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

Step-by-step 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* 10^(-11)`

`3.2* 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* 10^(-8) M`

The value of
`K_w` at at this temperature:

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

`K_w=[H^+][H^+]`

`K_w=(2.754* 10^(-8))^2=7.6* 10^(-16)`

The value of
`K_w` at at this temperature is
`7.6* 10^(-16)`.