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

## When solid silver chloride (MM=143.3) is added to 100 mL of ${H}_{2} O$, $1.9 \times {10}^{-} 4$ grams dissolves. What is the ${K}_{s p}$ for silver chloride? The answer should be $1.8 \times {10}^{-} 10$, but I'm not quite sure how, could someone please explain? Thanks

Feb 24, 2018

WE address the equilibrium...$A {g}^{+} + C {l}^{-} r i g h t \le f t h a r p \infty n s A g C l \left(s\right)$

#### Explanation:

For which K_"sp"=[Ag^+][Cl^-]=??

We are given that $\left[A g C l\right] = \frac{\frac{1.9 \times {10}^{-} 4 \cdot g}{143.32 \cdot g \cdot m o {l}^{-} 1}}{0.100 \cdot L} = 1.33 \times {10}^{-} 5 \cdot m o l \cdot {L}^{-} 1$...

But by the stoichiometry...

$\left[A {g}^{+}\right] = \left[C {l}^{-}\right] = 1.33 \times {10}^{-} 5 \cdot m o l \cdot {L}^{-} 1$

And thus ${K}_{\text{sp}} = {\left[1.33 \times {10}^{-} 5\right]}^{2} = 1.76 \times {10}^{-} 10$...the which is close enuff to your answer....

Happy?

Feb 24, 2018

$1.8 \cdot {10}^{- 10}$

#### 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.

${\text{AgCl"_ ((s)) rightleftharpoons "Ag"_ ((aq))^(+) + "Cl}}_{\left(a q\right)}^{-}$

Now, you are told that when you dissolve silver chloride in water, only $1.9 \cdot {10}^{- 4}$ $\text{g}$ per $\text{100 mL}$ of water will 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 $1.326 \cdot {10}^{- 6}$ moles of silver chloride for every $\text{100 mL}$ of water will dissociate to produce ions.

Since every mole of silver chloride that dissociates produces $1$ mole of silver(I) cations and $1$ mole of chloride anions, you can say that, at equilibrium, your solution will contain $1.326 \cdot {10}^{- 6}$ moles of silver(I) cations and $1.326 \cdot {10}^{- 6}$ moles of chloride anons per $\text{100 mL}$ of water.

To calculate the molarity of the ions, scale up the volume to $\text{1 } = {10}^{3}$ $\text{mL}$ of water, which, for all intents and purposes, will be 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

${K}_{s p} = \left[{\text{Ag"^(+)] * ["Cl}}^{-}\right]$

Plug in your values to find--I'll leave the value without added units!

${K}_{s p} = \left(1.326 8 {10}^{- 5}\right) \cdot \left(1.326 \cdot {10}^{- 5}\right) = 1.758 \cdot {10}^{- 10}$

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.

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{{K}_{s p} = 1.8 \cdot {10}^{- 10}}}}$