State briefly what a reversible reaction means.
State briefly what dynamic equilibrium means.
Water ionises only slightly according to the equation below. $\text{H}_2\text{O}(l) \rightleftharpoons \text{H}^+(aq) + \text{OH}^-(aq)$, $\Delta H = +58\,\text{kJ mol}^{-1}$ Write an expression for $K_c$ for this reaction.
Write the expression for $K_w$, the ionic product of water, and explain how it is obtained from your $K_c$ expression in (i).
State and explain how the value of $K_w$ for hot water differs from its value for cold water.
Use the value of $K_w$ from the Data Booklet to determine the pH of $0.050\,\text{mol dm}^{-3}$ $\text{NaOH}$.
Water ionises only slightly, as shown below: $\text{H}_2\text{O}(l) \rightleftharpoons \text{H}^+(aq) + \text{OH}^-(aq)\quad \Delta H = +58\,\text{kJ mol}^{-1}$.
$K_w$ can be used to calculate the pH of strong-base and weak-base solutions.
Calculate $[\text{OH}^-(aq)]$ in a $0.050\,\text{mol dm}^{-3}$ solution of $\text{NH}_3$. You may assume that only a small proportion of the $\text{NH}_3$ ionises, so $[\text{NH}_3]$ at equilibrium stays at $0.050\,\text{mol dm}^{-3}$. The equilibrium is: $\text{NH}_3(aq) + \text{H}_2\text{O}(l) \rightleftharpoons \text{NH}_4^+(aq) + \text{OH}^-(aq)$, with $[\text{H}_2\text{O}] \times K_c = \frac{[\text{NH}_4^+][\text{OH}^-]}{[\text{NH}_3]} = 1.8 \times 10^{-5}\,\text{mol dm}^{-3}$.
Use the Data Booklet value of $K_w$, together with your answer to (ii), to determine $[\text{H}^+(aq)]$ in $0.050\,\text{mol dm}^{-3}$ $\text{NH}_3(aq)$.
Calculate the pH of the solution.