PH / Topic 12

Question 1

PH equation

Answer 1

-log10[H+]

Question 2

Pka Equation

Answer 2

-log10(Ka)

Question 3

Derive Kw

Answer 3

Water is slightly conductive therefore it must ionise
H2O —> H+(aq) + OH-(aq)
applying equilibrium law: Kw = [H+][OH-]/[H2O]
…[H2O] is a constant at rtp hence can be incorporated into K
… hence K = [H+][OH-]
… as water is neutral [H+] approx = [OH-]
…hence K = [H+]^2 at 298K
…H+ at 298K = 1.0x10^-7 hence Kw = 1.0x10^-14

Question 4

Pkw equation

Answer 4

PKw = -log10(Kw)

Question 5

Using the indicator constant Kin

Answer 5

At the equivalence point [HIN]=[In-]
hence… Kin = [H+][In-]/[HIn]
…[In-] approx = [HIn] at equivalence point hence Kin = [H+] as given by the data sheet

Question 6

What is a buffer solution

Answer 6

A solution which is able to resist small changes in PH when small volumes of acid/base are added.

Question 7

How is an acidic buffer solution made

Answer 7

Solution of weak acid and conjugate base

Question 8

How is an alkaline buffer solution made

Answer 8

Solution of weak base and conjugate acid

Question 9

Ph of buffer solution

Answer 9

PH = Pka + log[salt]/[acid]