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A solution was prepared by dissolving 0.0005 mol of $Ba(OH)_2$ in 100 mL of the solution. If the base is assumed to ionise completely, the $pOH$ of the solution will be

(a) 12

(b) 10

(c) unpredictable

(d) 2

$Ba(OH)_2 \longrightarrow Ba^{2+} + OH^-$

Concentration of $Ba(OH)_2 = 0.0005 \times 10 = 0.005 M$
$OH^-$concentration$= 2 \times 0.005 = 0.01 = 10^{-2}$

$pOH = -log [OH^-] = - log 10^{-2} = 2$