a 50 mL sample consisting of 0.15 M nicotine C10H14N2 is tot rated with 0.25 M...

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a 50 mL sample consisting of 0.15 M nicotine C10H14N2 is totrated with 0.25 M HBr. The Ka of nicotine is 1.0*10^-6 calculatethe pH after adding 5.0 mL of HBr at the equivalence point andafter adding 5.0 mL HBr beyond the equivalence point.

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4.3 Ratings (676 Votes)

millimoles of nicotine = 50 x 0.15 = 7.5

Ka = 1.0 x 10^-6

At the equivalence point :

millimoles of base = millimoles of acid

7.5 = 0.25 x V

V = 30 mL

volume at equivalence point = 30 mL

after the addition of 5.0 mL HBr beyond equivalence point :

beyond the equivalence point :

volume = 30 + 5 = 35

millimoles of HBr = 35 x 0.25 = 8.75

C10H14N2 +Â Â HBrÂ Â ---------------->Â Â salt

Â Â 7.5Â Â Â Â Â Â Â Â Â Â Â Â Â Â 8.75Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0

Â Â Â 0Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 1.25Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 7.5

here strong acid remains. so

concentration of HBr = 1.25 / (50 + 35) = 0.015 M

pH = -log [H+] = -log [0.015]

Â Â Â Â Â = 1.83

pH = 1.83

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a 50 mL sample consisting of 0.15 M nicotine C10H14N2 is tot rated with 0.25 M...

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60.1K

Verified Solution

Question

Chemistry

a 50 mL sample consisting of 0.15 M nicotine C10H14N2 is totrated with 0.25 M HBr. The Ka of nicotine is 1.0*10^-6 calculatethe pH after adding 5.0 mL of HBr at the equivalence point andafter adding 5.0 mL HBr beyond the equivalence point.

Answer & Explanation Solved by verified expert

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