What is the #K_(sp)# for silver chloride?
When solid silver chloride (MM=143.3) is added to 100 mL of #H_2O# , #1.9 xx 10^-4# grams dissolves. What is the #K_(sp)# for silver chloride?
The answer should be #1.8 xx 10^-10# , but I'm not quite sure how, could someone please explain? Thanks
When solid silver chloride (MM=143.3) is added to 100 mL of
The answer should be
2 Answers
WE address the equilibrium...
Explanation:
For which
We are given that
But by the stoichiometry...
And thus
Happy?
Explanation:
In order to find the solubility product constant for silver chloride, you need to determine the equilibrium concentrations of the dissolved ions.
As you know, silver chloride is considered insoluble in water, and so when you dissolve this salt, an equilibrium is established in aqueous solution between the undissolved solid and the dissolved ions.
#"AgCl"_ ((s)) rightleftharpoons "Ag"_ ((aq))^(+) + "Cl"_ ((aq))^(-)#
Now, you are told that when you dissolve silver chloride in water, only
To convert this mass to moles, use the molar mass of silver chloride.
#1.9 * 10^(-4) color(red)(cancel(color(black)("g"))) * "1 mole AgCl"/(143.3color(red)(cancel(color(black)("g")))) = 1.326 * 10^(-6) quad "moles AgCl"#
So, you know for a fact that at the temperature at which you're performing your experiment, only
Since every mole of silver chloride that dissociates produces
To calculate the molarity of the ions, scale up the volume to
#10^3 color(red)(cancel(color(black)("mL solution"))) * (1.326 * 10^(-6) quad "moles Ag"^(+))/(100color(red)(cancel(color(black)("mL solution")))) = 1.326 * 10^(-5) quad "moles Ag"^(+)#
The calculation is the same for the chloride anions, which means that, at equilibrium, your solution contains
#["Ag"^(+)] = ["Cl"^(-)] = 1.326 * 10^(-5) quad "M"#
By definition, the solubility product constant for silver chloride is equal to
#K_(sp) = ["Ag"^(+)] * ["Cl"^(-)]#
Plug in your values to find--I'll leave the value without added units!
#K_(sp) = (1.326 8 10^(-5)) * (1.326 * 10^(-5)) = 1.758 * 10^(-10)#
Based on the values you have for the volume of water, the answer should be rounded to one significant figure, but if you go by the number of sig figs you have for the mass of silver chloride, you can round the answer to two sig figs.
#color(darkgreen)(ul(color(black)(K_(sp) = 1.8 * 10^(-10))))#