Question

If the solubility of ${Li}_3{Na}_3{\left(Al{F}_6\right)}_2$ is $xmol{L}^{-1}$, then its solubility product is equal to:

• A. $12x^3$
• B. $18x^3$
• C. $x^8$
• D. $2916x^8$

Answer 1

The correct option is D $2916x^8$

${Li}_3{Na}_3{\left({AlF}_6\right)}_2\rightleftharpoons {3Li}^{1+}+3{Na}^{+}+2{AlF}_6^{3-}$

$x$ $3x$ $3x$ $2x$

The solubility product $={\left[{Li}^{+}\right]}^3{\left[{Na}^{+}\right]}^3{\left[{AlF}_6^{3-}\right]}^2$

$={\left(3x\right)}^3{\left(3x\right)}^3{\left(2x\right)}^2$

$=2916x^8$