Acid–base Equilibria

Question

Calculate the pH of a 0.020 M Ba(OH)2 solution at 25°C (assume complete dissociation).

RankGuru · Accepted Answer

1. **Base Dissociation**: $Ba(OH)_2 	o Ba^{2+} + 2OH^-$.
2. **Stoichiometry**: Each mole of $Ba(OH)_2$ produces $2$ moles of $OH^-$.
3. **Concentration**: $[OH^-] = 2 	imes 0.020 M = 0.040 M$.
4. **pOH Calculation**:
   - $pOH = -\log(0.04) = -\log(4 	imes 10^{-2})$
   - $pOH = 2 - \log 4 = 2 - 0.602 = 1.398 \approx 1.40$.
5. **pH Relation**: $pH + pOH = 14$ at $25^\circ C$.
6. **Final Result**: $pH = 14 - 1.4 = **12.60$**.

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Acid–base Equilibria

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