Chemical and Ionic Equilibrium 2 Question 24
24. When equal volumes of the following solutions are mixed, precipitation of $\mathrm{AgCl}\left(K_{\mathrm{sp}}=1.8 \times 10^{-10}\right)$ will occur only with
$(1988,1 \mathrm{M})$
(a) $10^{-4} \mathrm{M}\left(\mathrm{Ag}^{+}\right)$and $10^{-4} \mathrm{M}\left(\mathrm{Cl}^{-}\right)$
(b) $10^{-5} \mathrm{M}\left(\mathrm{Ag}^{+}\right)$and $10^{-5} \mathrm{M}\left(\mathrm{Cl}^{-}\right)$
(c) $10^{-6} \mathrm{M}\left(\mathrm{Ag}^{+}\right)$and $10^{-6} \mathrm{M}\left(\mathrm{Cl}^{-}\right)$
(d) $10^{-10} \mathrm{M}\left(\mathrm{Ag}^{+}\right)$and $10^{-10} \mathrm{M}\left(\mathrm{Cl}^{-}\right)$
Show Answer
Solution:
- For precipitation to occur, $K_{\text {sp }}<Q_{\text {sp }}$.
$$ Q_{\mathrm{sp}}=\left(\frac{10^{-4}}{2}\right)\left(\frac{10^{-4}}{2}\right)=2.5 \times 10^{-9}>K_{\mathrm{sp}} $$
Hence, precipitate will be formed in this case. In all other case, $Q_{\mathrm{sp}}<K_{\mathrm{sp}}$ and no precipitation will occur.
