3.3.12.3 the ionic product of water, Kw

because water can act as both an acid + base, what does this mean?

that both hydrogen ions + hydroxide ions exist simultaneously in water according to the equilibrium

give the full + simplified equation of the equilibrium of hydrogen + hydroxide ions

as this is an equilibrium reaction, what can be derived?

an equilibrium constant, Kc

give the expression of Kc

rearrange the Kc expression

where does equilibrium lie to + why?

as water is only weakly dissociated, the equilibrium position lies far over to the left

explain what Kw is + how it is formed

• due to the equilibrium lying to the left, the concentration of water is very high + can be taken to be constant + its value is incorporated into Kc

• the new constant resulting is called the ionic product of water, Kw

give the Kw expression, its conditions + units

Kw = [H⁺][OH⁻]

• where at 298K (25℃), Kw is 1.00 ×10⁻¹⁴

• units: mol²dm⁻⁶

outline why pure water is neutral

because the hydrogen ion concentration is equal to the hydroxide ion concentration → for every hydrogen ion formed, there is a hydroxide ion formed as well

• [H⁺] = [OH⁻]

because in pure water [H⁺] = [OH⁻], what does this mean you can do to the Kw expression?

this means you can replace the [OH⁻] term in the Kw expression by another [H⁺]

so give the Kw expression used for pure water

Kw = [H⁺]²

example:

calculate the pH of pure water at 298K when Kw = 1.00 ×10⁻¹⁴ mol²dm⁻⁶

the value of Kw is _______ dependent

• temperature

how is Kw affected by temperature?

it increases with temperature

explain why Kw increases with temperature

what is meant by monoprotic + diprotic bases?

• monoprotic base: base that can accept one proton eg NaOH, KOH

• dirprotic base: base that can accept two protons eg Ba(OH)₂

give the steps for calculating the pH of a strong base

1. use concentration of base to find [OH⁻]

2. use Kw to find [H⁺]

   • [H⁺] = Kw/[OH⁻]

3. convert [H⁺] into pH

   • pH = -log[H⁺]

for example, calculate the pH go a 1.00moldm⁻³ solution of NaOH at 298K

for example, calculate the pH of [Ba(OH)₂] which has a concentration of 0.0500 moldm⁻³ at 298K

because barium hydroxide produces 2 OH- ions, the concentration of the base should be multiplied by 2 to find [OH⁻]

give the steps of finding the [OH⁻] from the pH (+ so the conc of the base)

1. use pH to find [H⁺]

   • [H⁺] = 10⁻pH

2. use Kw to find [OH⁻]

   • [OH⁻] = Kw/[H⁺]

3. use [OH⁻] to find the concentration of the base

for example, calculate the concentration of a solution of NaOH at 298K with pH = 12.80

for example, find the concentration of Ba(OH)₂ with pH 13.30 at 298K

[OH⁻] is divided by 2 to find concentration of base

calculate the pH of a solution formed by a dded 5.82g of NaOH to 250cm³ water

• find moles

• use moles/volume to calculate concentration of solution + so [OH⁻]

• find [H⁺]

• find pH

what does dilution refer to?

the addition of water

give the steps for calculating the pH of a strong acid after it has been diluted with water

1. calculate moles of H⁺

2. find new [H⁺] by dividing moles by new volume (volume of acid + water)

3. pH = -log[H⁺]

for example, calculate the pH of the solution when 50cm³ of water is added to 50cm³ of 0.20 moldm⁻³ HCl

give the steps for calculating the pH of a strong base after it has been diluted with water

1. calculate moles of OH⁻

2. find new [OH⁻] by dividing moles by new volume (volume of hydroxide + water)

3. [H⁺] = Kw/[OH⁻]

4. pH = -log[H⁺]

for example

outline the steps for working out the pH of a solution from a neutralisation reaction where either the acid or alkali is in excess (mixtures)

example 1

example 2

example 3