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  3. Chemistry
  4. 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

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Q. 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

Equilibrium

Solution:

$Ba(OH)_{2} \to Ba^{+2}+2OH^{-}$
Conc. of $Ba(OH)_{2}=0.0005\times10=0.005\,M$
$OH^{-}$ conc $=2 \times 0.005=0.01=10^{-2}$
$pOH=-log[OH^{-}]=-log\, 10^{-2}=2$