How do you get Kw=1.0×10*-14 at 25° in no 1 question

Hey Temi, good to see you again.

The value of $K_w = 1.0 \times 10^{-14}$ at $25^\circ \text{C}$ comes from the autoionization of water.

Step 1: Water undergoes a slight autoionization, meaning it reacts with itself to form hydronium ions ($\text{H}_3O^+$ or simply $\text{H}^+$) and hydroxide ions ($\text{OH}^-$). $H_2O(l) + H_2O(l) \rightleftharpoons H_3O^+(aq) + OH^-(aq)$ This can be simplified as: $H_2O(l) \rightleftharpoons H^+(aq) + OH^-(aq)$

Step 2: The equilibrium constant for this reaction is called the ion product of water, $K_w$. $K_w = [H^+][OH^-]$

Step 3: In pure water at $25^\circ \text{C}$, the concentrations of $\text{H}^+$ and $\text{OH}^-$ ions are equal because they are produced in a 1:1 ratio. Through experimental measurements (e.g., conductivity), it's found that the concentration of each ion is $1.0 \times 10^{-7} \text{ M}$. $[H^+] = 1.0 \times 10^{-7} M$ $[OH^-] = 1.0 \times 10^{-7} M$

Step 4: Substitute these concentrations into the $K_w$ expression. $K_w = (1.0 \times 10^{-7})(1.0 \times 10^{-7})$ $K_w = 1.0 \times 10^{(-7) + (-7)}$ $K_w = 1.0 \times 10^{-14}$

Thus, at $25^\circ \text{C}$, the ion product of water is $\boxed{K_w = 1.0 \times 10^{-14}}$.

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