Calculate the hydronium ion concentration and the pH when 50.0mL of 0.40M NH 3 is mixed with 50.0mL of 0.40 M HCl.

Interpretation:

The hydronium ion concentration and the pH has to be calculated when 50.0mL of 0.40M NH3 is mixed with 50.0mL of 0.40 M HCl.

Concept introduction:

Equilibrium constants:

The equilibrium constant is used to quantitative measurement of the strength of the acid and bases in the water.

Ka is acid constant for equilibrium reactions.

HA + H2O⇌H3O++ A-Ka= [H3O+][A-][HA]

Kb is base constant for equilibrium reaction.

BOH + H2O⇌B++ OH-Ka= [B+][OH-][BOH]

Ionic product constant for water  Kw= [H3O+][OH-]       =1.00×10-14 pH = -log[H3O+]pOH= -log[OH-]Reltion between pH and pOH pH + pOH =14

Given volume of 0.40MNH3 = 50mL

And it is mixed with 50mL of 0.40M HCl. Therefore, the hydronium ions from HCl neutralize ammonia a weak base to its conjugate acid. NH4+.

Hydronium ions reacts with ammonia to give NH4+.

The equilibrium chemical reaction is as follows.

NH3(aq) +H3O+(aq) ⇌NH4+(aq) + H2O(aq)

Let’s calculate the amount of consumed HCl:

= (50mL×1L1000mL) × (0.40molL)=  0.02 mol

Let’s calculate the consumed amount of NH3

= (50mL×1L1000mL) × (0.40molL)=  0.02 mol

Let’s calculate the produced amount of NH4+ after completion of reaction.

= 0.02mol NH3  × 1mol NH4+1mol NH3= 0.02 mol NH4+

The total volume of solution = 50.0mL + 50.0mL= 100.0mL × 1L1000mL= 0.1L

Concentration of NH4+ =0.02 mol0.1L= 0.2M

Therefore, 0.2M is the final volume of NH4+ after completion of the reaction.

The NH4+ ions react with water. The resulting solution is acidic in nature.

The equilibrium reaction is as follows.

NH4+(aq) + H2O(l) ⇌NH3(aq) + H3O+(aq)

The equilibrium expression:

Ka= [NH3][H3O+][NH4+]

Ka of NH4+ is 5