Whatare the molar concentrations of all ions in a 500 mL saturated solution of silver chloride at 25 degrees Celsius?

1 Answer
Mar 14, 2016

Answer:

#1.34 * 10^(-5)"M"#

Explanation:

Your strategy here will be to use the solubility product constant, #K_(sp)#, of silver chloride to find the compound's molar solubility at #25^@"C"#.

You can find the value for the #K_(sp)# of silver chloride here

http://bilbo.chm.uri.edu/CHM112/tables/KspTable.htm

So, you know that

#K_(sp) = 1.8 * 10^(-10)#

Silver chloride, #"AgCl"#, is considered insoluble in aqueous solution, which means that an equilibrium will be established between the undissolved solid and the dissolved ions.

Use an ICE table to find the molar solubility of silver chloride, #s#, at this temperature

#" ""AgCl"_text((s]) " "rightleftharpoons" " "Ag"_text((aq])^(+) " "+" " "Cl"_text((aq])^(-)#

#color(purple)("I")" " " "color(white)(a)-" " " " " " " " " "0" " " " " " " " " "0#
#color(purple)("C")" " " "color(white)(a)-" " " " " " " "(+s)" " " " " "(+s)#
#color(purple)("E")" " " "color(white)(a)-" " " " " " " " " "s" " " " " " " " " "s#

By definition, the solubility product constant will be equal to

#K_(sp) = ["Ag"^(+)] * ["Cl"^(-)]#

In your case, you will have

#1.8 * 10^(-10) = s * s = s^2#

This will give you

#s = sqrt(1.8 * 10^(-10)) = 1.34 * 10^(-5)#

This means that the concentrations of silver cations and of chloride anons in a saturated solution of silver chloride will be equal to

#["Ag"^(+)] = ["Cl"^(-)] = s = 1.34 * 10^(-5)"mol L"^(-1)#

Therefore, the concentration of both ions in your #"500-mL"# saturated silver chloride solution will be equal to

#["Ag"^(+)] = ["Cl"^(-)] = color(green)(|bar(ul(color(white)(a/a)1.34 * 10^(-5)"M"color(white)(a/a)|)))#

I'll leave the answer rounded to three sig figs.

Do not get confused by the fact that you're dealing with a #"500-mL"# solution!

The molar solubility represents concentration, not number of moles of solute. This means that the concentration of both ions in a saturated silver chloride solution will be equal to #1.34 * 10^(-5)"M"# regardless of the volume of said solution!