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What is the pH at the quivalence point when 50.0 mL of 0.10 M ethylamine (C2H5NH2)...

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Chemistry

What is the pH at the quivalence point when 50.0 mL of 0.10 Methylamine (C2H5NH2) is titratedwith 0.20 M HCLO4? (Kb ofC2H5NH2 = 5.6 x 10-4)(Hint: You will need to determine the volume ofHCLO4 solution used) Be sure to show ICE table.

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3.8 Ratings (554 Votes)

moles of ethylamine = 0.1 M x 50 ml = 5 mmol

At equivalence point moles of base = moles of acid added

So moles of HClO4 = 5 mmol

Volume of HClO4 added = 5 mmol/0.2 M = 25 ml

C2H5NH2 + HClO4 ---> C2H5NH3ClO4

concentration of salt formed [C2H5NH3+] = 5 mmol/75 ml = 0.067 M

           C2H5NH3+ + H2O <==> C2H5NH2 + H3O+

initial     0.067         excess              -                -    

change     -x            excess             +x             +x

final      0.067-x        excess              x               x

Ka = [C6H5NH2][H3O+]/[C6H5NH3+]

Kw/Kb = 1 x 10^-14/5.6 x 10^-4 = x^2/0.067 - x

let x be a small amount and can be ignored in the denominator

x = [H3O+] = 1.10 x 10^-6 M

pH = -log[H+] = 5.96
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