20.5 - pH and Strong Bases

Question 1

What occurs during the ionisation of water?

Answer 1

Question 2

Derivd Kw using Ka

Answer 2

Ka = [H⁺][OH^-]/[H₂O]

Water is mainly undissociated (very slightly)

1dm³ (1000g) of water…

n_H2O = 1000/18 = 55.6moldm^-3

n=cv, v = 1, therefore n=c, 55.6 = [H₂O]

Ka.[H₂O] = [H⁺][OH^-]

= Kw

Question 3

What is Kw?

Answer 3

The ionic product of water - [H⁺][OH^-]

Kw varies with temperature

At 25^oC or 298K, Kw is a constant at 1 x 10^-14 mol²dm^-6

Question 4

How is pH calculated with whole numbers?

Answer 4

Proton and hydroxide indices add up to -14