2.2 M solution of hydrocyanic acid, HCN, at 25°C. pKa=9.21 at 25°C. Find the concentrations of all species present in the solution at equilibrium.
First, we need to determine the initial concentration of HCN in the solution. Since the solution is 2.2 M, the initial concentration of HCN is also 2.2 M.
Next, we need to set up an ICE table to determine the equilibrium concentrations of the species present in the solution:
HCN ⇌ H+ + CN-
\[
\begin{array}{|c|c|c|c|}
\hline
& HCN & H^+ & CN^- \\
\hline
Initial & 2.2 & 0 & 0 \\
Change & -x & +x & +x \\
Equilibrium & 2.2-x & x & x \\
\hline
\end{array}
\]
Since hydrocyanic acid is a weak acid, we will use the equation for the dissociation of a weak acid to determine the concentration of H+:
\[
K_a = \frac{[H^+][CN^-]}{[HCN]}
\]
Substitute the given pKa value and the equilibrium concentrations into the equation:
\[
10^{-9.21} = \frac{x^2}{2.2-x}
\]
Solve for x using the quadratic formula:
\[
x^2 = 2.2 \times 10^{-9.21} - 10^{-9.21}x
\]
\[
x^2 + 10^{-9.21}x - 2.2 \times 10^{-9.21} = 0
\]
Solving this quadratic equation will give us the concentration of H+ and CN- in the solution at equilibrium.