Question

The solubility product at $298\mathrm{K}$is $6\times {10}^{-31}$, what is the concentration of hydroxide ions in a saturated solution?

• A. ${\left(18\times {10}^{-31}\right)}^{1/4}$
• B. ${\left(18\times {10}^{-31}\right)}^{1/2}$
• C. ${\left(2.22\times {10}^{-31}\right)}^{1/34}$
• D. ${\left(4.86\times {10}^{-29}\right)}^{1/4}$

Answer 1

The correct option is A

${\left(18\times {10}^{-31}\right)}^{1/4}$

Explanation for the correct options

(A)${\left(18\times {10}^{-31}\right)}^{1/4}$

Step 1:

• $$\begin{gathered} \mathrm{Cr}{\left(\mathrm{OH}\right)}_3\left(\mathrm{s}\right)\rightleftharpoons {\mathrm{Cr}}^{3+}\left(\mathrm{s}\right)+3{\mathrm{OH}}^{-}\left(\mathrm{s}\right) \\ \left(\mathrm{s}\right)\left(3\mathrm{s}\right) \end{gathered}$$

Step 2:

• Expression used to calculate the value of $\mathrm{s}$is given below:
• ${\mathrm{K}}_{\mathrm{sp}}=\mathrm{s}.{\left(3\mathrm{s}\right)}^3$
• putting values in the above expression:
• $\begin{array}{rcl}6\times {10}^{-31}& =& 27.{\mathrm{s}}^4\\ \mathrm{s}& =& {\left(\frac{6}{27}\times {10}^{-31}\right)}^{1/4}\end{array}$

Step 3:

• $\begin{array}{rcl}{\left(\mathrm{OH}\right)}^{-}& =& 3\mathrm{s}\\ & =& 3\times {\left(\frac{6}{27}\times {10}^{-31}\right)}^{1/4}\\ & =& {\left(18\times {10}^{-31}\right)}^{1/4}\mathrm{M}\end{array}$

Final Answer: Hence option (A) is correct., ${\left(18\times {10}^{-31}\right)}^{1/4}\mathrm{M}$.