Question

If equal volumes of $BaC{I}_2$ and $NaF$ solutions are mixed, which of these combination will not give a precipitate?

$(K_{sp}ofBa{F}_2=1.7\times {10}^{-7})$.

• A. ${10}^{-3}BaC{l}_2$ and $2\times {10}^{-2}MNaF$
• B. ${10}^{-3}MBaC{l}_2$ and $1.5\times {10}^{-2}MNaF$
• C. $1.5\times {10}^{-3}MBaC{l}_2$ and ${10}^{-2}MNaF$
• D. $2\times {10}^{-2}MBaC{l}_2$ and $2\times {10}^{-2}MNaF$

Answer 1

The correct option is C $1.5\times {10}^{-3}MBaC{l}_2$ and ${10}^{-2}MNaF$

In predicting the formation of a precipitate

Case I: When $Q_{sp}<K_{sp}$, then solution is unsaturated in which more solute can be dissolved. i.e., no precipitation.

Case II: When $Q_{sp}=K_{sp}$ , then solution is saturated in which no more solute can be dissolved but no ppt. is fomed.

Case III: When $Q_{sp}>K_{sp}$, then solution is supersaturated and precipitation takes place.

When the ionic product exceeds the solubility product, the equilibrium shifts towards left-hand side, i.e., increasing the concentration of undissociated molecules of the electrolyte. As the solvent can hold a fixed amount of electrolyte at a definite temperature, the excess of the electrolyte is thrown out from the solutions as precipitate.

Here, $1.5\times {10}^{-2}MBaC{l}_{2}and{10}^{-2}MNaF$

$Q_{sp}=\left[B{a}^{+2}\right][F^{-}{]}^2=1.5/2\times {10}^{-3}\times 1/2\times {10}^{-4}=1.5/4\times {10}^{-7}<K_{sp}$

So here ppt. will not form.