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Titrations...need urgent help! thanks

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Question:

Cramster Expert

Werner

(Cramster SME)

Moderator

Cramster In-House Subject Matter Expert

Date Posted: 5/16/2008 10:21:47 AM Status: Live

Asker's Rating: None Provided Moderator's Rating: N/A

Response:

The molarity of NH3 = 0.9M

The volume of NH3 = 75.0 ml

The molarity of HCl = 1.5 M

The volume of HCl = 45.0 ml

The concentration of NH₄⁺ = 0.9M*75ml / 120ml

C = 0.5625 M

Given that

Ka of NH₄⁺ = 5.6x10^-10
We know that

[H₃O⁺]= √Ka*C

= √ 5.6x10^-10 * 0.5625

= 1.775 * 10^-5M

pH = -log([H₃O⁺])

= - log (1.775 * 10^-5M )

= 4.75

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