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?

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

mol HCl=75 mlmol

Volume total :

mol NH₄Cl=mol NH₃=75 mlmol=0.075 mol

Dissociation of NH₄Cl at water to find [H₃O⁺]

ICE at equilibrium :

0.41-x            x        x

Ka(Kw:Kb)= 10⁻¹⁴ : 1.8.10⁻⁵=5.6.10⁻¹⁰

[H₃O⁺]=x :

pH=-log[H₃O⁺]

in the given question, the partial pressure is calculated by using the ideal gas equation, that is, pv = nrt

here, p is pressure, v is volume = 25 l, n is number of moles = 0.0104 moles, r is gas constant = -0.0821 l.atm / mol.k

t is temperature = 273 k

p = nrt / v

p = (0.0104 mol × 0.0821 l.atm / mol.k × 273 k) / 25 l

= 0.0093 atm

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