MCQ
What is the $pH$ of $Ba{(OH)_2}$ if normality is $10$
- A$4$
- ✓$10$
- C$7$
- D$9$
✓
Answer
Correct option: B.
$10$
(b) $Ba{(OH)_2} ⇌ B{a^{2 + }} + 2O{H^ - }$
One molecule on dissociation furnishes $2O{H^ - }$ ions.
So, $[O{H^ - }] = 2 \times {10^{ - 4}}N$
$N = M \times 2$ ; $M = \frac{N}{2} = \frac{{2 \times {{10}^{ - 4}}}}{2} = {10^{ - 4}}$
$pOH = - \log [O{H^ - }] = - \log (1 \times {10^{ - 4}}) = - 4$
$pH + pOH = 14$; $pH = 14 - 4 = 10$.
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