STRUCTURE DETERMINES FUNCTION – the notes … · Web viewACIDS & BASES Chapter Ten | 20 Dena K....

A C I D S & B A S E S C h a p t e r T e n | 1

NAME___________________________________________ PERIOD________

Chapter 10 ACID-BASE CHEMISTRY & EQUILIBRIUMAll states are aqueous unless indicated otherwise.

I. STRUCTURE DETERMINES FUNCTION – the notes below are covered in the Acid/Base Inquiry.

ACID STRUCTURE & STRENGTHTREND DIAGRAM EXPLANATION

BINARY ACIDS: INCREASE in strength LR across a period.Binary acids INCREASE in strength DOWN a group.

The acid strength is related to the bond strength (down a group) as well as the electronegativity of the element bonded to the “H” (across a period)


OXYACIDS:INCREASE in strength as the number of oxygen atoms on the central atoms INCREASES

Nitric acid has one more oxygen.

More electron density is pulled from the O-H bond.

H+ leaves more readily HNO3 is thus a stronger acid.

OXYACID: BE CAREFUL! Not all “H” atoms are acidic!

The element bonded directly to the “H” must be sufficiently electronegative to weaken the bond to the “H”. A bond between P-H is not acidic. Therefore H3PO4 is TRI-protic and H3PO3 is DI-protic

Dena K. Leggett, Ph.D., Allen High School, Allen, TX 2015-2016

Acidic Hydrogen



OXYACIDS: Acid strength increases UP the group rather than down as seen in the binary acids. This is because electronegativity is the key factor since the halogen is not bonded directly to the “H”

As the electronegativity of the halogen increases, electron density is pulled from the H-O bond, making the H more acidic – i.e. more readily ionizes as H+ in water!

Pictures: http://wps.prenhall.com/wps/media/objects/602/616516/index.html

LEGGETT APIB CHEM ACIDS BASES REVIEW ()

II. Review of pH, pOH, etc

You will be reviewing these calculations in your daily work, but I want to help you learn how to estimate since you will not be able to use your calculators on the Multiple Choice portion of the AP exam. BONUS – knowing how to estimate will help you catch mistakes on the Free Response!

LETS DO IT 1.

pH [H+], M Estimating guidelines2.002.252.753.00


A NOTE ON SIF FIGS: when you take the “log” of a number, the number(s) to the left of the decimal place indicate magnitude and are hence NOT significant. The numbers to the right of the decimal place are significant.

[H+] > [OH─] ACIDIC [H+] < [OH─] BASIC

http://wps.prenhall.com/wps/media/objects/602/616516/index.html


LEGGETT APIB CHEM ACIDS BASES 2 ()

III. Naming

Naming: The name of an acid is derived from the name of the anion (the cation is always H+). So…If you don’t know your polyatomic ions, then your are definitely ….. at a huge disadvantage! GO STUDY!

LETS DO IT 2.

NAME ANION FORMULAPermanganic acid

Sulfuric acid

Nitrous acid

Hypoiodous acid

Hydrobromic acid

LEGGETT APIB CHEM ACIDS BASES CONJUGATES ()

IV. Conjugates Reactants are always described as either an acid or a base The products are described as either a conjugate base (if reactant was an acid) or a conjugate acid (if the reactant was a

base). Reversible reactions: there are acids & bases going both directions! The “conjugate” terminology is not always used

stringently. The “modified form” of the acid is the form without an H+ - ie do what acids do best and take away an H+ to find the

conjugate. The “modified form” of a base is the form with an H+ added – ie do what bases do best and add an H+ to find the conjugate. IT IS VITAL THAT YOU ARE ABLE TO RECOGNIZE A CONJUGATE IN AN EQUILIBRIUM PROBLEM!

Conjugate acids and bases are often added in the form of salts. THEY ARE TYPICALLY PRODUCTS – THEY DO NOT REACT WITH THE ORIGINAL ACID OR BASE!

LETS DO IT 3. Fill in the following tables.

ACID CONJUGATE BASE POSSIBLE SALT OF CBH3PO3

H2C2O4


I “ate” something “ic”-yAll n”ite” I was naus”ous”Cause I took a r”ide” on a

“hydro” l “ic” plane


BASE CONJUGATE ACID POSSIBLE SALT OF CANH3

CH3CH2NH2

NOTE: do not use strong/weak comparisons use stronger/weaker. In other words, the STRONGER the acid, the WEAKER its conjugate base. Weak acids do not necessarily have strong conjugate bases, and in fact will not likely have a strong conjugate base.

Acid Formula KaCONJUGATE Kb

Hydrochloric HCl Strong Cl−¿¿ VERY WEAK

Nitric HNO3 Strong NO3−¿¿ VERY WEAK

Iodic HIO3 1.70E-01 IO3−¿ ¿ 5.88E-14

Oxalic HOOCCOOH 5.60E-02 HOOCOO−¿¿ 1.79E-13

Phosphorous H3PO3 3.00E-02 H 2 PO3−¿ ¿ 3.33E-13

Phosphoric H3PO4 7.11E-03 H 2 PO4−¿ ¿ 1.41E-12

Formic HCOOH 1.80E-04 HCOO−¿¿ 5.56E-11

Hydrazoic HN3 2.20E-05 N 3−¿ ¿ 4.55E-10

Hypochlorous HOCl 3.00E-08 OCl−¿ ¿ 3.33E-07

Trimethyl Ammonium Ion (CH3)3NH+ 1.58E-10 (CH 3 )3 N 6.33E-05

LETS DO IT 4. Predicting relative strength.

HOCl+ IO3−¿⇌HIO3+OCl−¿ Keq≪ 1¿ ¿

Which acid is stronger?

Which base is stronger?


K a×Kb=Kw


LEGGETT APIB CHEM ACIDS&BASES POLYPROTIC & AMPHOTERIC ()

V. Polyprotic acids & Amphoteric substancesPolyprotic mean many protons. When we write ionization reactions, remove only one proton at a time.

REMEMBER H+ = H3O+

H2SO4 H+ + HSO4─ BECOMES H2SO4 + H2O H3O+ + HSO4

─

H2SO4 + H2O H3O+ + HSO4─ H3PO4 + H2O ⇌ H2PO4

─ + H3O+

HSO4─ + H2O ⇌ H3O+ + SO4

2─ H2PO4─ + H2O ⇌ HPO4

2─ + H3O+

HPO42─ + H2O ⇌ PO4

3─ + H3O+

The second and third protons are much harder to remove. For a PURE H3PO4 system, the original H3PO4 is present in the greatest concentration and the PO4

3─ is present in the least concentration.

Amphoteric (amphiprotic) substances can act as both an acid and a base depending on the circumstances. We will see that WATER is amphoteric.

LETS DO IT 5. Circle the amphoteric substances in the equations above.

VI. Some Problem Solving Tools/Acronyms – I am not going to do a video – this is here for reference. I will use it a lot!

STOICHIOMETERY – THE “BSA” METHOD.

B BEFORE – write your starting values under the reaction. Moles or Molarity will work (must do dilution first!0S Shift – shift amounts from reactants to products (according to mole ratios) until the limiting runs out.A AFTER (= B + S) – write your final amounts remaining after shift (or reacting completely)

EQUILIBRIUM – THE “RICE” METHOD

R Reaction – write the balanced reaction remember to use “⇄”I Initial – Write initial values under the correct species in the balanced equation. If no products are explicitly given assume

value of zero. BE CAREFUL – salts of conjugates and the “stop” values of a stoichiometry often give initial concentrations.C Change – the amounts by which species change as they approach equilibrium (according to mole ratios)E Equilibrium (=I + C) – The amounts remaining when equilibrium has been reached. Given pH and pOH values are typically

considered equilibrium concentrations.

COMBINED – THE DSE or “DOC Saves Everyone” METHOD

D Dilution – if you added “volume to volume” you must VooM=VooM!S Stoichiometry – If you have (1) strong acid (2) strong base (3) any acid + any base – you must first do a stoichiometry.

NOTE: do not react acids or bases with their conjugates! AFTER amounts remaining will typically become INITIAL amounts in the equilibrium.

E Equilibrium – Find your weak acid or weak base equilibrium. Always start with the parent acid or base as the reactant unless the ONLY thing present in the solution is the conjugate. REMEMBER – you must add water as a reactant when working with a weak base!



Leggett APIB Strong Acid Strong Base (7 :44) http://youtu.be/H34yiAjP8kA http://vimeo.com/87632304

VII. Acid-Base Calculations – Four Beakers

Some of this work is overkill for some of you. It is okay to simplify as long as the mole ratio is clearly shown!Make sure you note the color and approximate pH of each of the solutions!

LETS DO IT 6. Calculate the pH of the four solutions. What is the percent ionization for each of these substances?

(a) 0.023 M HCl

Dilution?

HCl H+ + Cl ー

B

S

A

(b) 0.023 M Ca(OH)2

Dilution?

Ca(OH)2 Ca2+ + 2OH ー

BSA


0.023 M Ca(OH)2 0.023 M HCl 0.023 M NH3 0.023 M HC2H3O2

http://youtu.be/H34yiAjP8kA


Leggett APIB Weak Acid Weak Base (8 :32) http://youtu.be/rbU_WhfuN0s http://vimeo.com/87632306

(c) 0.023 M NH3

Dilution?Stoichiometry?

R NH3 + H2O ⇌ NH4+ OH ー

I

C

E

(d) 0.023 M HC2H3O2

Dilution?Stoichiometry?

R HC2H3O2 + H2O ⇌ C2H3O2ー H3O+

I

C

E



Leggett APIB Strong Acid Weak Base (10:7) http://youtu.be/lOR45bnJDVs http://vimeo.com/87632305

LETS DO IT 7. Mixing it up

(a) Calculate the pH when 20. mL of the 0.023 M solution of HCl is added to 50. mL of 0.023 M NH3

Dilution?

Stoichiometry?

BSA

Equilibrium?

R

I

C

E



Leggett APIB Weak Acid Strong Base (7:45) http://youtu.be/Q9Q-UBbbO2U http://vimeo.com/87632307

(b) Calculate the pH when 25 mL of the 0.023 M solution of Ca(OH)2 is added to 50. mL of 0.023 M HC2H3O2

Dilution?

Stoichiometry?

BSA

Equilibrium?

R

I

C

E



LEGGETT APIB CHEM ACIDS BASES INDICATORS ()

VIII. TITRATION:

Titration is a technique where a ________________________________________ (one of the reactants) is placed in a buret and slowly added to an ________________________________________ (the other reactant that is being analyzed). It is commonly used to determine the unknown molarity of the analyte, but in acid base reactions we can learn much more!

A. INDICATORSMost acid & base solutions are clear and colorless. Monitoring the pH using a pH probe shows the changes in pH as solutions are mixed, but this takes time. Often we just need to determine the volume of base (or acid) required to exactly neutralize an acid (or base). The occurs at the ________________________________________ point (moles titrant = moles analyte).

An acid base indicator changes color with pH, helping us to visualize an equivalence point. The following shows the structural changes involved in de-protonating thymol blue.

RED YELLOW BLUE

Let’s simplify: HIn+H2O⇌ ¿−¿+H3O+¿¿ ¿ K a=¿¿ ;

where HIn is the protonated form and In─ is the de-protonated form (sometimes called the modified form).

Indicators change color as soon as the ¿ is large enough for our eyes to detect the color change. For most indicators the APPROXIMATE range is:

0.1≤¿¿ SO K a=0.1× ¿ to K a=10× ¿

In terms of pH, the range becomes: pH=p K a±1 ; where p Ka=−log (K a)

The point at which the color changes is called the ________________________________________. We want the endpoint to be as close to the equivalence point as possible, so we pick an indicator where the p Ka is as close as possible to the pH at the equivalence point.

LETS DO IT 8. The acid dissociation constant for methyl orange is about 2 x 10-4. Is this a good indicator choice for and acid with an equivalence point at pH = 3.65?



LEGGETT APIB CHEM ACIDS BASES TITRATION ()

B. TITRATING A WEAK ACID WITH A STRONG BASE

A 40.00 mL sample of a 0.100 M solution of propanoic acid, C2H5COOH, is titrated with 0.200 M solution of NaOH. Calculate the pH at the following points. Ka = 1.3 x 10-5

(A) 0 mL (B) 10.00 mL (C) 15.00 mL (D) 20.00 mL (E) 30.00 mL

BEFORE YOU START: Determine the volume required to reach the equivalence point – EVEN IF NOT REQUESTED IN QUESTION. Use modified Voom-Voom. A titration deals with one equivalence point at a time, therefore #H+ = 1 and #OH─ = 1. YOU MUST SHOW THE “1” IN THE FORMULA TO CONVINCE THE READER THAT YOU ARE INCLUDING THE MMR!

(A) 0 mL base added. The only thing present is 40.00 mL of 0.100 M weak acid. No volume has been added and no base has been added therefore we do not need dilution and stoichiometry steps.

R C2H5COOH + H2O ⇌ C2H5COO─ + H3O+

I

C

E



(B) 10.00 mL – Half Equivalence Point

DILUTION

STOICHIO-METRY

C2H5COOH + OH─ C2H5COO─ + H2O

B WHO CARES!

S

A

EQUILIBRIUM


I

C

E

(C) 15.00 mL – in the BUFFER ZONE!

DILUTIONV total=40mL¿original acid+15mL¿ the added base=55mLV A1 M A1=V A 2 M A2 ; (40 ) (0.100 )=(55 ) M A2 M A2=0.0727V B 1MB1=V B2M B2 ; (15 ) (0.200 )= (55 ) MB 2 M B2=0.0545



STOICHIO-METRY


B WHO CARES!

S

A

EQUILIBRIUM


I

C

E



(D) 20.00 mL – since this is the equivalence point, the equilibrium is for the weak conjugate base! Two things to remember: The equilibrium is written for the conjugate and you have to calculate its K using Ka x Kb = Kw

DILUTIONV total=40mL¿original acid+20mL ¿the added base=60mLV A1 M A1=V A 2M A2 ; (40 ) (0.100 )=(60 ) M A2 M A2=0.0667V B 1 MB1=V B2 M B2 ; (20 ) (0.200 )=(60 )M B2 M B2=0.0667

STOICHIO-METRY


B WHO CARES!

SA

EQUILIBRIUM

R C2H5COO─ + H2O ⇌ C2H5COOH + OH─

Kb=Kw

Ka= 1 x10−14

1 . 3x 10−5

I

C

E



(E) 30.00 mL

DILUTIONV total=40mL¿original acid+30mL¿ the added base=70mLV A1 M A1=V A 2 M A2 ; (40 ) (0.100 )=(70 ) M A2 M A2=0.0571V B 1MB1=V B2M B2 ; (30 ) (0.200 )= (70 )M B2 M B2=0.0857

STOICHIO-METRY


B WHO CARES!

S

A

NOEQUILIBRIUM

NEEDED!

[OH─]TOTAL = [OH─]SB from stoich + [OH─]WCB from weak conj. base equil.

SINCE [OH─]SB >>> [OH─]WCB

[OH─]TOTAL ≈ [OH─]SB

The contribution the weak conjugate base makes to the pOH and hence the pH is NEGLIGIBLE compared to the excess strong base!

ALTERNATIVE TO EQUILIBRIUM WHEN BOTH A WEAK AND ITS CONJUGATE I PRESENT (AKA – THE BUFFER ZONE):

SHORTCUT!IF: You were able to identify that an acid and its conjugate base are present initially. (or base + conjugate acid)AND: initial concentrations > 100 x Ka (always true for AP Questions)THEN: You can skip RICE and use the Henderson-Hasselbach equation! (simply Ka rearranged)

pH=pK a+ log( [CB ][Acid ] )

pOH=pK B+log ( [CA ][base] )


CB = molarity of conjugate baseCA = molarity of the conjugate acidpKa = -log(Ka)pKb = -log(Kb)


C. Comparing Titration Curves1

Stronger acids start at a lower pH ________________________________________ ________________________ ________________________________________

Strong acids are much ________________________________________ before equivalence point and do not have a ________________________________________ region.

As the strength of the acid ________________________________________, the strength of the conjugate base ________________________________________, so the pH at the equivalence point ________________________________________, becomes more basic.

As the strength of the acid ________________________________________, the jump to the buffer zone ________________________________________

As the strength of a weak acid ________________________________________, the pH at the half equivalence point ________________________________________, the pKa ________________________________________ and the Ka ________________________________________.

For all of the curves, the pH after the equivalence point is dependent only on the molarity of the strong base.

Stronger bases start at a higher pH ________________________________________ ________________________ ________________________________________Strong bases are much ________________________ before equivalence point and do not have a ________________________ region.

As the strength of the bases ________________________, the strength of the conjugate acid ________________________, so the pH at the equivalence point ________________________, becomes more acidic.

As the strength of the base ________________________, the jump down to the buffer zone ________________________

As the strength of a weak base ________________________, the pH at the half equivalence point ________________________, the pKb creases and the Kb ________________________.

For all of the curves, the pH after the equivalence point is dependent only on the molarity of the strong acid.

1 http://2012books.lardbucket.org/books/principles-of-general-chemistry-v1.0/section_20/9e679e00ac436bc1b485369c9db8190f.jpgDena K. Leggett, Ph.D., Allen High School, Allen, TX 2015-2016


IX. Buffers

A. General Concepts A Buffer is a solution designed to resist changes in pH upon the addition of small amounts of acid or base. A buffer system is always an equilibrium.

HA+H 2O⇌ H 3O+¿+A−¿¿ ¿ B+H2O⇌OH−¿+HB+¿ ¿¿

Add acid, H+

Add base, OH−¿¿

A good buffer has [WEAK ACID] close to the [CB] OR [WEAK BASE] close to the [CA]. The concentrations do not need to be EQUAL to one another, but in a good buffer, they are CLOSE TO one another

K a=¿¿ ; K a≅ ¿ ; pKa≅ pH K b=¿¿; K b≅ ¿ ; pK b≅ pOH

The buffer capacity refers to the amount of H+ or OH─ a buffer system can absorb before the pH begins to change dramatically. Buffers containing high concentrations of the buffer agents typically have a high buffering capacity.

B. Four ways to make a buffer solution:


Weak acid + Salt of the conjugate base

Weak acid + strong base ~ 2:1 ratio

Weak base+ Salt of the conjugate acid

Weak base + strong acid~ 2:1 ratio


WEAK + OPPOSITE STRONG: Roughly 2:1 ratio is IDEAL, although not required.

H+ + HONH2 HONH3+

B 0.55 M 1.0 M 0S -0.55 -0.55 +0.55A 0 0.45 0.55

LETS DO IT 9. Check the pairs that can be used to make a buffer solution. Justify your choice.

PAIR: JUSTIFICATION

HNO3 with Sr(OH)2

C6H5COOH (benzoic acid) with C6H5COONa

C6H5NH2 with C6H5NH3Br

HCl with NaCl

LETS DO IT 10. Which of the following would be the best buffer for pH 4?

Hypochlorous HOCl 3.0x10─8

Iodic HIO3 1.7x10─1

Lactic CH3CHOHCOOH 1.38x10─4

RESOURCE: http://www.mhhe.com/physsci/chemistry/essentialchemistry/flash/buffer12.swf

LEGGETT APIB CHEM ACIDS BASES 8(9) http://youtu.be/D5Iwzmg74BA http://vimeo.com/35832201

LETS DO IT 11.(a) What ratio of molarities of conjugate acid and base would you use if you wanted to prepare a solution buffered at pH 10.0 using dimethylamine ((CH3)2NH). Kb = 5.4 x 10─4

(b) If you wanted to use 0.750 M dimethylamine, what molarity of the conjugate acid would you use?

(c) If you needed 1.000 L of solution, how many grams of diamethylamine and (CH3)2NH2Cl would you need?


OH─ + HC2H3O2 H2O + C2H3O2─

B 0.45 M 1.0 M 0S -0.45 -0.45 +0.45A 0 0.55 0.45

http://vimeo.com/35832201

http://youtu.be/D5Iwzmg74BA

http://www.mhhe.com/physsci/chemistry/essentialchemistry/flash/buffer12.swf


LEGGETT APIB CHEM ACIDS BASES 10 (14:20) http://youtu.be/sFgRybgrMpw http://vimeo.com/35832757

X. Salt Hydrolysis – Psych! Salts aren’t always neutral!NOTE: The equilibrium calculation is just like the equivalence point on a titration curve!

LETS DO IT 12. Write the full hydrolysis reactions and predict whether the following salts would be acidic, neutral, or basic.

SALT REACTION(s) PREDICTIONNH4NO3

SALT REACTION(s) PREDICTIONKCHO2

Potassium formateCa(ClO4)2

C2H5NH3F Ka(HF) = 6.8x10─4 Kb(C2H5NH2) = 6.4 x 10─4

LETS DO IT 13. Determine the pH of a solution prepared when 0.288 g of potassium cyanide is added to 0.25 L of water. Ka (HCN)= 4.9 x 10─10

✔D:

✔ S:

✔ E:

R CN─ + H2O ⇌ HCN + OH─

I

C

E


http://vimeo.com/35832757

http://youtu.be/sFgRybgrMpw


LETS DO IT 14. Calculate the pH of a 0.392 M CH3NH3NO3. Kb(CH3NH2) = 4.4 x 10─4

PREDICTION:

✔ D: Do you need to do any dilutions?✔ S: Are there any stoichiometry calculations?

✔ E: Equilibrium - Now find the acid or base and do the equilibrium!

R K expression:

I

C

E


STRUCTURE DETERMINES FUNCTION – the notes … · Web viewACIDS & BASES Chapter Ten | 20 Dena K....

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Transcript of STRUCTURE DETERMINES FUNCTION – the notes … · Web viewACIDS & BASES Chapter Ten | 20 Dena K....