The solubility product of Cr(OH)3 at 298 K is 6.0 × 10–31. The concentration of hydroxide ions in a saturated solution of Cr(OH)3 will be :
Solution
<table class="tg">
<tbody><tr>
<th class="tg-nrix">Cr(OH)<sub>3</sub></th>
<th class="tg-nrix">⇌</th>
<th class="tg-nrix">Cr<sup>+3</sup></th>
<th class="tg-nrix">+</th>
<th class="tg-nrix">3OH<sup>-</sup></th>
</tr>
<tr>
<td class="tg-nrix"></td>
<td class="tg-nrix"></td>
<td class="tg-nrix">S</td>
<td class="tg-nrix"></td>
<td class="tg-nrix">3S</td>
</tr>
</tbody></table>
<br><br>K<sub>sp</sub> = [Cr<sup>3+</sup>] [OH<sup>–</sup>
]<sup>3</sup>
<br><br>$\Rightarrow$ 6 × 10<sup>–31</sup> = S × (3S)<sup>3</sup>
<br><br>$\Rightarrow$ 6 × 10<sup>–31</sup> = 27 S<sup>4</sup>
<br><br>S = ${\left( {{6 \over {27}} \times {{10}^{31}}} \right)^{{1 \over 4}}}$
<br><br>As [OH<sup>–</sup>] = 3S
<br><br>= $3{\left( {{6 \over {27}} \times {{10}^{31}}} \right)^{{1 \over 4}}}$
<br><br>= (18 × 10<sup>–31</sup>)<sup>1/4</sup> M
About this question
Subject: Chemistry · Chapter: Equilibrium · Topic: Chemical Equilibrium and Kc, Kp
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