CHEM 212 Extra-Credit: DUE: Friday, October 27 by the END of lecture Consider th

Question

CHEM 212 Extra-Credit: DUE: Friday, October 27 by the END of lecture Consider the mixing of 25.00 mL of 0.2000 M Pyridine (CsHsN) with 0.5000 M HCI. Pyridine is a weak base with Kb = 1.4 × 10-9, Determine the pH after the following volumes of 0.5000 M HCl been added to the Pyridine solution. A) pH before any HCI is added. B) pH after the addition of 5.00 mL of HCI is added C) pH after the addition of 10.00 mL of HCI is added. D) pH after the addition of 15.00 mL of HCI is added Be sure to use enough significant digits (we used four in class). I want you to write out any relevant equilibrium reactions (double arrow 'back and forth') or complete reactions (single 'one way' arrow) that you use (pay attention to charges and phases). Keep your work in logical order, and please BE NEATI Please staple multiple pages together

Explanation / Answer

As HcL is added pyridine acts as a buffer due to the formation of pyridinium ion conjugate acid.

Py + H+ ------------------> PyH+1

Kb of pyridine = 1.4x10-9

thus pKb = 8.8538

A) before HCl is added.

  Py + H+ ------------------> PyH+1

25x0.20M 0 0 initial mmoles

=5

the pH of weak base = 14- pOH

and pOH = 1/2 [pkb -logc]

= 1/2 [8.8538 - log 0.2]

= 4.776

and pH = 14-4.776

= 9.224

b) after 5 mL of 0.500M HCl

Py + H+ ------------------> PyH+1

25x0.2=5.0 5x0.5=2.5 0 initial mmoles

2.5 0 2.5 after reaction

It acts as a buffer.

The pH of buffer is given by Hendersen equation

pOH = pKb + log [conjugate acid]/[base]

= 8.8538 + log 2.5/2.5

= 8.8538

c)afteradding 10mL of 0.50M Hcl

Py + H+ ------------------> PyH+1

25x0.2=5.0 10x0.5=5 0 initial mmoles

0 0 5 after reaction

[salt] = mmoles/ volume

= 5/35

=0.1428 M

the pH of a salt of weak base and strong acid is acidic and is given by

pH = 1/2 [pKw -pkb -logC]

= 1/2 [14 - 8.8358 -log 0.1428]

= 3.00

d) after adding 15mL of 0.5M HCl

  

Py + H+ ------------------> PyH+1

25x0.2=5.0 15x0.5=7.5 0 initial mmoles

0 2.5 5.0 after reaction

thus the solution now contains a salt and strong acid , so the ph is decided only by the strong acid.

thus [strong acid] = 2.5/40=0.0625 M

and pH = -log 0.0625

=1.2041

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