Question

The pH value of 0.005 M Ba(OH)_2, assume Ba(OH)_2 is completely ionized, is:

• A. 5
• B. 3
• C. 10
• D. 12

Hint:

When calculating pH or pOH, remember that the sum of pH and pOH always equals 14 in aqueous solutions. Also, for a base like Ba(OH)_2, the concentration of \( OH^{-} \) ions is directly related to the concentration of the base.

Answer 1

The Correct Option is D

Ba(OH)_2 dissociates completely into \( Ba^{2+} \) and \( 2OH^{-} \). The concentration of hydroxide ions, \( [OH^{-}] \), is double the concentration of Ba(OH)_2 because each formula unit of Ba(OH)_2 provides two hydroxide ions.
- The concentration of hydroxide ions \( [OH^{-}] = 2 \times 0.005 \, \text{M} = 0.010 \, \text{M} \).
Now, we calculate the pOH using the formula: \[ pOH = -\log[OH^{-}] \] Substitute the value of \( [OH^{-}] \): \[ pOH = -\log(0.010) = 2 \] Since \( pH + pOH = 14 \), we can calculate the pH: \[ pH = 14 - pOH = 14 - 2 = 12 \] Thus, the pH value of the solution is 12.