Water ionizes by the equation H2O(l)H+(aq)+OH(aq) The extent of the reaction is

Question

Water ionizes by the equation

H2O(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.

Part A

What is the H+ concentration for an aqueous solution with pOH = 3.19 at 25 C?

Express the molar concentration numerically using two significant figures.

Part B

Arrange the following aqueous solutions, all at 25 C, in order of decreasing acidity.

Rank from most acidic to most basic.

0.0018 M KOH

pH = 5.45

pOH = 8.55

0.0023 M HCl

Part C

At a certain temperature, the pH of a neutral solution is 7.51. What is the value of Kw at that temperature?

Express your answer numerically using two significant figures.

0.0018 M KOH

pH = 5.45

pOH = 8.55

0.0023 M HCl

Explanation / Answer

Solution- pH = -log[H+]
pOH = 3.19 => pH = 14-3.19 = 10.81
10.81 = -log[H+] => [H+] = 1.55*10^-12M

b) pOH of KOH = 2.63 => pH = 11.25

pH = 5.45
pOH = 8.55 => pH = 5.45
0.0018 M => pH = -log(0.0018) = 2.74
so acidity order
4> 2 = 3 > 1
c) pH = pOH= 7.51 => [H+] = 3.09*10^-8
Kw = p[H] P[OH] = (3.09*10^-8)^2 = 9.55*10^-16

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