Chemistry - Ionic Equilibrium Question with Solution | TestHub

$\text{100 mL}$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

$\text{= 10}\text{×}\text{0}\text{.005 mole of Ba}{\left(\text{OH}\right)}_{\text{2}}$

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${\text{OH}}^{-}\left(\text{per L}\right)=2\times 0.05=0.1$

$\left(\text{iii}\right)\because \text{pOH}\text{=}-\text{log}\left[{\text{OH}}^{-}\right]$

$\text{pOH}\text{=}-\text{log}\left[\text{0}\text{.1}\right]\text{=}-{\text{log10}}^{-\text{1}}$

$\text{pOH}\text{=}\text{log}\text{10}\text{=}\text{1}$