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please help- short buffer calculation will rate high!

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Question:

Cramster Expert

Werner

(Cramster SME)

Moderator

Cramster In-House Subject Matter Expert

Date Posted: 5/16/2008 10:04:02 AM Status: Live

Asker's Rating: Lifesaver

Response:

Given that

The molarity of C₂H₅NH₂ = 0.10 M

The volume of C₂H₅NH₂ = 100.0 ml

Kb = 5.6 * 10^-4

The molarity of HNO3 = 0.2 M

At equivalence point all the C₂H₅NH₂ will be converted as C₂H₅NH₃⁺

The volume of HNO3 required for this point is 50.0 ml because the molarity of acid is double to base.

The molarity of C₂H₅NH₃⁺ = 100.0 ml * 0.10 M / 150 ml

C = 0.067 M

The Ka for C₂H₅NH₃⁺ = Kw / Kb

= 1*10^-14/5.6 * 10^-4

= 1.786*10^-11

∴ [H⁺] = √Ka*C

= √1.786*10^-11 *0.067 M

= 1.093*10^-6 M

pH = - log([H⁺])

= - log(1.093*10^-6 M)

= 5.961

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