing point depression.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Molarity of the acid solution, \(m = 0.025 \, \text{mol/kg}\)
- Freezing point depression, \(\Delta T_f = 0 - (-0.060) = 0.060 \, ^\circ C\)
- Freezing point constant for water, \(K_f = 1.86 \, ^\circ C/\text{molal}\)
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