If the #Cr_3^+# concentration in a saturation solution of #Cr(OH)_3# is #4.0xx10^-6 M#. How do you calculate the Ksp?

Question

If the #Cr_3^+# concentration in a saturation solution of #Cr(OH)_3# is #4.0xx10^-6 M#.  How do you calculate the Ksp?

Noah Aldrich · Accepted Answer

#K_(sp) = 6.9 * 10^(-21)#

Chromium(III) hydroxide, #"Cr"("OH")_3#, is made up of chromium(III) cations, #"Cr"^(3+)#, and hydroxide anions, #"OH"^(-)#.

Chromium(III) hydroxide is considered to be insoluble in water, which means that when the salt dissolves, only very, very small amounts will actually dissociate to produce ions in solution.

#"Cr"("OH")_ (color(red)(3)(s)) rightleftharpoons "Cr"_ ((aq))^(3+) + color(red)(3)"OH"_ ((aq))^(-)#

Notice that for every mole of chromium(III) hydroxide that dissociates you get #1# mole of chromium cations and #color(red)(3)# moles of hydroxide anions.

This means that at equilibrium, an aqueous solution of chromium(III) hydroxide will contain three times as many moles of hydroxide anions than of chromium cations.

Therefore, for any concentration of chromium(III) cations you have in solution, you will also have

#["OH"^(-)] = color(red)(3) xx ["Cr"^(3+)]#

In your case, you know that

#["Cr"^(3+)] = 4.0 * 10^(-6)"M"#

which means that this saturated solution contains

#["OH"^(-)] = color(red)(3) xx 4.0 * 10^(-6)"M" = 1.2 * 10^(-5)"M"#

Now, the solubility product constant for this dissociation equilibrium is equal to

#K_(sp) = ["Cr"^(3+)] * ["OH"^(-)]^color(red)(3)#

Plug in your values to find

#K_(sp) = 4.0 * 10^(-6)"M" * (1.2 * 10^(-5)"M")^color(red)(3)#

#K_(sp) = 6.9 * 10^(-21)"M"^4#

You'll usually see the #K_(sp)# expressed without added units, so you can say that you have

#K_(sp) = color(green)(|bar(ul(color(white)(a/a)color(black)(6.9 * 10^(-21))color(white)(a/a)|)))#

The answer is rounded to two sig figs.

Cameron Andrews · Answer

undefined $$ K_{sp} = [Cr^{3+}][OH^-]^3 $$
$$ K_{sp} = (4.0 \times 10^{-6}) \times (3 \times 4.0 \times 10^{-6})^3 $$

Olivia Anton · Answer

undefined To calculate the $ K_{sp} $ (solubility product constant) for Cr(OH)₃, use the formula:

$$ K_{sp} = [Cr^{3+}][OH^-]^3 $$

Given that the concentration of Cr³⁺ in the saturation solution is $ 4.0 \times 10^{-6} $ M, and assuming that Cr(OH)₃ fully dissociates into Cr³⁺ and OH⁻ ions, we can substitute the concentration of Cr³⁺ into the formula. Since Cr(OH)₃ dissociates to produce three OH⁻ ions for every Cr³⁺ ion, we use the concentration of Cr³⁺ to determine the concentration of OH⁻ ions.

$$ [OH^-] = 3 \times [Cr^{3+}] $$

$$ K_{sp} = [Cr^{3+}][OH^-]^3 $$

$$ K_{sp} = (4.0 \times 10^{-6}) \times (3 \times 4.0 \times 10^{-6})^3 $$

$$ K_{sp} = (4.0 \times 10^{-6}) \times (3^3 \times 4.0^3 \times 10^{-18}) $$

$$ K_{sp} = 3^3 \times 4.0^4 \times 10^{-24} $$

$$ K_{sp} = 108 \times 256 \times 10^{-24} $$

$$ K_{sp} = 27648 \times 10^{-24} $$

$$ K_{sp} = 2.7648 \times 10^{-19} $$

So, the solubility product constant ($ K_{sp} $) for Cr(OH)₃ is $ 2.7648 \times 10^{-19} $.