buozone x M Final exam in Science ds X Faculty of Science l Unive X C chegg stud

Question

buozone x M Final exam in Science ds X Faculty of Science l Unive X C chegg study IGuided sol x 3 x x M Assignment 9 C ezto.mheducation.com /hm.tpx EE Apps N Netflix Member Lo... uozone M sign In IMcGraw-Hii... D Mat 1341A YouTube to mp3 Co... A Google Maps D teaching subjects c... t Log in ITumblr 4.00 points Question Assistance Be sure to answer all parts. View Hint Find the pH and the volume (mL) of 0.527 M HNO3 needed to reach the equivalence View Qu point in the titration of 2.65 L of 0.0750 M pyridine (C5H5N). Show Me Guided Solution Volume Question Help mL HNO3 Report a Problem pH References Difficulty: 2 Multipart Answer ENG 34 AM Ask me anything US 2015-12-14

Explanation / Answer

number of moles of pyridine = V x M ( volume x molarity)

2.65 L( 0.0750 M) = 0.199 mole pyridine

To reach the equivalent point we need 0.199 mole H+

so volume of HNO3 (V) = number of moles / malarity
V (mL) = 0.199 /0.447 = 0.4452 L of HNO3 = 445.2 mL HNO3

H+ + C5H5N >>>C5H5NH+

Total volume = 2.65 L + 0.4452 L = 3.1 L

Concentration C5H5NH+ = 0.199 / 3.1 =0.0642 M

For the dissociation reaction

C5H5NH+ <> C5H5N + H+
initial concentration
0.0642.. ... .... .. .. .0...... ..0
at equilibrium
0.0642-x.... ..... .... ..x... ... ..x
(C5H5N Kb = 1.7 10-9)
The equilibrium constant is Kw/Ka = 10^-14/1.7 10^-9=

=5.88 10^-6

5.88 10^-6 = (x)(x) /0.0642-x ( x <<0.0642 so we neglect x)

x= 0.000614 M

pH = 3.21

Related Questions

Navigate

• Browse (All)
• Browse B
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.