Question 18.p.7: Write a balanced equation for the dissociation of HOCN in wa......
Write a balanced equation for the dissociation of HOCN in water. Using the given information, construct a table that describes the initial and equilibrium concentrations. Construct an equilibrium expression and solve the quadratic expression for x, the concentration of \text{H}_3\text{O}^+.
In this example, the dissociation of HOCN is not negligible in comparison to the initial concentration:
Therefore, [HOCN]_\text{eq} = 0.10 – x and the equilibrium expression is solved using the quadratic formula.
\text{x}^2= 3.5 \times 10^{-4} \ (0.10 – \text{x}) \\ \text{x}^2 = 3.5 \times 10^{-5} – 3.5 \times 10^{-4} \text{ x} \\ \text{x}^2 + 3.5 \times 10^{-4}\text{ x} – 3.5 \times 10^{-5} = 0 \quad\quad (\text{a x}^2 + \text{b x + c} = 0) \\ \text{a = 1}\quad\text{b} = 3.5 \times 10^{-4} \quad \text{c} = -3.5 \times 10^{-5} \\ \text{x}=\frac{-\text{b}\pm \sqrt{\text{b}^2-4\text{ac}} }{2\text{a}} \\ \text{x}=\frac{-3.5\times 10^{-4}\pm \sqrt{(3.5\times 10^{-4})^2-4(1)(-3.5\times 10^{-5})} }{2(1)} \\ \text{x}=5.7436675\times 10^{-3}=5.7\times 10^{-3} \ M\text{ H}_3\text{O}^+ \\ \text{pH}=-\log \ [\text{H}_3\text{O}^+]=-\log \ [5.7436675\times 10^{-3}]=2.2408=2.24