Acid-Base Equilibria: If youve seen one equilibrium problem youve - - PDF document

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Acid-Base Equilibria: If youve seen one equilibrium problem youve - - PDF document

Acid-Base Equilibria: If youve seen one equilibrium problem youve seen em all What is the pH of 1 10 8 M HNO 3 ( aq )? A. 6.0 B. 7.0 C. 8.0 D. 9.0 1 Rank the following solutions in order of increasing pH: K a =


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Acid-Base Equilibria:

If you’ve seen one equilibrium problem you’ve seen ‘em all

What is the pH of 1  108 M HNO3(aq)?

  • A. 6.0
  • B. 7.0
  • C. 8.0
  • D. 9.0
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Rank the following solutions in order of increasing pH: 0.1 M HNO2, 0.1 M HNO3, 0.1 M NaNO2, 0.1 M NaNO3, 0.1 M NaOH

  • A. HNO2 < HNO3 < NaNO2 < NaNO3 < NaOH
  • B. HNO3 < HNO2 < NaNO2 < NaNO3 < NaOH
  • C. HNO2 < HNO3 < NaNO3 < NaNO2 < NaOH
  • D. HNO3 < HNO2 < NaNO3 < NaNO2 < NaOH
  • E. HNO3 < HNO2 < NaOH < NaNO3 < NaNO2

Ka = x2/(0.1‒x); x = [H+] pH = ‒log(0.1) Kb = x2/(0.1‒x); x = [OH‒] pH  7.0 pH = 14.00 ‒ [‒log(0.1)]

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You can calculate the pH of 0.10 M NH4Cl(aq) using the equation given on the right where K and x are A. Ka [H+] B. Ka [OH] C. Kb [H+] D. Kb [OH]

2

0.10 x K x  

The pH of 0.10 M HNO3 in 1.00. HNO2 is a weak acid with Ka = 4.5  104

.

The pH of 0.10 M HNO2 is...

  • A. < 1.00
  • B.  1.00
  • C. > 1.00
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0.10 M HNO2 (Ka = 4.5  104) is pH 2.19 What if the solution is 0.10 M HNO2 and 0.10 M NaNO2? Will the pH be:

  • A. < 2.19
  • B.  2.19
  • C. > 2.19
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Which acid / conjugate base pair would be best to prepare a 6.85 pH buffer? Three practical methods to prepare a buffer:

1- First Method : By the Titration, in the presence of one of the two buffer forms with strong base or acid:

 Prepare a buffer composed of an acid and its salt by adding a

strong base(e.g. NaOH) to a weak acid (e.g. Acetic acid) until the required pH is obtained

 If the other form of buffer is available (in this case sodium

acetate), a strong acid is added (e.g. HCl) until the required pH is

  • btained.

CH3COONa+HClCH3COOH+NaCl

 So acetate buffer is formed(CH3COOH/CH3COONa)

Source: fac.ksu.edu.sa/sites/default/files/BUFFER_0.ppt

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Advantages: Easy to understand. Useful when only one form of the buffer is available (in this case acetic acid)

Disadvantages:

1.

Slow.

2.

May require lots of base (or acid). 2- Second Method: Using the buffer pKa , calculate the

amounts (in moles) of acid/salt or base/salt present in the buffer at the desired pH.

 If both forms (i.e., the acid and the salt) are available, convert the

amount required from moles to grams ,using the molecular weight of that component, and then weigh out the correct amounts of both forms. Or convert moles to volume if the stock is available in the liquid form.

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Advantages:

1.

Fast.

2.

Easy to prepare.

3.

Additional pH adjustment is rarely necessary, and when necessary, the adjustment is small.

Disadvantages:

1.

Requires the buffer pKa

2.

and solving two equations.

A 50 mL aliquot of 0.10 M HNO2(aq) is titrated with 0.10 M NaOH(aq) HNO2(aq) + OH(aq)  NO2

 (aq) + H2O(l)

The pKa of HNO2 is 3.15. After 15 mL of OH(aq) is added, the pH will be…

  • A. less than 3.15
  • B. 3.15
  • C. greater than 3.15, but less than 7
  • D. greater than 7, but less than 10.85
  • E. greater than 10.85
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A 50 mL aliquot of 0.10 M HNO2(aq) is titrated with 0.10 M NaOH(aq) HNO2(aq) + OH(aq)  NO2

 (aq) + H2O(l)

The pKa of HNO2 is 3.15. After 70 mL of OH(aq) is added, the pH will be…

  • A. less than 13.00
  • B. 13.00
  • C. greater than 13.00

1 2 3 4 5 6 7 8 9 10 11 12 13 14 10 20 30 40 50 60 70 80 90 100 110

pH volume of 0.1 M NaOH(aq) added

2.19 48 4.73 49 5.04 49.9 6.05 49.99 7.04 49.999 7.82 50 8.01 50.001 8.20 50.01 8.97 50.1 9.97 51 10.96 52 11.26 65 12.08 70 12.18 75 12.26 100 12.48 1000 12.90 10000 12.94

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Acid-Base Models

Arrhenius Theory acids increase the concentration of H+ in water bases increase the concentration of OH‒ in water Brønsted-Lowry Theory acids are proton (H+) donors bases are proton (H+) acceptors Lewis Theory acids are lone pair acceptors bases are lone pair donors

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SLIDE 11

11 acidity increases with increasing charge acidity increases with decreasing size

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http://www.chemguide.co.uk/inorganic/group7/properties.html

Which of the following species is more acidic?

A B C D

Co2+ or Co3+ Co2+ Co2+ Co3+ Co3+ PH3 or SH2 PH3 SH2 PH3 SH2 MgO or SO2 MgO MgO SO2 SO2

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Which of the following species is more acidic?

A B C D

Ba2+ or Be2+ Ba2+ Be2+ Be2+ Ba2+ SH2 or SeH2 SH2 SeH2 SH2 SeH2 HClO3 or HClO2 HClO3 HClO3 HClO2 HClO2

To determine if 0.10 M HCO3

 is acid or basic,

you must compare Ka2 for HCO3

 against the Kb

for HCO3

 which is given by

  • A. Kb = Kw/Ka1
  • B. Kb = Kw/Ka2
  • C. Kb = Ka1/Kw
  • D. Kb = Ka2/Kw
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  • Fig. 13.10
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Which of the following possible totals for “rolling” two dice have the most ways of getting that total? A. 5 B. 6 C. 7 D. 8 E. 9

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How many ways are there to distribute six identical objects such that there are four on the left and two on the right? A. 6 B. 9 C. 12 D. 15 E. 20

Chemical Research News Items Use ScienceDaily (or some other digest for science research) to find an article describing recent work (January 2017 to present) involving a topic we covered since the last exam: zero-order kinetics; acid-base equilibrium, including buffers and titrations; entropy. Reply-to-all to this message with the following information.

  • 1. Title of the article.
  • 2. Link to the article.
  • 3. Topic(s) from the article we covered in class.
  • 4. Question(s) based on the research described in the article that would constitute
  • ne problem on the exam worth around 20 points. [In other words, the question(s)

would easily fit on one page of the exam including enough room for the answers.]

  • 5. The information you would need to work the problem.

The above information should be sent (remember to reply-to-all) by Monday evening, April 9. Be sure to check the thread before you begin to make sure no

  • ne already used that article.
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How does entropy change for the following processes? A B C D inc T S>0 S>0 S<0 S<0 liqgas S>0 S>0 S<0 S<0 inc ?? S<0 ?? S>0 complexity go into S>0 ?? S>0 ?? solution

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Note to students: A student sent me this song parody several years ago, and I thought it was so clever that I wanted to share it with the class. The person who wrote it really understands the language of thermodynamics, common over- simplifications, and implications of entropy. Given the genre of music, there is some profanity, but others had to point that out to me. The song is supposedly by "MC Hawking." Stephen Hawking is Einstein of your generation. You probably know that he recently succumbed to Amyotrophic Lateral Sclerosis, commonly known in this country as Lou Gehrig’s disease, a horrible illness that confined him to a wheel chair and forced him to use an electronic synthesizer for a speaking voice. The student who sent me this parody wanted to make sure that we accepted it as an imitation of Stephen Hawking's voice, and not as making fun of his physical condition. This student clearly has a high regard for Dr. Hawking, as do I. I wonder if 18-year olds even know who MC Hammer is.

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Entropy by "MC Hawking" selected lyrics Verse 1 Entropy, how can I explain it? I'll take it frame by frame it, to have you all jumping, shouting saying it. Let's just say that it's a measure of disorder, in a system that is closed, like with a border. It's sorta, like a, well a measurement of randomness, proposed in 1850 by a German, but wait I digress. . . . You ever drop an egg and on the floor you see it break? You go and get a mop so you can clean up your mistake. But did you ever stop to ponder why we know it's true, if you drop a broken egg you will not get an egg that's new. . . . Order from disorder is a scientific rarity, allow me to explain it with a little bit more clarity. Did I say rarity? I meant impossibility, at least in a closed system there will always be more entropy. . . . Verse 2 Defining entropy as disorder's not complete, 'cause disorder as a definition doesn't cover heat. So my first definition I would now like to withdraw, and offer one that fits thermodynamics second law. First we need to understand that entropy is energy, energy that can't be used to state it more specifically. In a closed system entropy always goes up, that's the second law, now you know what's up.

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You can't win, you can't break even, you can't leave the game, 'cause entropy will take it all 'though it seems a shame. The second law, as we now know, is quite clear to state, that entropy must increase and not dissipate. Creationists always try to use the second law, to disprove evolution, but their theory has a flaw. The second law is quite precise about where it applies,

  • nly in a closed system must the entropy count rise.

The earth's not a closed system' it's powered by the sun,

. . .