To determine the pH of a 0.320 M solution of [tex]$ \text{Ca(NO}_2\text{)}_2 $[/tex], where the [tex]$K_a$[/tex] of [tex]$ \text{HNO}_2 $[/tex] is [tex]$4.5 \times 10^{-4}$[/tex], follow the steps below:
1. Set up the problem:
We start by considering the dissociation of [tex]$ \text{HNO}_2 \rightarrow \text{H}^+ + \text{NO}_2^- $[/tex]. However, we need to consider the stoichiometry involved with [tex]$ \text{Ca(NO}_2\text{)}_2 $[/tex]:
- Each mole of [tex]$ \text{Ca(NO}_2\text{)}_2 $[/tex] dissociates to produce two moles of [tex]$ \text{NO}_2^- $[/tex].
2. Initial Concentration:
Since the solution is 0.320 M of [tex]$ \text{Ca(NO}_2\text{)}_2 $[/tex], the concentration of [tex]$ \text{NO}_2^- $[/tex] will be twice that because each mole of [tex]$ \text{Ca(NO}_2\text{)}_2 $[/tex] produces two moles of [tex]$ \text{NO}_2^- $[/tex].
- Initial concentration of [tex]$ \text{NO}_2^- $[/tex] is [tex]$2 \times 0.320 = 0.640 \text{ M} $[/tex].
3. ICE Table:
We will set up the ICE table for the dissociation of [tex]$ \text{HNO}_2 $[/tex]:
[tex]\[
\begin{array}{ c|c|c|c }
& \text{HNO}_2 & \text{H}^+ & \text{NO}_2^- \\
\hline
\text{Initial} & 0.320 & 0 & 0.640 \\
\text{Change} & -x & +x & +x \\
\text{Equilibrium} & 0.320 - x & x & 0.640 + x \\
\end{array}
\][/tex]
4. Expression for [tex]$ K_a $[/tex]:
[tex]\[
K_a = [\text{H}^+][\text{NO}_2^-] / [\text{HNO}_2]
\][/tex]
Substitute the equilibrium concentrations into the expression:
[tex]\[
K_a = \left( x \right) \left( 0.640 + x \right) / \left( 0.320 - x \right)
\][/tex]
5. Simplification and Assumption:
Assume that [tex]$ x $[/tex] is very small compared to 0.320 and 0.640 which allows us to simplify the expression:
[tex]\[
K_a \approx x (0.640) / 0.320
\][/tex]
Solving for [tex]$ x $[/tex]:
[tex]\[
x = \sqrt{K_a \cdot 0.320} = \sqrt{4.5 \times 10^{-4} \cdot 0.320 }
\][/tex]
6. Calculate [tex]$ x $[/tex]:
[tex]\[
x \approx 0.012
\][/tex]
This [tex]$ x $[/tex] represents the concentration of [tex]$ \text{H}^+ $[/tex] ions in the solution.
7. Calculate pH:
\\
[tex]\[
pH = - \log [ \text{H}^+ ]
\][/tex]
Substituting in the value of [tex]$ x $[/tex]:
[tex]\[
pH = - \log (0.012)
\][/tex]
[tex]\[
pH \approx 1.92
\][/tex]
Thus, the pH of the 0.320 M solution of [tex]$ \text{Ca(NO}_2\text{)}_2 $[/tex] where the [tex]$ K_a $[/tex] of [tex]$ \text{HNO}_2 $[/tex] is [tex]$ 4.5 \times 10^{-4}$[/tex] is approximately [tex]$1.92$[/tex].
