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

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

$\tt 2.5\times 30=75~mlmol$

mol HCl=75 mlmol

• Volume HCl :

$\tt \dfrac{75}{0.5}=150~ml$

Volume total :

$\tt 150+30=180~ml$

• molarity of salt(NH₄Cl)

mol NH₄Cl=mol NH₃=75 mlmol=0.075 mol

$\tt M=\dfrac{0.075}{0.180}= 0.42$

• 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

$\tt Ka=\dfrac{NH_3.H_3O}{NH_4}=\dfrac{x^2}{0.41}$

[H₃O⁺]=x :

$\tt \sqrt{5.6.10^{-10}\times 0.41}=1.515.10^{-5}$

pH=-log[H₃O⁺]

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