Determine the solubilities of silver chromate, barium chromate, ferric hydroxide,

(1) Silver chromate:

$\mathrm{Ag}_{2} \mathrm{CrO}_{4} \longrightarrow 2 \mathrm{Ag}^{+}+\mathrm{CrO}_{4}{ }^{2-}$

Then,

$K_{x p}=\left[\mathrm{Ag}^{+}\right]^{2}\left[\mathrm{CrO}_{4}{ }^{2-}\right]$

Let the solubility of $\mathrm{Ag}_{2} \mathrm{CrO}_{4}$ be s.

$\Rightarrow\left[\mathrm{Ag}^{+}\right] 2 s$ and $\left[\mathrm{CrO}_{4}{ }^{2-}\right]=s$

Then,

$K_{s p}=(2 s)^{2} \cdot s=4 s^{3}$

$\Rightarrow 1.1 \times 10^{-12}=4 s^{3}$

$.275 \times 10^{-12}=s^{3}$

$s=0.65 \times 10^{-4} \mathrm{M}$

Molarity of $\mathrm{Ag}^{+}=2 \mathrm{~s}=2 \times 0.65 \times 10^{-4}=1.30 \times 10^{-4} \mathrm{M}$

Molarity of $\mathrm{CrO}_{4}{ }^{2-}=s=0.65 \times 10^{-4} \mathrm{M}$

(2) Barium chromate:

$\mathrm{BaCrO}_{4} \longrightarrow \mathrm{Ba}^{2+}+\mathrm{CrO}_{4}{ }^{2-}$

Then, $K_{s p}=\left[\mathrm{Ba}^{2+}\right]\left[\mathrm{CrO}_{4}{ }^{2-}\right]$

Let $s$ be the solubility of $\mathrm{BaCrO}_{4}$.

Thus, $\left[\mathrm{Ba}^{2+}\right]=s$ and $\left[\mathrm{CrO}_{4}{ }^{2-}\right]=s$

$\Rightarrow K_{S P}=s^{2}$

$\Rightarrow 1.2 \times 10^{-10}=s^{2}$

$\Rightarrow s=1.09 \times 10^{-5} \mathrm{M}$

Molarity of $\mathrm{Ba}^{2+}=$ Molarity of $\mathrm{CrO}_{4}{ }^{2-}=s=1.09 \times 10^{-5} \mathrm{M}$

(3) Ferric hydroxide:

$\mathrm{Fe}(\mathrm{OH})_{3} \longrightarrow \mathrm{Fe}^{2+}+3 \mathrm{OH}^{-}$

$K_{s p}=\left[\mathrm{Fe}^{2+}\right]\left[\mathrm{OH}^{-}\right]^{3}$

Let $s$ be the solubility of $\mathrm{Fe}(\mathrm{OH})_{3}$.

Thus, $\left[\mathrm{Fe}^{3+}\right]=s$ and $\left[\mathrm{OH}^{-}\right]=3 s$

$\Rightarrow K_{s p}=s \cdot(3 s)^{3}$

$=s .27 s^{3}$

$K_{s p}=27 s^{4}$

$.037 \times 10^{-38}=s^{4}$

$.00037 \times 10^{-36}=s^{4} \quad \Rightarrow 1.39 \times 10^{-10} \mathrm{M}=\mathrm{S}$

Molarity of $\mathrm{Fe}^{3+}=s=1.39 \times 10^{-10} \mathrm{M}$

Molarity of $\mathrm{OH}^{-}=3 s=4.17 \times 10^{-10} \mathrm{M}$

(4) Lead chloride:

$\mathrm{PbCl}_{2} \longrightarrow \mathrm{Pb}^{2+}+2 \mathrm{Cl}^{-}$

$K_{S P}=\left[\mathrm{Pb}^{2+}\right]\left[\mathrm{Cl}^{-}\right]^{2}$

Let $K_{S P}$ be the solubility of $\mathrm{PbCl}_{2}$.

$\left[\mathrm{PB}^{2+}\right]=s$ and $\left[\mathrm{Cl}^{-}\right]=2 s$

Thus, $K_{s p}=s \cdot(2 s)^{2}$

$=4 s^{3}$

$\Rightarrow 1.6 \times 10^{-5}=4 s^{3}$

$\Rightarrow 0.4 \times 10^{-5}=s^{3}$

$4 \times 10^{-6}=s^{3} \Rightarrow 1.58 \times 10^{-2} \mathrm{M}=\mathrm{S} .1$

Molarity of $\mathrm{PB}^{2+}=s=1.58 \times 10^{-2} \mathrm{M}$

Molarity of chloride $=2 s=3.16 \times 10^{-2} \mathrm{M}$

(5) Mercurous iodide:

$\mathrm{Hg}_{2} \mathrm{I}_{2} \longrightarrow \mathrm{Hg}^{2+}+2 \mathrm{I}^{-}$

$K_{s p}=\left[\mathrm{Hg}_{2}^{2+}\right]^{2}\left[\mathrm{I}^{-}\right]^{2}$

Let $s$ be the solubility of $\mathrm{Hg}_{2} \mathrm{I}_{2}$.

$\Rightarrow\left[\mathrm{Hg}_{2}^{2+}\right]=s$ and $\left[\mathrm{I}^{-}\right]=2 s$

Thus, $K_{s p}=s(2 s)^{2} \Rightarrow K_{s p}=4 s^{3}$

$4.5 \times 10^{-29}=4 s^{3}$

$1.125 \times 10^{-29}=s^{3}$

$\Rightarrow s=2.24 \times 10^{-10} \mathrm{M}$

Molarity of $\mathrm{Hg}_{2}^{2+}=s=2.24 \times 10^{-10} \mathrm{M}$

Molarity of $\mathrm{I}^{-}=2 s=4.48 \times 10^{-10} \mathrm{M}$