Wrapping Up pH Calculations with a Surefire Stepwise Process | CH 302, Study notes of Chemistry

Material Type: Notes; Professor: Laude; Class: PRINCIPLES OF CHEMISTRY II; Subject: Chemistry; University: University of Texas - Austin; Term: Unknown 1989;

Typology: Study notes

Pre 2010

Uploaded on 08/30/2009

koofers-user-zkl
koofers-user-zkl 🇺🇸

10 documents

1 / 3

Toggle sidebar

This page cannot be seen from the preview

Don't miss anything!

bg1
Lecture 14: Wrapping up pH calculations with a surefire stepwise process for working problems
If we make the right assumptions about K values (far apart) and concentrations (large) then we can apply approximations
so that there are only six equations needed to solve acid base problems. Even better, there are only five variables to
consider in using these equations: Ka, Kb, [H+], [OH-], C acid, C base
Strong acid [H+] = Ca
Strong base [OH-] = Cb
Weak acid [H+] = (Ka Ca)1/2
Weak base [OH-] = (Kb Cb)1/2
Acid buffer [H+] = Ka Ca/Cb
Basic buffer [OH-] = KbCb/Ca
So there isn't a lot of complexity at the bottom of this. The hard part is figuring out which equation to use and what each
of the variables is. To accomplish this task, we use the following procedure: 1) strip away all the extraneous information
(spectator ions), 2) identify strong acids and bases, 3) identify weak acids and bases, 4) determine if you should
neutralize, 5) perform neutralization calculation, 6) decide whether to work the problem as an acid or a base. Once
these steps are done, the problem is greatly simplified to the point that you can use the table above to work a calculation.
The back of this page shows every possible type of starting conditions and how they reduce to one of the problems above.
1) Getting rid of spectator ions. Always eliminate the ions that do nothing: all alkali metals and alkali earths (Na+, K+,
Ca++) and all conjugate bases of strong acids (Cl-, NO3-, ClO4-, I-, Br-). Thus
NH4Cl is NH4+ NaOH is just OH- KCOOH is just COOH-
2) Identify strong acids and bases. Strong acids are HCl, HNO3, H2SO4, HClO4, HBr, Hl. Strong bases are NaOH,
KOH, Mg(OH)2, Ba(OH)2 and other alkali metal or earth hydroxides. Notice what happens when you get rid of
spectator ions for strong acids and bases.
HCl become H+ HNO3 becomes H+ NaOH becomes OH- Mg(OH)2 becomes 2OH-
In other words, all strong acids are H+. All strong bases are OH-.
3) Identify weak acids and weak bases. Hint: this done by looking for the words: weak acid or weak base; it is also
done by looking for a small Ka or small Kbvalues, (numbers like 1.4 x 10-5 or 6.3 x 10-9, it is also done by looking for
the word acid in a compound that is not strong acid; it is also done by looking for the suffix ate.
Thus formic acid is a weak acid and sodium malonate is a weak base.
And how do you represent a weak acid? HA (instead of HCH3CH2COO which only serves to confuse you).
And how do you represent a weak base: A- (instead of NaCH3CH2COO which only serves to confuse you).
By the time you are through with step 3, you will have identified the presence of all acids and bases. You should have
only six possible symbols representing them:
H+ or OH- for strong acids and bases
HA or BH+ for weak acids
B or A- for weak bases
Any other terminology is a waste of time on a test without much time.
4) If possible, NEUTRALIZE. You neutralize if:
you have both an acid and a base present
one or both of the acid or base are strong
for example:
HCl and Sodium Acetate are H+ and A- so neutralize
Acetic acid and NaOH are HA and OH- so neutralize
HCl and NaOH are H+ and OH- so neutralize
Acetic acid and sodium acetate are HA and A- so do not neutralize
pf3

Partial preview of the text

Download Wrapping Up pH Calculations with a Surefire Stepwise Process | CH 302 and more Study notes Chemistry in PDF only on Docsity!

Lecture 1 4 : Wrapping up pH calculations with a surefire stepwise process for working problems If we make the right assumptions about K values (far apart) and concentrations (large) then we can apply approximations so that there are only six equations needed to solve acid base problems. Even better, there are only five variables to consider in using these equations: Ka, Kb, [H+], [OH-], C acid, C base

Strong acid [H+] = Ca Strong base [OH-] = Cb Weak acid [H+^ ] = (Ka Ca ) 1/ Weak base [OH -^ ] = (Kb Cb ) 1/ Acid buffer [H+^ ] = Ka Ca /Cb Basic buffer [OH -^ ] = Kb Cb /Ca

So there isn't a lot of complexity at the bottom of this. The hard part is figuring out which equation to use and what each of the variables is. To accomplish this task, we use the following procedure: 1) strip away all the extraneous information (spectator ions), 2) identify strong acids and bases, 3) identify weak acids and bases, 4) determine if you should neutralize, 5) perform neutralization calculation, 6) decide whether to work the problem as an acid or a base. Once these steps are done, the problem is greatly simplified to the point that you can use the table above to work a calculation. The back of this page shows every possible type of starting conditions and how they reduce to one of the problems above.

  1. Getting rid of spectator ions. Always eliminate the ions that do nothing: all alkali metals and alkali earths (Na+^ , K +^ , Ca++^ ) and all conjugate bases of strong acids (Cl-^ , NO 3 -^ , ClO 4 - , I -^ , Br -^ ). Thus NH 4 Cl is NH 4 +^ NaOH is just OH-^ KCOOH is just COOH-

  2. Identify strong acids and bases. Strong acids are HCl, HNO 3 , H 2 SO 4 , HClO 4 , HBr, Hl. Strong bases are NaOH, KOH, Mg(OH) 2 , Ba(OH) 2 and other alkali metal or earth hydroxides. Notice what happens when you get rid of spectator ions for strong acids and bases.

HCl become H +^ HNO 3 becomes H +^ NaOH becomes OH-^ Mg(OH) 2 becomes 2OH -

In other words, all strong acids are H +^. All strong bases are OH-^.

  1. Identify weak acids and weak bases. Hint: this done by looking for the words: weak acid or weak base; it is also done by looking for a small Ka or small Kb values, (numbers like 1.4 x 10 -5^ or 6.3 x 10 -9^ , it is also done by looking for the word acid in a compound that is not strong acid; it is also done by looking for the suffix ate. Thus formic acid is a weak acid and sodium malonate is a weak base.

And how do you represent a weak acid? HA (instead of HCH 3 CH 2 COO which only serves to confuse you). And how do you represent a weak base: A-^ (instead of NaCH 3 CH 2 COO which only serves to confuse you).

By the time you are through with step 3, you will have identified the presence of all acids and bases. You should have only six possible symbols representing them: H+^ or OH-^ for strong acids and bases HA or BH+^ for weak acids B or A-^ for weak bases

Any other terminology is a waste of time on a test without much time.

  1. If possible, NEUTRALIZE. You neutralize if:
  • you have both an acid and a base present
  • one or both of the acid or base are strong

for example:

  • HCl and Sodium Acetate are H+^ and A-^ so neutralize
  • Acetic acid and NaOH are HA and OH-^ so neutralize
  • HCl and NaOH are H+^ and OH-^ so neutralize
  • Acetic acid and sodium acetate are HA and A-^ so do not neutralize
  1. To neutralize, you convert both acid and base into moles. Then create a neutralization reaction into which you place the initial mole amounts. Identify the limiting reagent and then calculate the final mole amounts. Convert back to molarity by dividing by total volume if necessary. Examples:

• 5 moles H +^ and 5 moles A-^ → 5 moles of HA plus 0 moles of H+^ and A -

• 1 moles of H+^ and 2 mole of A-^ → 1 mole of HA with one mole of A-^ left over.

• 0.03 moles of OH -^ and 0.01 moles of HA → 0.01 moles A -^ with 0.02 moles OH-^ left over

Note that after neutralization, you can still have a weak base problem, a weak acid problem, a buffer, a

strong acid problem or a strong base problem. In other words, you have to do a neutralization to find out

what kind of problem you have.

6) Decide on your calculation terrain. Do you work with acids: calculate with pH, H+^ and Ka. Want to work

with bases? Calculate with pOH, OH -^ and Kb. It doesn't matter what you choose but remember to give the

answer they ask for (H+, OH-, pH or pOH). How do you move between acid and base terrain? Use:

• to move from a Ka to a Kb : Kw = KaKb =10-14^ or pKw = pKa + pKb =

• to move from a pH to a pOH: Kw =[H +] [OH-] = 10 -14^ or pKw = pH + pOH = 14

Examples of Acid/Base Problems Using Different Starting Materials

in calculations use Ka for acetic acid = 1.8 x 10-5^ and Kb for ammonia = 1.8 x 10-

Starting Materials Materials after neutralization

Equation to use Sample problem Calculate pH

Examples that use the strong acid equation

Strong acid alone H +^ pH = -log [H+] 0.2 M HNO 3

Strong acid and weak acid H+^ and HA (ignore HA)

pH =-log [H +^ ] 0.2 M HNO 3 and 0.4 M acetic acid Strong acid and weak base H+^ and HA (ignore HA)

pH = -log [H +^ ] 0.2 M HNO 3 and 0.1 M sodium acetate

Examples that use the strong base equation

Strong base alone OH -^ alone pOH = -log [OH-^ ] 0.1 M Ba(OH) 2

Strong base and weak base

OH -^ and A- (ignore A-)

pOH = -log [OH -^ ] 0.1 M Ba (OH) 2 and 0.1M sodium acetate Strong base and weak acid OH-^ and A- (ignore A -^ )

pOH =-log [OH -^ ] 0.4 M Ba(OH) 2 and 0.1M ammonium chloride

Examples that use the weak acid equation

Weak acid HA or BH+^ [H +] = (K (^) aC (^) a)1/2^ 0.3 M acetic acid

Equivalent strong acid and weak base

HA or BH +^ H +^ ] = (K a Ca ) 1/2^ 0.1M HCl and 0.1 M

ammonia

Examples that use the weak base equation

Weak base A-^ or B [OH - ] = (K (^) bC (^) b) 1/2^ 0.2 M NH (^3)

Equivalent strong base and weak acid

A -^ or B [OH -^ ] = (K (^) bCb)1/2^ 0.1M NaOH and 0.1M acetic acid

Examples that use the acid buffer equation

Weak acid and conjugate weak base

HA and A [H +^ ] = K (^) a Ca /Cb 0.2 M acetic acid and 0.1M sodium acetate Strong acid and weak base HA and A [H+^ ] =K (^) a Ca /Cb 0.2 M HCl and 0.4 M sodium acetate

Examples that use the basic buffer equation

Weak base and conjugate weak acid

B and BH +^ [OH -^ ] = K b Cb /Ca 0.2 M ammonia and 0.3 M

ammonium chloride

Strong base and weak acie B and BH +^ [OH -^ ] = K b Cb /Ca 0.3 M Ba(OH) 2 and 0.4 M

ammonium chloride