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

Question:

Determine the solubilities of silver chromate, barium chromate, ferric hydroxide, lead chloride and mercurous iodide at 298K from their solubility product constants given in Table 7.9 (page 221). Determine also the molarities of individual ions.

Solution:

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

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

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

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