Question

Just enough 0.500 M HCl is added to 30.0 mL of 2.5 M NH3 to reach the equivalence point. The Kb of NH3 = 1.8 X 10-5

Write the balanced equation for this reaction.




What volume of 0.500 M HCl solution was added?



What is the molarity of the salt produced from the neutralization reaction?




What is the pH of the solution at the equivalence point?

Answer 1

a.NH₃+HCl⇒NH₄Cl

b.volume HCl=150 ml

c. pH=4.82

Further explanation

Reaction

NH₃+HCl⇒NH₄Cl

The equivalence point⇒mol NH₃=HCl

Titration formula :

M₁V₁n₁=M₂V₂n₂(n=acid base valence, NH₃=HCl=1)

mol NH₃

`\tt 2.5* 30=75~mlmol`

mol HCl=75 mlmol

• Volume HCl :

`\tt (75)/(0.5)=150~ml`

Volume total :

`\tt 150+30=180~ml`

• molarity of salt(NH₄Cl)

mol NH₄Cl=mol NH₃=75 mlmol=0.075 mol

`\tt M=(0.075)/(0.180)= 0.42`

• pH of solution

Dissociation of NH₄Cl at water to find [H₃O⁺]

`\tt NH_4+H_2O\rightarrow NH_3+H_3O^+`

ICE at equilibrium :

0.41-x x x

Ka(Kw:Kb)= 10⁻¹⁴ : 1.8.10⁻⁵=5.6.10⁻¹⁰

`\tt Ka=(NH_3.H_3O)/(NH_4)=(x^2)/(0.41)`

[H₃O⁺]=x :

`\tt \sqrt{5.6.10^(-10)* 0.41}=1.515.10^(-5)`

pH=-log[H₃O⁺]

`\tt pH=5-log~1.515=4.82`