Answer
a. The autoionization of water is endothermic.
b. $[H^+]$ and $[OH^-] = 2.34\times 10^{- 7}M$
Work Step by Step
a. According to the Le Chatelier principles, if the increase of temperature favors the products side of the reaction, this reaction is endothermic.
- Let's imagine "heat" as a product/reactant:
- If the increase of heat favors the right side of the reaction, it should be on the left side:
$H_2O(l) + heat \lt -- \gt H^+(aq) + OH^-(aq)$
- If the "heat" is on the reactants side; the reaction is endothermic because it needs heat to occur.
b. - Pure water solution : $[H^+] = [OH^-]$
- Create a unknown called 'x' that has the value of $[H^+]\ and\ [OH^-]$
- Write the $K_w$ expression, and calculate the x value:
$[H^+] * [OH^-] = K_w = 5.47\times 10^{- 14}$
$x * x = 5.47\times 10^{- 14}$
$x^2 = 5.47\times 10^{- 14}$
$x = \sqrt { 5.47\times 10^{- 14}}$
$x = 2.34\times 10^{- 7}$
Therefore: $[H^+]\ and\ [OH^-] = 2.34\times 10^{- 7}M$
