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Aniline hydrochloride, [C6H5NH3]Cl, is a weak acid. Its conjugate base is the weak base aniline C6H5NH2....

Aniline hydrochloride, [C6H5NH3]Cl, is a weak acid. Its conjugate base is the weak base aniline C6H5NH2. Assume 50.0 mL of 0.100 M C6H5NH3 is titrated with 0.125M NaOH. (C6H5NH3)Cl Ka = 2.40 X 10-5. What is the ph at 40.0ml Calculate the pH after 100.0 ml of NaOH has been added

Expert Solution

1)

millimoles of C6H5NH3 = 50 x 0.1 = 5

millimoles of NaOH = 0.125 x 40 = 5

C6H5NH3 + NaOH -----------------------> C6H5NH2 + H2O

5 5 0 0

0 0 5

C6H5NH2 concentration = 5 / (50 + 40) = 0.056 M

pOH = 1/2 [pKb - log C]
pOH = 1/2 [9.38 - log 0.056]

pOH = 5.32

pH + pOH = 14

pH = 8.68

2)

millimoles of C6H5NH3 = 50 x 0.1 = 5

millimoles of NaOH = 0.125 x 100 = 12.5

C6H5NH3 + NaOH -----------------------> C6H5NH2 + H2O

5 12.5 0 0

0 7.5 5

NaOH strong base its concentration = 7.5 / (50 + 100) = 0.05M

pOH = -log OH-
pOH = 1.30

pH + pOH = 14

pH = 12.70

queen_honey_blossom answered 2 years ago

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