Calculate the pH for each of the cases in the titration of 25.0 mL of 0.220 M pyridine, C5H5N(aq) with 0.220 M HBr(aq) . The Kb of pyridine is 1.7×10−9.
Solved
Show answers
More tips
- H Health and Medicine Angina: Causes, Symptoms, and Treatment...
- H Health and Medicine What vaccines do children need?...
- H Health and Medicine Reasons for the Appearance of Warts: Everything You Need to Know...
- A Art and Culture How to Learn Screaming: Step-by-Step Guide for Beginners...
- H Health and Medicine Contraceptive Pills After 35: The Importance Of Choosing The Right Medication...
- C Computers and Internet How to Choose a Monitor?...
- H Horoscopes, Magic, Divination Where Did Tarot Cards Come From?...
- S Style and Beauty How to Make Your Lips Fuller? Ideas and Tips for Beautiful Lips...
- C Computers and Internet How to Learn to Type Fast?...
Answers on questions: Chemistry
- C Chemistry A is an insect which looks like butterfly. A is not formed as such from its eggs directly. The hatching of eggs produces a stage called B or C and then an encased from...
- C Chemistry Which types of energy are generated when a light bulb is turned on? A.Pontential and electrical B.Kinetic and electrical...
- C Chemistry In each case tell which sn2 reaction will proceed faster. 1. the displacement on 2-bromopropane by (a) ch3ch2o- or (b) cn-. 2. the displacement by oh- on (a) h2c=chbr...
- C Chemistry The form of heat transfer taking place in the crust of the earth would be...
- C Chemistry Which of the following elements will produce the same spectrum? carbon and oxygen hydrogen and helium gold and silver no two elements produce the same spectrum....
- C Chemistry When al(no3)3 is separated into its ions, how is it written?...
- C Chemistry What is an extensive property of marker ink?...
- C Chemistry Water is classified as an inorganic compound because it...
- C Chemistry How is carbon dioxide released by animals? * O during inhalation O during exhalation O during urination O during eating...
- C Chemistry E grantcosd.owschools.com/owsoo/studentAssignment/index?en=264406005 Z Zimbra: Inbox SKSRT Equipment. P Pandora Plus - Liste... S SPANS MDPAW Root Group | PRT... * Apps...
Ответ:
[A]. 9.26
[B]. 5.2
[C].4.9.
[D]. 3.1
[E].1.70
Explanation:
So, we are given the following parameters:
=> 25.0 mL of 0.220 M pyridine, C5H5N(aq), 0.220 M HBr(aq) and the Kb of pyridine = 1.7×10−9.
CASE ONE: Before the addition of HBr to pyridine.
C5H5N(aq) + H2O <> C5H5NH^+(aq) + OH^-
Kb = [C5H5NH^+][OH^-]/[C5H5N].
1.7×10^−9. = x^2/.190 - x.
x = 1.80 × 10^-5.
Therefore, pOH = - log[OH^-] = - log [1.80 × 10^-5] = 4.74.
Thus, pH = 14 - 4.74 = 9.26.
CASE TWO: after addition of 12.5 mL of HBr.
Number of moles of H^+ present= (The initial concentration of H^+) ×( addition of 12.5 mL of HBr) = 0.220 × 12.5 × 10^-3= 2.7 × 1O^-3 mol.
Number of moles of C5H5N present = 25 × 10^-3 × 0.220. = 5.5 × 10^-3 mol.
Therefore at equilibrium;
=> we have that 5.5 × 10^-3 mol - 2.7 × 1O^-3 mol = 2.8 × 10^-3 for C5H5N.
=> We have 2.7 × 1O^-3 mol - 2.7 × 1O^-3 mol = 0 for H^+ and 2.7 × 1O^-3 mol for C5H5NH^+.
Therefore, pH = - log (1.0 × 10^-14/ 1.7 × 10^-9) + log (2.8 × 10^-3/ 2.7 × 10^-3) = 5.23.
CASE THREE:
Number of moles of H^+ = 17 × 10^-3 × 0.220 = 3.74 ×10^-3 mol.
At equilibrium we have;
=> for C3H5N= 5.5 × 10^-3 mol - 3.74 ×10^-3 mol = 1.76 × 10^-3 and 3.74 ×10^-3 mol for C5H5NH^+.
pH = 5.23 + log (1.76 × 10^-3/3.74 × 10^-3) = 4.9.
CASE FOUR:
Number of moles of H^+ = 25 × 10^-3 × 0.220 = 5.5 × 10^-3 moles.
Thus, at equilibrium we have;
=> For C3H5N= 5.5 × 10^-3 mol - 5.5 × 10^-3 mol = 0, H^+ = 0 and C5H5NH^+ = 5.5 × 10^-3 mol.
C5H5NH^H+ + H20 > C5H5N + H3O^+.
Thus, the molarity of C5H5NH^H+ = 5.5 × 10^-3 mol/0.05 = 0.11M. At equilibrium, C5H5NH^H+ = 0.11 - x. Where C5H5 and H3O^+ are x respectively.
Therefore, ka = x^2/ 0.11 -x. (a).
Thus, we have ka = 1.0 × 10^-14/7.4 × 10^-9 = 5.8× 10^-6.
Slotting in the ka value into the equation (a) above.
x = 6.47 × 10^-7.
pH = - log { 6.47 × 10^-7} = 3.09.
CASE FIVE:
The concentration of H^+ = 1.1 × 10^-3/ 0.055L = 0.02
pH = - log { 0.02}
pH = 1.70
Ответ:
3.55 × 10n M n = -10