For the reaction of hydrazine \(\left(\mathrm{N}_{2} \mathrm{H}_{4}\right)\) in water, $$ \mathrm{H}_{2} \mathrm{NNH}_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{2} \mathrm{NNH}_{3}^{+}(a q)+\mathrm{OH}^{-}(a q) $$ \(K_{\mathrm{b}}\) is \(3.0 \times 10^{-6} .\) Calculate the concentrations of all species and the pH of a \(2.0-M\) solution of hydrazine in water.

$$\mathrm{H}_{2} \mathrm{NNH}_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{2} \mathrm{NNH}_{3}^{+}(a q)+\mathrm{OH}^{-}(a q) $$

An ICE table helps us keep track of the changes in the concentrations of the species in the reaction. The table should look like this: | | \(\mathrm{H}_{2} \mathrm{NNH}_{2}(a q)\) | \(\mathrm{H}_{2} \mathrm{NNH}_{3}^{+}(a q)\) | \(\mathrm{OH}^{-}(a q)\) | |-------|-----------------------------------|-------------------------------------|---------------------| | Initial | 2.0 M | 0 M | 0 M | | Change | -x | +x | +x | | Equilibrium | 2.0 M - x | x | x |

The expression for \(K_\mathrm{b}\) is given by: \(K_\mathrm{b} = \frac{[\mathrm{H}_{2}\mathrm{NNH}_{3}^{+}][\mathrm{OH}^{-}]}{[\mathrm{H}_{2}\mathrm{NNH}_{2}]}\)

Using the equilibrium concentrations from the ICE table, we substitute them into the \(K_\mathrm{b}\) expression: \(3.0 \times 10^{-6} = \frac{x(x)}{2.0 - x}\) Since the value of \(K_\mathrm{b}\) is very small, it's reasonable to assume that the change in concentration (x) is very small compared to 2.0 M, and can be ignored: \(3.0 \times 10^{-6} ≈ \frac{x^2}{2.0}\) Now, we can solve for x: \(x^2 = 2.0 \times 3.0 \times 10^{-6}\) \(x = \sqrt{6.0 \times 10^{-6}}\) \(x \approx 2.45 \times 10^{-3}\) Now we know the equilibrium concentrations: \([\mathrm{H}_{2}\mathrm{NNH}_{2}] = 2.0 - x \approx 2.0\M\) \([\mathrm{H}_{2}\mathrm{NNH}_{3}^{+}] = x \approx 2.45 \times 10^{-3}\ M\) \([\mathrm{OH}^{-}] = x \approx 2.45 \times 10^{-3}\ M\)

To find the pH, first calculate the pOH using the \(\mathrm{OH}^{-}\) concentration: \(pOH = -\log{[\mathrm{OH}^{-}]} = -\log{(2.45 \times 10^{-3})} \approx 2.61\) Next, use the relationship between the pH and pOH: \(pH + pOH = 14\) Finally, calculate the pH: \(pH = 14 - pOH \approx 14 - 2.61 \approx 11.39\) The concentrations of all species in the solution are: - \([\mathrm{H}_{2}\mathrm{NNH}_{2}] \approx 2.0\ M\) - \([\mathrm{H}_{2}\mathrm{NNH}_{3}^{+}] \approx 2.45 \times 10^{-3}\ M\) - \([\mathrm{OH}^{-}] \approx 2.45 \times 10^{-3}\ M\) And the pH of the solution is approximately 11.39.