Question

If the solubility of lithium sodium hexafluoride aluminate, $L{i}_{3}N{a}_3(Al{F}_6{)}_2$ is 's' mol l${}^{-1}$. Its solubility product is equal to:

• A. $729s^8$
• B. $12s^8$
• C. $3900s^8$
• D. $2916s^8$

Answer 1

The correct option is D $2916s^8$

${}^{\prime }{s}^{\prime }$ moles of $L{i}_{3}N{a}_3(Al{F}_6{)}_2$ gives $3s$ moles of $L{i}^{+}$, $3s$ moles of $N{a}^{+}$, $2s$ moles of $\left[Al\right(F_6){]}^{3-}$
$K_{sp}=\left[L{i}^{+}{]}^3\right[N{a}^{+}{]}^3\left[\right[Al\left(F_6\right){]}^{3-}{]}^2$

$=(3s{)}^3\times (3s{)}^3\times (2s{)}^2=3^6\times 2^2\times s^8$

$=2916S^8$

Hence, option $D$ is correct.