• 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:

• 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

Strong bases

• 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+]

• 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+

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

Answer

The correct option is?D.

• • Since?Kw?= [H+] [OH–], rearranging gives [H+] ?=?Kw?÷ [OH–]

The concentration of ?[H+] is (1 × 10?14) ÷ (1.0 × 10?3) = 1.0 × 10?11?mol dm?3

[H+]= 10-pH

So the?pH = 11.00