pH of a saturated solution of Ba(OH)^ 2 is 12 . The value of its K...

Given that,

pH of $Ba\left(OH\right)^{2} = 12$

$∵ pH + pOH = 14$

$∴ pOH = 14 - pH = 14 - 12 = 2$

Now, $pOH = - \log \left[{OH}^{-}\right]$

$⇒ 2 = - \log \left[OH\right]</p><p>\left[{OH}^{-}\right] =${10}^{- 2}M</p><p>\mathop{Ba\left(OH\$\right)^{2}\left(s\right)_{{SmolL}^{- 1}} ⇌ \mathop{Ba2 +}_{S} + \mathop{2{OH}^{-}}_{2S}

\left[{OH}^{-}\right]=2S = {10}^{- 2}M$</p><p>$∴ S = 0. 5 $\times {10}^{- 2} =$5 $\times {10}^{- 3}$ M$</p><p>${K}_{sp} = \left[{Ba}^{2 +}\right]\left[{OH}^{-}\right]2$</p><p>$⇒ ( 5 \times {10}^{- 3} ) {( {10}^{- 2} )}^{2} =5 \times {10}^{- 7}M^{3}$</p><p>Therefore, the value of${K}_{sp}$is$5. 0 \times {10}^{- 7}M^{3}$.</p>$