# Quiz | Solubility and Precipitation Reactions

1

What is the solubility-product constant expression for the dissolution of tin(II) hydroxide?

$\left[{\mathrm{Sn}}^{2+}\right]{\left[{\mathrm{HO}}^{-}\right]}^{2}$

$\frac{1}{\left[{\mathrm{Sn}}^{2+}\right]{\left[{\mathrm{HO}}^{-}\right]}^{2}}$

${\left[{\mathrm{Sn}}^{2+}\right]}^{2}\left[{\mathrm{HO}}^{-}\right]$

$\frac{1}{{\left[{\mathrm{Sn}}^{2+}\right]}^{2}\left[{\mathrm{HO}}^{-}\right]}$

Solubility-product constant Ksp: equilibrium-constant of a dissolution reaction
Dissolution of tin(II) hydroxide: Sn(OH)2 (s)  Sn2+ (aq) + 2 HO- (aq)

2

What is the molar solubility of BaCO3?

s = Ksp1/3

s = Ksp1/2

s = Ksp2

s = ( ${\mathrm{K}}_{\mathrm{sp}}}{4}$ )1/3

Dissolution of BaCO3: BaCO3 (s) $⇌$ Ba2+ (aq) + CO32- (aq)
According to the reaction stoichiometry, molar solubility s of BaCO3 (s) = [Ba2+] = [CO32-]
Ksp = [Ba2+] [CO32-] = s2    ⇒   s = Ksp1/2

3

What is the molar solubility of Ca3(PO4)2?

s = Ksp1/5

s = ( ${\mathrm{K}}_{\mathrm{sp}}}{27}$ )1/5

s = ( ${\mathrm{K}}_{\mathrm{sp}}}{36}$ )1/5

s = ( ${\mathrm{K}}_{\mathrm{sp}}}{108}$ )1/5

Dissolution of Ca3(PO4)2: Ca3(PO4)2 (s) $⇌$ 3 Ca2+ (aq) + 2 PO43- (aq)
According to the reaction stoichiometry, molar solubility s of Ca3(PO4)2$\frac{\left[{\mathrm{Ca}}^{2+}\right]}{3}$ = $\frac{\left[{{\mathrm{PO}}_{4}}^{2-}\right]}{2}$
Ksp = [Ca2+]3 [PO43-]2 = (3s)3 x (2s)2 = 27s3 x 4s2 =108s5   ⇒ s =  ${\left(\frac{{K}_{sp}}{108}\right)}^{1}{5}}$

4

In which conditions would Ag3PO4 be least soluble?

Pure water

0.1 M AgNO3

0.1 M HNO3

Solubility does not change

AgNO3 and Ag3PO4 have a common ion: Ag+
Common ion effect: the solubility of an ionic solid decreases when a common ion is present in solution

5

Which of the following sentences is incorrect?

The solubility of an ionic solid can be decreased by the formation of a soluble complex ion

The solubility of an ionic solid decreases when a common ion is present in solution

When equilibrium is disturbed and Qsp > Ksp, more precipitate forms until Qsp = Ksp

None of the above

The solubility of an ionic solid is increased by the formation of a soluble complex ion

6

Why Al(OH)3 is soluble in basic solution but insoluble in neutral solution?

Because of the common ion effect

Because of the acid-base neutralization

Because of the formation of a soluble hydroxy complex ion [Al(OH)4]-

Because of the solubility-product constant Ksp

Al(OH)3 is an amphoteric metal hydroxides:
Neutral solution: insoluble
Basic solution: Al(OH)3 (s) + HO- (aq) ⇌ [Al(OH)4](aq)     [formation of a soluble hydroxy complex ion]
Acidic solution: Al(OH)3 (s) + 3 H3O+ (aq) ⇌ Al3+ (aq) + 6 H2O (l)     [acid-base neutralization]