Question

Calculate the molar solubility of $\text{ Pb(}{{\text{N}}_{\text{3}}}{{\text{)}}_{\text{2}}}\text{ }$ in a buffer solution with$\text{ pH = 2}\text{.65}$ , $\text{ }{{\text{K}}_{\text{a}}}\text{ }$ of $\text{ H}{{\text{N}}_{\text{3}}}\text{ }$ is $\text{ 2 }\times \text{1}{{\text{0}}^{-5}}\text{ }$ & $\text{ }{{\text{K}}_{\text{sp}}}\text{ }$ of $\text{ Pb(}{{\text{N}}_{\text{3}}}{{\text{)}}_{\text{2}}}\text{ }$ is $\text{ 2}\text{.5 }\times \text{ 1}{{\text{0}}^{-9}}\text{ }$.

Answer 1

Hint: The molar solubility is the number of moles of a substance that can be dissolved per litre of the solution after achieving saturation. The molar solubility for salt $\text{ AxBy }$ is written as,
$\text{ AxBy }\rightleftharpoons \text{ x}{{\text{A}}^{\text{+}}}\text{ (aq) + y}{{\text{B}}^{-}}\text{(aq) }$
Solubility product is,
$\text{ }{{\text{K}}_{\text{sp}}}\text{ = }{{\left[ {{\text{A}}^{\text{+}}} \right]}^{\text{x}}}{{\left[ {{\text{B}}^{-}} \right]}^{\text{y}}}\text{ }$
The presence of the same ion in the solution reduces the solubility of the salt. This effect is well known as the common ion effect. However, for insoluble salt, the effect can be the opposite.

Complete Solution :
> We have given the following data:
> The hydrogen concentration of the solution is given as $\text{ pH = 2}\text{.65}$
> The dissociation constant $\text{ }{{\text{K}}_{\text{a}}}\text{ }$ for $\text{ H}{{\text{N}}_{\text{3}}}\text{ }$ is $\text{ 2 }\times \text{1}{{\text{0}}^{-5}}\text{ }$
Solubility product $\text{ }{{\text{K}}_{\text{sp}}}\text{ }$ $\text{ Pb(}{{\text{N}}_{\text{3}}}{{\text{)}}_{\text{2}}}\text{ }$ is $\text{ 2}\text{.5 }\times \text{ 1}{{\text{0}}^{-9}}\text{ }$
We are interested to find out the molar solubility of the $\text{ Pb(}{{\text{N}}_{\text{3}}}{{\text{)}}_{\text{2}}}\text{ }$.

- The $\text{ Pb(}{{\text{N}}_{\text{3}}}{{\text{)}}_{\text{2}}}\text{ }$ in the solution dissociated into the ions. The equilibrium reaction for $\text{ Pb(}{{\text{N}}_{\text{3}}}{{\text{)}}_{\text{2}}}\text{ }$ is given as follows:
$\text{ Pb(}{{\text{N}}_{\text{3}}}{{\text{)}}_{\text{2}}}\text{ (s})\text{ }\rightleftharpoons \text{ P}{{\text{b}}^{\text{2+}}}\text{ (aq) + 2N}_{3}^{-}\text{ }$

The solubility product $\text{ Pb(}{{\text{N}}_{\text{3}}}{{\text{)}}_{\text{2}}}\text{ }$is given as follows,
$\text{ }{{\text{K}}_{\text{sp}}}\text{ = }\left[ \text{P}{{\text{b}}^{\text{2+}}} \right]{{\left[ \text{N}_{\text{3}}^{-} \right]}^{2}}\text{ }$ (1)

- The $\text{ H}{{\text{N}}_{\text{3}}}\text{ }$ dissociates into its corresponding $\text{ }{{\text{H}}^{\text{+}}}\text{ }$and $\text{N}_{3}^{-}$ ions. The dissociation constant is terms as the $\text{ }{{\text{K}}_{\text{a}}}\text{ }$. The dissociation of $\text{ H}{{\text{N}}_{\text{3}}}\text{ }$ as follows,
 $\text{ H}{{\text{N}}_{\text{3}}}\text{+ }{{\text{H}}_{\text{2}}}\text{O }\rightleftharpoons \text{ }{{\text{H}}_{\text{3}}}{{\text{O}}^{\text{+}}}\text{ + }{{\text{N}}^{-}}_{\text{3}}\text{ }$

- Therefore, the dissociation constant $\text{ H}{{\text{N}}_{\text{3}}}\text{ }$ can be written as:
$\text{ }{{\text{K}}_{\text{a}}}\text{= }\dfrac{\left[ {{\text{H}}^{\text{+}}} \right]\left[ \text{N}_{\text{3}}^{-} \right]}{\left[ \text{H}{{\text{N}}_{\text{3}}} \right]}\text{ = 2}\text{.0 }\times \text{1}{{\text{0}}^{-5}}\text{ }$ (2)
Here, the $\text{ }{{\text{H}}^{\text{+}}}\text{ }$ reacts with the azide ion $\text{N}_{3}^{-}$ (from $\text{ Pb(}{{\text{N}}_{\text{3}}}{{\text{)}}_{\text{2}}}\text{ }$salt), the more solid will form and dissolve so the solubility product keeps fulfilling.

- Therefore, the total number of azide ion in the solution would be equal to the half of the sum of the azide ion from $\text{ Pb(}{{\text{N}}_{\text{3}}}{{\text{)}}_{\text{2}}}\text{ }$ and $\text{ H}{{\text{N}}_{\text{3}}}\text{ }$. Since each $\text{P}{{\text{b}}^{\text{2+}}}$corresponds to the two azide ions. It can be written as,
$\text{ }\left[ \text{P}{{\text{b}}^{\text{2+}}} \right]\text{ = }\dfrac{1}{2}\left\{ \left[ \text{N}_{3}^{-} \right]+\left[ \text{H}{{\text{N}}_{3}} \right] \right\}\text{ }$ (3)

- The concentration of $\text{ H}{{\text{N}}_{\text{3}}}\text{ }$ can be deduced from the dissociation constant of $\text{ H}{{\text{N}}_{\text{3}}}\text{ }$(equation (2)) as follows,
$\text{ }\left[ \text{H}{{\text{N}}_{\text{3}}} \right]\text{= }\dfrac{\left[ {{\text{H}}^{\text{+}}} \right]\left[ \text{N}_{\text{3}}^{-} \right]}{{{\text{K}}_{\text{a}}}}\text{ = }\dfrac{2.238\times {{10}^{-3}}\left[ \text{N}_{\text{3}}^{-} \right]}{\text{2}\text{.0 }\times \text{1}{{\text{0}}^{-5}}}\text{ = 111}\text{.9}\left[ \text{N}_{\text{3}}^{-} \right]\text{ }$ (4)
Since ,$\text{ }\left[ {{\text{H}}^{+}} \right]\text{ = 1}{{\text{0}}^{-\text{pH}}}=\text{ 1}{{\text{0}}^{-2.65\text{ }}}=2.238\text{ }\times \text{1}{{\text{0}}^{-3}}\text{ }$ .

- Let's substitute the obtained concentration of the $\text{ H}{{\text{N}}_{\text{3}}}\text{ }$ in equation (3) to determine the concentration of lead ion in terms of concentration of azide ion as follows,
$\text{ }\left[ \text{P}{{\text{b}}^{\text{2+}}} \right]\text{ = }\dfrac{1}{2}\left\{ \left[ \text{N}_{3}^{-} \right]+111.9\left[ \text{N}_{3}^{-} \right] \right\}\text{ = }\dfrac{1}{2}\left\{ 112.9\left[ \text{N}_{3}^{-} \right] \right\}=56.45\left[ \text{N}_{3}^{-} \right]\text{ }$ (5)

- Now, substitute the value of the concentration of lead ion obtained from equation (4) in the equation (1) we have,
$\begin{align}
  & \text{ 2}\text{.5}\times \text{1}{{\text{0}}^{-9}}\text{ = }\left[ \text{P}{{\text{b}}^{\text{2+}}} \right]{{\left[ \text{N}_{\text{3}}^{-} \right]}^{2}}\text{ } \\
 & \Rightarrow \text{2}\text{.5}\times \text{1}{{\text{0}}^{-9}}\text{ = 56}\text{.45}\left[ \text{N}_{\text{3}}^{-} \right]{{\left[ \text{N}_{\text{3}}^{-} \right]}^{2}}\text{ }\because \text{56}\text{.45}\left[ \text{N}_{\text{3}}^{-} \right]=\left[ \text{P}{{\text{b}}^{\text{2+}}} \right] \\
 & \Rightarrow \text{2}\text{.5}\times \text{1}{{\text{0}}^{-9}}\text{ = 56}\text{.45}{{\left[ \text{N}_{\text{3}}^{-} \right]}^{3}} \\
 & \Rightarrow \left[ \text{N}_{\text{3}}^{-} \right]\text{ = }\sqrt[3]{\dfrac{\text{2}\text{.5}\times \text{1}{{\text{0}}^{-9}}}{\text{56}\text{.45}}}=\sqrt[3]{4.428\times {{10}^{-11}}} \\
 & \therefore \left[ \text{N}_{\text{3}}^{-} \right]\text{ = 3}\text{.538}\times \text{1}{{\text{0}}^{-4}}\text{ M } \\
\end{align}$

- Now, we substitute the value of the concentration of the $\left[ \text{N}_{\text{3}}^{-} \right]$ in the relation of it with the lead ion. Substitute in the equation (5), we have,
$\text{ }\left[ \text{P}{{\text{b}}^{\text{2+}}} \right]=56.45\left[ \text{N}_{3}^{-} \right]\text{ }\Rightarrow \text{ }56.45\times 3.53\times {{10}^{-4}}\text{ = 1}\text{.99 }\times \text{1}{{\text{0}}^{-2}}\text{M }$
Thus, the molar solubility of the lead azide in the buffer solution of hydrogen azide $\text{ pH = 2}\text{.65}$ is equal to \[\text{1}\text{.99 }\times \text{1}{{\text{0}}^{-2}}\text{M }\].

So, the correct answer is “Option B”.

Note: The common ion effect is the addition of the salt to a solution such that the two slats have the common ion.for example, lead azide, and hydrogen azide. Both slats have the common ion of azide. The solubility of the salt is measuredly affected by the common ion. It decreases the solubility of the salt in the solution. The lead azide $\text{ Pb(}{{\text{N}}_{\text{3}}}{{\text{)}}_{\text{2}}}\text{ }$ is insoluble in water, but here the hydrogen azide is soluble in water, thus dissolves all the solid of $\text{ Pb(}{{\text{N}}_{\text{3}}}{{\text{)}}_{\text{2}}}\text{ }$ as the $\text{ H}{{\text{N}}_{\text{3}}}\text{ }$ and the solubility is only due to the lead ion as it does not further undergo the reaction.