a.NH₃+HCl⇒NH₄Cl
b.volume HCl=150 ml
c. pH=4.82
Reaction
NH₃+HCl⇒NH₄Cl
The equivalence point⇒mol NH₃=HCl
Titration formula :
M₁V₁n₁=M₂V₂n₂(n=acid base valence, NH₃=HCl=1)
mol NH₃
[tex]\tt 2.5\times 30=75~mlmol[/tex]
mol HCl=75 mlmol
[tex]\tt \dfrac{75}{0.5}=150~ml[/tex]
Volume total :
[tex]\tt 150+30=180~ml[/tex]
mol NH₄Cl=mol NH₃=75 mlmol=0.075 mol
[tex]\tt M=\dfrac{0.075}{0.180}= 0.42[/tex]
Dissociation of NH₄Cl at water to find [H₃O⁺]
[tex]\tt NH_4+H_2O\rightarrow NH_3+H_3O^+[/tex]
ICE at equilibrium :
0.41-x x x
Ka(Kw:Kb)= 10⁻¹⁴ : 1.8.10⁻⁵=5.6.10⁻¹⁰
[tex]\tt Ka=\dfrac{NH_3.H_3O}{NH_4}=\dfrac{x^2}{0.41}[/tex]
[H₃O⁺]=x :
[tex]\tt \sqrt{5.6.10^{-10}\times 0.41}=1.515.10^{-5}[/tex]
pH=-log[H₃O⁺]
[tex]\tt pH=5-log~1.515=4.82[/tex]
