A 22.7 mL sample of 0.208 M diethylamine, (C2H3)2NH, is titrated with 0.263 M nitric acid....

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Question

Answer 1

Answer #1

(C2H5)2NH + HNO3 ----------> (C2H5)2NH⁺ + NO3^-

millimoles of base = 22.7 x 0.208 = 4.7216

4.7216 millimoles HNO3 must be added

4.7216 = V x 0.263

V = 17.95 mL HNO3 must be added

total volume = 17.95 + 22.7 = 40.65 mL

[salt] = 4.7216 / 40.65 = 0.116 M

for salt of weak base and strong acid

pOH = 1/2 [pKw + pKb + log C]

pKb of diethyl amine = 3.0

pOH = 1/2 [14 + 3.0 + log 0.116]

pOH = 8.03

pH = 14 - 8.03

pH = 5.97

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Answer 2

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