Question

How much KBr should be added to 1 L of 0.05 M $AgN{{O}_{3}}$ solution just to start precipitation of AgBr? ${{K}_{sp}}$ of AgBr = $5\times {{10}^{-13}}$ .

Answer 1

Hint: The ${{K}_{sp}}$ is the simplified equilibrium constant which is defined as the solubility product constant for equilibrium between a solid and its respective ions in a solution. The ionisation and their relations between the respective solids can help solve this problem better.

Complete step-by-step answer: Let us solve the illustration in detail;
${{K}_{sp}}$of AgBr means;
${{K}_{sp}}$ = $\left[ A{{g}^{+}} \right]\left[ B{{r}^{-}} \right]=5\times {{10}^{-13}}$
Now, we have given concentration of $A{{g}^{+}}=0.05mol/L$; as, one mole of $A{{g}^{+}}$ is there per mole of $AgN{{O}_{3}}$.
So, now;
$ 5\times {{10}^{-2}}\times \left[ B{{r}^{-}} \right]=5\times {{10}^{-13}} \\
 \left[ B{{r}^{-}} \right]=1\times {{10}^{-11}}mol/L \\
$
So, if we consider a plane stoichiometry i.e. mole ratio as 1 : 1 and thus, the KBr solution as the concentrated one the, we can say that we will require $1\times {{10}^{-11}}mol$ KBr for precipitation as the volume won’t change with addition of one or two drops of KBr.

Note: Do note the definition of ${{K}_{sp}}$ and hence the solution we solved right now is similar to each other, just we haven’t mentioned again about the general facts. Secondly, use units carefully as molarity is denoted by M and also as mol/L.
During precipitation, if we exceed one more drop than as described above the solution will turn into cloudy white colour which on exposure to sunlight can form a photographic film (different concept).