Question

Water ionizes by the equationH2O(l)⇌H+(aq)+OH−(aq)The extent of the reaction is small in pure water and dilute aqueous solutions. This reaction creates the following relationship between [H+] and [OH−]:Kw=[H+][OH−]Keep in mind that, like all equilibrium constants, the value of Kw changes with temperature.a. What is the H+ concentration for an aqueous solution with pOH = 3.51 at 25°C? Express your answer to two significant figures and include the appropriate units.b. At a certain temperature, the pH of a neutral solution is 7.56. What is the value of Kw at that temperature?Express your answer numerically using two significant figures.

Answer 1

Explanation:

$H_2O(l)\rightleftharpoons H^+(aq)+OH^-(aq)$

The value of $K_w$ :

$K_w=[H^+][OH^-]$

a. pOH = 3.51

The sum of pH and pOH is equal to 14.

pH + pOH = 14 (at 25°C)

pH = 14 - 3.51 = 10.49

The pH of the solution is defined as negative logarithm of hydrogen ion concentration in solution.

$pH=-\log[H^+]$

$10.49=-\log[H^+]$

$[H^+]=3.2\times 10^{-11}$

$3.2\times 10^{-11} M$is the $H^+$ concentration for an aqueous solution with pOH = 3.51 at 25°C.

b.

At a certain temperature, the pH of a neutral solution is 7.56.

Neutral solution means that concentration of hydrogen ion and hydroxide ions are equal.

$[H^+]=[OH^-]$

$7.56=-\log[H^+]$

$[H^+]=2.754\times 10^{-8} M$

The value of $K_w$ at at this temperature:

$K_w=[H^+][OH^-]$

$K_w=[H^+][H^+]$

$K_w=(2.754\times 10^{-8})^2=7.6\times 10^{-16}$

The value of $K_w$ at at this temperature is $7.6\times 10^{-16}$.