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.