What you know:
-the strong acids
-the weak acids
-the weak (molecular) bases - like NH3
KaKb = Kw
cations = +ve ions
anions = -ve ions
Consider the acid HNO3 (aq), the nitrate ion (NO3-) is the anion from a strong acid
Consider the acid HClO(aq), the hypochlorite ion (ClO-) is the anion from a weak acid
Consider the weak molecular base NH3, NH4+ is the conjugate acid of the weak molecular base
the charts on page 803 can help the identification process
General Salt Rules
anions
-anions of strong acids are neutral (if you know the strong acids then you know the six neutral anions)
-anions of weak acids are basic (If it didn't come from a strong acid, then it came from a weak acid....)
cations
-conjugate acids of weak molecular bases are acidic
-alkali/alkaline earth ions are neutral
except Be2+ = acidic and high charge density ions like Al3+; also acidic
Using the above rules, salts can be identified as acidic basic or neutral.
Consider KCl (K+ & Cl-)
K+ is neutral (alkali metal); Cl- = neutral (anion of strong acid)
-salt is neutral
Consider NH4NO3
NH4+ = acidic (conjugate of weak molec base); NO3- = neutral (anion of strong acid)
-salt is acidic
NaClO = a basic salt (due to the ClO- anion; from a weak acid)
Question: Determine the pH of a 0.100M NaClO solution:
llbm equn: ClO-(aq) + H2O <=> HClO(aq) + OH-(aq)
use ICE chart to organize info
Kb expression** {it is a base afterall....:<}
-solve for x where x = [OH-]
-solve for pOH, then pH
-simple, right.....?
**since the question requires Kb - it must be available somewhere....
back to the top - for a weak and its conjugate, their respective Ka & Kb values are connected: KaKb=Kw
-Kw=1.0x10^-14
-Ka for HClO is on the p803 chart
-via KaKb=Kw, the value for Kb can be determined
Follow-up Questions P 588(1bc,2,3,4)
-now about the acidity/basicity of a metal oxide...., what about a nonmetal oxide...hmmm - Google anyone?
With all of this examination of K, a knock-knock joke: Knock, knock Who's There? Cow-go
Cow-go Who? No, Cow go MOO!!!
Thursday, November 26, 2009
Wednesday, November 25, 2009
Weak, weak day 2
Another day, more weak acid and weak base
The following information is intended to assist you with one
(or two) of the questions assigned from day 2 of A/B llbm
The K value gives an indication about the amount of product.
The larger the K value, the great the amount of product.
In the world of acids and bases, the larger the K,
the greater [H3O+] and/or [OH-]
The greater the [H3O+] the lower the pH.
The greater the [OH-], the lower the pOH, and higher pH.
Ready for more acid base llbm questions?
Assigned Questions:
A few textbook questions to answer:
Page 579:
Weak acids: 4,5,6,7ab(for part 7b – calculate using only HNO3 and HF)
Weak bases: 9ac, 10ab(for part 10b-calculate using only atropine), 16
Relationship between Ka and Kb ? – read pages 559 to 563, then answer:
Page 563 (6) Page 579 (14)
Finished all of the above? Begin the next organic handout – reactions of alkanes/enes/ynes. Don’t be afraid to look in the textbook.
A few weak jokes. What is the best way to carve wood? [Ans: Whittle by whittle] Don't be afraid of the next couple. If fruit comes from a fruit tree, what kind of tree does a chicken come from? [Ans: A poul-tree] Why did Beethoven get rid of his chickens? [Ans: Because the kept saying "Bach, Bach, Bach."]
The following information is intended to assist you with one
(or two) of the questions assigned from day 2 of A/B llbm
The K value gives an indication about the amount of product.
The larger the K value, the great the amount of product.
In the world of acids and bases, the larger the K,
the greater [H3O+] and/or [OH-]
The greater the [H3O+] the lower the pH.
The greater the [OH-], the lower the pOH, and higher pH.
Ready for more acid base llbm questions?
Assigned Questions:
A few textbook questions to answer:
Page 579:
Weak acids: 4,5,6,7ab(for part 7b – calculate using only HNO3 and HF)
Weak bases: 9ac, 10ab(for part 10b-calculate using only atropine), 16
Relationship between Ka and Kb ? – read pages 559 to 563, then answer:
Page 563 (6) Page 579 (14)
Finished all of the above? Begin the next organic handout – reactions of alkanes/enes/ynes. Don’t be afraid to look in the textbook.
A few weak jokes. What is the best way to carve wood? [Ans: Whittle by whittle] Don't be afraid of the next couple. If fruit comes from a fruit tree, what kind of tree does a chicken come from? [Ans: A poul-tree] Why did Beethoven get rid of his chickens? [Ans: Because the kept saying "Bach, Bach, Bach."]
Tuesday, November 24, 2009
Weak acid, weak base
Acids = proton donors.
Weak acids - those that are not strong!
Weak acids do not ionize 100%.
Weak acids set up equilibrium with water to produce the acid's anion and hydronium ion.
If the general representation of a weak acid is HA, then in llbm:
HA(aq) + H2O <=> H3O+(aq) + A-(aq)
Generally two question types are asked
(1) Determine the Ka value for a particular weak acid
-given llbm information about [H3O+] via pH and/or %ionization information
-where pH can be used to determine [H3O+] @ llbm
-and where %ionization = [H3O+]llbm/[HA] x 100%
(2) Determine pH (or % ionization) given specific acid(useKa chart to determine Ka value)
Complete mathematical solution would include:
-llbm equation
-Ka expression
-ICE chart
-solve for whatever it is that is being asked
-looking for pH or %ionizn?
-solve for [H3O+]llbm, then calculate to get requested info
-looking for Ka, use pH or %ionzn to determine [H3O+]
Weak bases - similar set up as with weak acids
Difference, weak bases involve [OH-], which leads to pOH
If the geneal formula for a weak base is 'Base' and bases are proton acceptors
Generally: Base(aq) + H2O <=> BaseH+(aq) + OH-(aq)
Again given two basic question types
(1) Determine pH of a stated basic solution.
(2) Determine Kb of a stated base.
Complete mathematical solution would include:
-llbm equation
-Kb expression
-ICE chart
-solve for whatever it is that is being asked
-looking for pH?
-solve for [OH-]llbm , then determine pOH, then pH
@25C pH + pOH = 14.00
-looking for Kb use given pH (or pOH) to get to [OH-]llbm
Homework:
p568(8) p570(10) p574(12,13) p579(1,3,13)
Next up?
Using the Ka/Kb chart to calculate K values of conjugate ions
KaKb = Kw
Cute joke coming, ready? Why did the chicken cross the playground? [Ans: To get to the other slide.]
Weak acids - those that are not strong!
Weak acids do not ionize 100%.
Weak acids set up equilibrium with water to produce the acid's anion and hydronium ion.
If the general representation of a weak acid is HA, then in llbm:
HA(aq) + H2O <=> H3O+(aq) + A-(aq)
Generally two question types are asked
(1) Determine the Ka value for a particular weak acid
-given llbm information about [H3O+] via pH and/or %ionization information
-where pH can be used to determine [H3O+] @ llbm
-and where %ionization = [H3O+]llbm/[HA] x 100%
(2) Determine pH (or % ionization) given specific acid(useKa chart to determine Ka value)
Complete mathematical solution would include:
-llbm equation
-Ka expression
-ICE chart
-solve for whatever it is that is being asked
-looking for pH or %ionizn?
-solve for [H3O+]llbm, then calculate to get requested info
-looking for Ka, use pH or %ionzn to determine [H3O+]
Weak bases - similar set up as with weak acids
Difference, weak bases involve [OH-], which leads to pOH
If the geneal formula for a weak base is 'Base' and bases are proton acceptors
Generally: Base(aq) + H2O <=> BaseH+(aq) + OH-(aq)
Again given two basic question types
(1) Determine pH of a stated basic solution.
(2) Determine Kb of a stated base.
Complete mathematical solution would include:
-llbm equation
-Kb expression
-ICE chart
-solve for whatever it is that is being asked
-looking for pH?
-solve for [OH-]llbm , then determine pOH, then pH
@25C pH + pOH = 14.00
-looking for Kb use given pH (or pOH) to get to [OH-]llbm
Homework:
p568(8) p570(10) p574(12,13) p579(1,3,13)
Next up?
Using the Ka/Kb chart to calculate K values of conjugate ions
KaKb = Kw
Cute joke coming, ready? Why did the chicken cross the playground? [Ans: To get to the other slide.]
Thursday, November 19, 2009
Acid- Base: pH/pOH Calculations
Strong Acids
If it is not a stong acid, it is a weak acid.
Know the strong acids (monoprotic and diprotice)
Strong acids ionize virtually 100%
HCl(aq) + H2O => H3O+(aq) + Cl-(aq)
pH = -log[H3O+]; 10^-pH = [H3O+];
Question: Determine the pH of a 0.150M HCl solution
(since strong acid ionizes 100%, [HCl] = [H3O+]=0.150M
pH = -log(0.150M) = 0.824
Notice: 3 sgfigs in concentration 0.150M; 3 decimal places
in pH value 0.824
Determine the concentration of a monoprotic acid solution if the pH
of the solution is 4.35
10^-4.35 =4.5x10^-5 M
Notice: 2 decimal places n pH value, 2 sigfigs in concentration answer
Only the decimal places of pH values are significant.
In a similar way that pH/[H3O+] connect, so can
pOH and [OH-]
pOH = -log[OH-] and 10^-pOH = [OH-]
Connecting pH and pOH?
@25C: pH + pOH = 14.00
Question: Determine the pH of a 0.250M NaOH solution
Strong bases dissociate virtually 100%
NaOH is a strong base, dissociates 100%
NaOH(aq) => Na+(aq) + OH-(aq)
So, [NaOH] = [OH-] = 0.250M
When working with a base, one cannot determine pH directly from the given base concentration . Determine pOH, then determine pH
pOH = -log(0.250M) = 0.602
pH = 14.00-0.602 = 13.398
there is another way to carry out this calculation
- saving that for another time
A few basic jokes to end this post: What happened when the lion ate the comedian? Ans: It felt funny. Where do sheep go to get a haricut? Ans: To the Baa Baa shop. They are simple, basic jokes. Get it? This post dealt with bases, so it ended with some basic jokes!
If it is not a stong acid, it is a weak acid.
Know the strong acids (monoprotic and diprotice)
Strong acids ionize virtually 100%
HCl(aq) + H2O => H3O+(aq) + Cl-(aq)
pH = -log[H3O+]; 10^-pH = [H3O+];
Question: Determine the pH of a 0.150M HCl solution
(since strong acid ionizes 100%, [HCl] = [H3O+]=0.150M
pH = -log(0.150M) = 0.824
Notice: 3 sgfigs in concentration 0.150M; 3 decimal places
in pH value 0.824
Determine the concentration of a monoprotic acid solution if the pH
of the solution is 4.35
10^-4.35 =4.5x10^-5 M
Notice: 2 decimal places n pH value, 2 sigfigs in concentration answer
Only the decimal places of pH values are significant.
In a similar way that pH/[H3O+] connect, so can
pOH and [OH-]
pOH = -log[OH-] and 10^-pOH = [OH-]
Connecting pH and pOH?
@25C: pH + pOH = 14.00
Question: Determine the pH of a 0.250M NaOH solution
Strong bases dissociate virtually 100%
NaOH is a strong base, dissociates 100%
NaOH(aq) => Na+(aq) + OH-(aq)
So, [NaOH] = [OH-] = 0.250M
When working with a base, one cannot determine pH directly from the given base concentration . Determine pOH, then determine pH
pOH = -log(0.250M) = 0.602
pH = 14.00-0.602 = 13.398
there is another way to carry out this calculation
- saving that for another time
A few basic jokes to end this post: What happened when the lion ate the comedian? Ans: It felt funny. Where do sheep go to get a haricut? Ans: To the Baa Baa shop. They are simple, basic jokes. Get it? This post dealt with bases, so it ended with some basic jokes!
Wednesday, November 18, 2009
llbm Review
-complete the llbm & solubility review handout
-check moodle for introduction to acid/base equilibrum handouts
-check moodle for upcoming quiz/test dates
-check moodle for introduction to acid/base equilibrum handouts
-check moodle for upcoming quiz/test dates
Tuesday, November 17, 2009
Will a precipitate occur?
Will 35.0mL of 0.0100M NaCl form a precipitate when combined
with 65.0 mL of 0.0150M AgNO3?
with 65.0 mL of 0.0150M AgNO3?
- write the double displacement to identify the slightly soluble salt (the precipitate)
- silver chloride in the example
- determine the concentration of the ions of the slightly soluble salt using c1V1=c2V2
- Cl and Ag in this example
- for Cl: (0.0350L)(0.0100M)=c2(0.100L*);
- * 35mL + 65mL = 100mL = 0.100L
- Write the llbm equation for the slightly soluble salt then the Ksp expression
- Use the calculated concentrations, sub into Ksp to determine a "trial K"
- compare "trial K" to real K (from Ksp chart)
- if trial K less than real K = no precipitate (not enough of the ions present to cause a ppt to form)
- if trial K greater than real K = ppt (too great of an amount of the ions in the solution - excess ppts out)
Monday, November 16, 2009
Lab Follow-Up
The lab involved neutralizing 25.00mL of an unknown concentration of a calcium hydroxide solution with a determined volume (via the buret) of 0.100M hydrochloric acid.
The solution is already at equilibrium when the titrating begins.
With the titration data it is possible to:
The titration data is all that is needed to complete the calculations.
Next up - Acid-Base llbm. Two introductory sheets are on moodle. Notice the moodle calendar for upcoming quiz dates.
Talent alone won't make you a success. Neither will being in the right place at the right time, unless you are ready. The most important question is: 'Are your ready?' Johnny Carson Are you ready for acid-base solution equilibrium?
The solution is already at equilibrium when the titrating begins.
With the titration data it is possible to:
- determine molar solubility of the calcium hydroxide in pure water
- determine Ksp value for calcium hydroxide
- determine molar solubility of calcium hydroxide in a calcium chloride solution
The titration data is all that is needed to complete the calculations.
Next up - Acid-Base llbm. Two introductory sheets are on moodle. Notice the moodle calendar for upcoming quiz dates.
Talent alone won't make you a success. Neither will being in the right place at the right time, unless you are ready. The most important question is: 'Are your ready?' Johnny Carson Are you ready for acid-base solution equilibrium?
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