# What is the #K_(sp)# for silver chloride?

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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 **dissociate** into ions to give a **saturated solution** of silver chloride.

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 **moles** of silver chloride for every **will dissociate** to produce ions.

Since every mole of silver chloride **that dissociates** produces **mole** of silver(I) cations and **mole** of chloride anions, you can say that, at equilibrium, your solution will contain **moles** of silver(I) cations and **moles** of chloride anons **per**

To calculate the **molarity** of the ions, scale up the volume to *equal* to the volume of the solution.

#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))))#