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Question: When equal volumes of the following solutions are mixed, precipitation of AgCl ($K_{sp}$ = 1.8×10$^{...

When equal volumes of the following solutions are mixed, precipitation of AgCl (KspK_{sp} = 1.8×10−10^{-10}) will occur only with

A

10−4^{-4}M (Ag+^+) and 10−4^{-4} M (Cl−^-)

B

10−5^{-5} M (Ag+^+) and 10−5^{-5} M (Cl−^-)

C

10−6^{-6} M (Ag+^+) and 10−6^{-6} M (Cl−^-)

D

10−10^{-10} M (Ag+^+) and 10−10^{-10} M (Cl−^-)

Answer

(a)

Explanation

Solution

When equal volumes of two solutions are mixed, the concentration of each solute is halved. Precipitation of AgCl occurs when the ionic product (QspQ_{sp}) exceeds the solubility product (KspK_{sp}). The ionic product is calculated as the product of the concentrations of Ag+^+ and Cl−^- ions after mixing. For option (a), initial concentrations are 10−410^{-4} M for both ions. After mixing equal volumes, concentrations become 0.5×10−40.5 \times 10^{-4} M for each. The ionic product is (0.5×10−4)×(0.5×10−4)=0.25×10−8=2.5×10−9(0.5 \times 10^{-4}) \times (0.5 \times 10^{-4}) = 0.25 \times 10^{-8} = 2.5 \times 10^{-9}. The given KspK_{sp} is 1.8×10−101.8 \times 10^{-10}. Since 2.5×10−9>1.8×10−102.5 \times 10^{-9} > 1.8 \times 10^{-10}, precipitation occurs in this case. For other options, the calculated ionic product is less than the KspK_{sp}, so precipitation does not occur.