Q. $0.005$ mol of $\text{Ba} \left(\text{OH}\right)_{\text{2}} \, $ is dissolved in $\text{100 mL}$ of water. Assuming complete ionisation of $\text{Ba} \left(\text{OH}\right)_{\text{2}}$ the $\text{pOH}$ of the solution will be

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A

1

B

5

C

2

D

14

Solution:

$\text{100 mL}$ of solution contain 0.005 moles of $\text{Ba} \left(\text{OH}\right)_{\text{2}} \, $
$\therefore \, \text{1} \, \text{L } \left(\text{1000 mL}\right)$ of solution contain
$\text{= 10} \, \text{\times } \, \text{0} \text{.005 mole of Ba} \left(\text{OH}\right)_{\text{2}}$
Concentration of $\text{Ba} \left(\text{OH}\right)_{\text{2}}$ (i.e. moles on litres)
$\text{= 0} \text{.05 M}$
Each $\text{Ba} \left(\text{OH}\right)_{\text{2}}$ give $\text{2OH}^{-}$ ions
Thus, moles of $\left(\text{OH}\right)^{-} \left(\text{per L}\right) = 2 \times 0.05 = 0.1$
$\left(\text{iii}\right) \, \because \text{pOH} \, \text{=} \, - \text{log} \left[\left(\text{OH}\right)^{-}\right]$
$\text{pOH} \, \text{=} \, - \text{log} \left[\text{0} \text{.1}\right] \, \text{=} \, - \text{log10}^{- \text{1}}$
$\text{pOH} \, \text{=} \, \text{log} \, \text{10} \, \text{=} \, \text{1}$

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