Question

Hydrogen ion concentration of acid is 1 x10 ^-2mol/i what is its ph

Answer 1

The concentration of hydrogen ions1, H+(aq), in an aqueous solution of a strong Arrhenius base can be calculated if we know :

(i) the temperature of the solution

and

(ii) either 
    (a) the pOH of the solution 
    or 
    (b) the concentration of hydroxide ions in solution.

For an aqueous basic (alkaline) solution at 25°C2

[H+] = 10-14 ÷ [OH-] 
where [H+] = concentration of hydrogen ions3 in mol L-1 
and [OH-] = concentration of hydroxide ions in mol L-1 
and 10-14 is the dissociation constant for water4 at 25°C

For an aqueous basic (alkaline) solution at 25oC

[H+] = 10-14 ÷ [OH-] 
but [OH-] = 10-pOH

So, [H+] = 10-14 ÷ [10-pOH] 
where pOH = the pOH of the solution 
and [H+] = concentration of hydrogren ions in mol L-1 
and 10-14 is the dissociation constant for water at 25°C
Calculating the Hydrogen Ion Concentration of Strong Arrhenius Bases

Before the base is added to water, water molecules are in equilibrium with hydrogen ions and hydroxide ions:

waterhydrogen
ions+hydroxide
ionsH2OH+(aq)+OH-(aq)

Only a very small number of water molecules dissociate into H+(aq) and OH-(aq).

At 25°C, [H+(aq)] = [OH-(aq)] ≈ 10-7 mol L-1 (a very low concentration) 
and Kw = [H+(aq)][OH-(aq)] = 10-7 × 10-7 = 10-14

When a strong Arrhenius base is added to water, the base dissociates completely to form OH-(aq) and hydrated metal cations:

base→hydroxide
ions+metal cationsGroup 1metal hydroxide:MOH→OH-(aq)+M+(aq)Group 2metal hydroxide:M(OH)2→2OH-(aq)+M2+(aq)

Adding the base to the water disturbs the water dissociation equilibrium:

H2O  H+(aq) + OH-(aq)


By Le Chatelier's Principle, adding more OH-(aq) to the water will shift the equilibrium position to the left. 
The water dissociation equilibrium system responds to the addition of more OH-(aq) by reacting some of the OH-(aq) with some of the H+(aq) in order to re-establish equilibrium. 
So, increasing the concentration of OH-(aq) in the water, reduces the concentration of H+(aq), but, the value of the dissociation constant for the water does not change5, Kw is still 10-14. 
So, Kw = [H+(aq)][OH-(aq)] = 10-14

At 25°C[H+(aq)]
mol L-1[OH-(aq)]
mol L-1Kwpure water
(before base added)10-710-710-14aqueous solution
(after base added)< 10-7> 10-710-14

We can use the value of Kw and [OH-(aq)] to calculate [H+(aq)] at a given temperature:

At 25°C: Kw = [H+(aq)][OH-(aq)] = 10-14


By rearranging this equation (formula) we can determine the concentration of hydrogen ions in the aqueous solution:

[H+(aq)] = 10-14 ÷ [OH-(aq)]


Both [H+(aq)] and [OH-(aq)] must be in units of mol L-1 (mol/L or M)

If we know the pOH of the basic solution, we know the concentration of hydroxide ions in the solution

because pOH = -log10[OH-(aq)]


By rearranging this equation (formula) we can find the concentration of hydroxide ions in mol L-1:

[OH-(aq)] = 10-pOH


This value of [OH-(aq)] can then be used to calculate [H+(aq)] in mol L-1 using the appropriate value for Kw.