Topic 14: Acids & Bases : Topic 14: Acids & Bases Zumdahl 7e, Chapter 14 & 15
I. Definitions of Acids and Bases : I. Definitions of Acids and Bases Arrhenius
Acids produce H+ ions in water (e.g. HCl, CH3COOH, H2SO4)
Bases produce OH- ions in water (e.g. NaOH, NH3)
Bronsted-Lowry
Acids are proton donors (e.g. HCl gives away H+ to become Cl-)
Bases are proton acceptors (e.g. NH3 accepts H+ to become NH4+)
Lewis
Acids are electron pair acceptors (e.g. BF3)
Bases are electron pair donors (e.g. NH3)
Conjugate Acid-Base Pairs : Conjugate Acid-Base Pairs Conjugate acid-base pairs are related by a hydrogen ion on either side of the equation.
E.g. HCl ? H+ + Cl-
HCl is an acid; Cl- is the conjugate base (difference of one H+)
E.g. NH3 + H+ ? NH4+
NH3 is a base, NH4+ is the conjugate acid (difference of one H+)
What is the conjugate base of perchloric acid? Of the bicarbonate ion? What is the conjugate acid of the bicarbonate ion?
The conjugate base of a strong acid is always weak (Cl- is an extremely weak base). The conjugate base of a weak acid is always strong (CH3COO- is a strong base; it strongly attracts protons). Same goes for bases.
d. Amphoteric Substances : d. Amphoteric Substances These compounds can act as either acids (donate H+) or bases (accept H+)
E.g. HCO3- : can become H2CO3 or CO32-
Water and most di- and triprotic acids can make amphoteric substances
E.g. H2O can become H3O+ or OH-
Phosphoric acids yields two amphoteric substances: H2PO4- and HPO4-
e. Strength vs. Concentration : e. Strength vs. Concentration Percent Ionization
The greater the degree of ionization/dissociation, the stronger the acid
E.g. HCl dissociates entirely, CH3COOH doesn’t
The greater the number of moles in a solution, the higher the concentration
E.g. 1.9 M CH3COOH is more concentrated, but weaker, than 0.001 M HCl
It is possible to have a dilute strong acid and a concentrated weak acid
Danger is unrelated to strength and concentration!!!
HF, which is classified as a weak acid because it doesn’t dissociate, is still extremely corrosive (attacks glass)
ii. Prediction of acidic/basic strength : ii. Prediction of acidic/basic strength Oxides of elements on the left side of the periodic table are strong bases (the X-O bond is ionic, and breaks easily)
Na – O – H will break between Na and O
Oxides of elements on the right side of the periodic table are strong acids (the X-O bond is covalent and doesn’t break easily)
H – O – Cl will break between H and O
The more oxygen atoms on an oxyanion, the stronger the acid (H2SO4 vs. H2SO3) due to decreased electron density on the O-H bond
HClO4 > HClO3 > HClO2 > HClO
Do Examples 6 & 7
iii. Ka and Kb : iii. Ka and Kb The equilibrium constant for acids and bases expresses how strong the acid/base is (small K means weaker acid, since less dissociation
For HA (aq) + H2O (l) ? H3O+ (aq) + A- (aq)
Ka = [H3O+] [A-] / [HA]
This usually turns into [H3O+]2 / [HA]
For B (aq) + H2O (l) ? HB+ (aq) + OH- (aq)
Kb = [OH-] [HB+] / [B]
This usually turns into [OH-]2 / [B]
% ionization = 100 x Mdissociated / Minitial
Weak acids will dissociate more in a dilute solution
Find Ka for a 1.00 M HF solution with 2.7% ionization.
Do Example 3
f. The pH scale : f. The pH scale pH ranges from 0-14, where pH < 7 is acidic, pH = 7 neutral, and pH > 7 is basic
pH = -log [H+]
pOH = log [OH-] (the opposite of pH)
pH + pOH = 14
Kw (the ionization constant of water) is 1.0 x 10-14
Kw = [H+] [OH-]
Kw = (Ka) (Kb)
pKa = -log Ka
pKb = -log Kb
Do Examples 1 & 2
Not the whole story! : Not the whole story! Cannot use pH = -log [M of acid] for weak acids, because not all of it has dissociated into actual H+ ions.
1.5 M HCl has 1.5 M H+ because of total dissociation, but 1.5 M CH3COOH doesn’t have 1.5 M H+ because it’s a weak acid
Need to use Ka expression to calculate actual [H+]
For polyprotic acids (e.g. H2SO4), there are multiple Ka values for the successive ionizations
1st ionization: H2SO4 ? H+ + HSO4- Ka1
2nd ionization: HSO4- ? H+ + SO42- Ka2
Ka1 >> Ka2
Calculate the pH of a 1.13 M oxalic acid solution if Ka1 = 6.5 x 10-2 and Ka2 = 6.1 x 10-5. Find the oxalate ion and hydroxide ion concentrations.
Do Example 4
II. Acid-Base Neutralization Reactions : II. Acid-Base Neutralization Reactions Acid + Base ? Salt + H2O
pH ~ 7, but not always, due to electrolysis of salts (more on that later)
Indicators: chemicals that change color depending on whether they are in an acidic/basic sol’n; used to tell when a titration is complete
Usually a weak acid: HIn ? H+ + In-, shifts according to Le Chatelier’s principle
See Titrations Handout for a list of indicators, colors, and pH ranges
Want to choose an indicator whose pH change happens within the pH range of your titration
b. Titrations : b. Titrations Calculations
MV = MV, adjusted for di- and triprotic acids
Stoichiometry, with the appropriate ratio
Sample curves
SA + SB WA + SB SA + WB WA + WB
Most indicators phenolphthalein methyl orange no sharp change in pH
c. Hydrolysis of salts : c. Hydrolysis of salts Conjugates of weak acids/bases are strong bases/acids that react with ions left in sol’n after the titration, giving a non-neutral final pH
E.g. SB + WA final pH > 7
NaOH + CH3COOH ? Na+ + CH3COO- + H2O
BUT! CH3COO- + H2O ? CH3COOH + OH-
Strong conjugate base reacts with water, stripping off the proton and leaving hydroxide ions
E.g. SA + WB final pH < 7
HCl + NH3 ? NH4+ +Cl-
BUT! NH4+ + H2O ? NH3 + H3O+
Strong conjugate acid reacts with water, giving away proton and leaving hydronium ions
Slide 13 : E.g. WA + WB final pH depends
Kb > Ka then final pH > 7
Kb < Ka then final pH < 7
Kb = Ka then final pH = 7
Predict the pH of these salt solutions:
Na3PO4
KI
NH4F
LiNO3
Calculate the pH of a 0.5 M NaNO2 sol’n if the Ka of HNO2 is 4.0 x 10-4.
Do Example 5
III. Buffer Systems : III. Buffer Systems Definition of a buffer solution
A sol’n that resists pH change upon the addition of strong acids/bases (up to a certain point)
Usually a weak acid/base and a salt of its conjugate
E.g. NH3 and NH4Cl, or CH3COOH and NaCH3COO
The base part of the buffer reacts with any added acids
The acid part of the buffer reacts with any added base
pHbuffer = pKa + log ([salt]/[acid])
E.g. Calculate the pH of a buffer made from 0.250 M formic acid, HCOOH, and 0.100 M potassium formate. (Ka = 1.8 x 10-4)
Do Example 8
b. Application of ICE Strategy to calculations : b. Application of ICE Strategy to calculations 10 mL of 6.00 M NaOH is added to 500 mL of the formic acid buffer described in the previous example. Find the final pH of the solution.
HCOOH + OH- ? HCOO- + H2O
Do Example 9
Ultra Topic 13-14 Problem : Ultra Topic 13-14 Problem Consider the titration of 50.0 mL of 0.10 M acetic acid with 0.10 M of NaOH. Calculate the pH of the solution at the following points in the titration process:
Before any NaOH has been added
10.0 mL of the 0.10 M NaOH has been added
25.0 mL of the 0.10 M NaOH has been added
40.0 mL of the 0.10 M NaOH has been added
50.0 mL of the 0.10 M NaOH has been added
60.0 mL of the 0.10 M NaOH has been added