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6.2. Equilibrium - II (icon Equilibrium) — Chemistry STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 ScienceChemistry6.2. Equilibrium - II (icon Equilibrium)1 Mark
MCQ
What will be the $pH$ value of $0.05 \,M$ $Ba{(OH)_2}$ solution
  • A
    $12$
  • ✓
    $13$
  • C
    $1$
  • D
    $12.96$
✓

Answer

Correct option: B.
$13$
(b) $Ba{(OH)_2}\; \to \;\mathop {B{a^{ + 2}}}\limits_{.05\,M} + \mathop {2O{H^ - }}\limits_{2 \times 0.5\,M} $

$pOH = \log \frac{1}{{{{[OH]}^ - }}} = \log \frac{1}{{.1}} = 1$

$pH + pOH = 14$; $pH + 1 = 14$; $pH = 14 - 1 = 13$

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