Edexcel A Level Chemistry:复习笔记5.2.4 Ionic Product of Water

Ionic Product of Water, Kw

 

  • In all aqueous solutions, an equilibrium exists in water where a few water molecules dissociate into protons and hydroxide ions
  • We can derive an equilibrium constant for the reaction:

5.5.3-Deriving-Kw

  • This is a specific equilibrium constant called the ionic product for water
  • The product of the two ion concentrations is always 1 x 10-14 mol2 dm-6
  • This makes it straightforward to see the relationship between the two concentrations and the nature of the solution:

[H+] & [OH–] Table

8.1.8-H-and-OH-table

The relationship between Kw and pKw is given by the following equation:

pKw = -logKw

pKa

  • The range of values of Ka is very large and for weak acids, the values themselves are very small numbers

Table of Ka values

5.6.2-Table-of-Ka-values

  • For this reason it is easier to work with another term called pKa
  • The pKa  is the negative log of the Ka value, so the concept is analogous to converting [H+] into pH values

pKa = -logKa

  • Looking at the pKa values for the same acids:

Table of pKa values

5.6.2-Table-of-pKa-values

  • The range of pKa values for most weak acids lies between 3 and 7

pH Calculation of a Strong Base

  • Strong bases are completely ionised in solution

BOH (aq) → B+ (aq) + OH- (aq)

  • Therefore, the concentration of hydroxide ions [OH-] is equal to the concentration of base [BOH]
    • Even strong alkalis have small amounts of H+ in solution which is due to the ionisation of water
  • The concentration of OH- in solution can be used to calculate the pH using the ionic product of water
  • Once the [H+] has been determined, the pH of the strong alkali can be founding using pH = -log[H+]

5.5.3-Finding-pH-of-strong-bases

  • Similarly, the ionic product of water can be used to find the concentration of OH- ions in solution if [H+] is known, simply by dividing Kw by the [H+]

Worked Example

pH calculations of a strong alkali

Question 1:

Calculate the pH of 0.15 mol dm-3 sodium hydroxide, NaOH

Question 2:

Calculate the hydroxide concentration of a solution of sodium hydroxide when the pH is 10.50

Answer

Sodium hydroxide is a strong base which ionises as follows:

NaOH (aq) → Na+ (aq) + OH- (aq)

Answer 1:

The pH of the solution is:

[H+] = Kw  ÷ [OH-]

[H+] = (1 x 10-14) ÷ 0.15 = 6.66 x 10-14

pH = -log[H+]

= -log 6.66 x 10-14  = 13.17

Answer 2

Step 1: Calculate hydrogen concentration by rearranging the equation for pH

pH = -log[H+]

[H+]= 10-pH

[H+]= 10-10.50

[H+]= 3.16 x 10-11 mol dm-3

Step 2: Rearrange the ionic product of water  to find the concentration of hydroxide ions

Kw = [H+] [OH-]

[OH-]= Kw  ÷  [H+]

Step 3: Substitute the values into the expression to find the concentration of hydroxide ions

Since Kw is 1 x 10-14 mol2 dm-6,

[OH-]= (1 x 10-14)  ÷  (3.16 x 10-11)

[OH-]= 3.16 x 10-4 mol dm-3

 

 

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