Chemistry: Acids & Bases Flashcards
Strong Acids
Strongest: HCl, H2SO4, HNO
HBr, HI, HClO4 (HPO4?), HClO3
Strong acid Formula Perchloric acid HClO4 Hydroiodic acid HI Hydrobromic acid HBr Sulfuric acid H2SO4 Hydrochloric acid HCl Nitric acid HNO3 Hydronium ion H3O+ or H+
Strong Bases
Strong bases Formula Lithium hydroxide LiOH Sodium hydroxide NaOH Potassium hydroxide KOH Rubidium hydroxide RbOH Cesium hydroxide CsOH Calcium Hydroxide Ca(OH)2 Strontium hydroxide Sr(OH)2 Barium hydroxide Ba(OH)2
What is the ionization constant of H2S?
much less than 1
WEAK ACID!
-SH vs. -OH
Because of charge density (charge per volume):
- SH: weak
- OH: strong
Share ~same charge but S has larger volume it has lower charge density… means more stable and therefore less reactive
charge density= charge/volume
pH
-log[H+]
Assuming 25 degrees Celsius:
pH + pOH =
14
What is the relationship between the variables on the L and R sides of the equations? pH=-log[H+] pOH=-log[OH] pKa = -log[Ka] pKb = -log[Kb]
inverse:
ie: if pH goes down, [H+] goes up
a) What is pH if [H+] = 1.0 x 10^-1?
b) What is pOH if 1.0 x 10^8
(@25 degrees C)
a) 1
b) -8
Steps to convert concentrations of H+ and OH- into pH and pOH (@25 degrees C)
Examples
A) H+: 6.4 x 10^-3
B) OH-: 2.1 x 10^-8
A) H+: 6.4 x 10^-3
1) For ones value of answer (pH from [H+]): Lock in on the power of 10. Flip the sign and subtract 1.
(10^-3) –> -3 –> +2
2) For decimal of answer: Subtract the digits (before the power of 10) from 10.
(6. 4 ~ 6) –> 10-6 = 4
3) Put it together: 2.4 = pH
_______________
B) OH-: 2.1 x 10^-8
pOH=7.8
(In 3 steps)
Convert pH into [H+]
a) pH: 13.3
b) pH: 1.8
(also works for pOH–>OH, pKa–>Ka, pKb–Kb, @25 degrees C)
A)pH: 13.3
1) 13 --> 10^-14 2) .3 -->10-3 = 7 3) 7 x 10^-14 = [H+]
A)pH: 1.8
1) 1 --> 10^-2 2) .8 -->10-8 = 2 3) 2 x 10^-2 = [H+]
Based on the pKa, which of the following species is the strongest base? pKa of ammonium ion for: *NH3: 9.26 *(CH3CH2)2NH: 10.64 *(Ch3CH2)2NH: 10.98 *(CH3CH2)3N: 10.76
Answer: (CH3CH2)2NH:
pKa –> pKb of ammonium ion: (14-pKa=pKb)
So,
- NH3: 9.26 (pKb: 4.74)
- (CH3CH2)2NH: 10.64 (pKb: 3.36)
- (CH3CH2)2NH: 10.98 (pKb: 3.02)
- (CH3CH2)3N: 10.76 (pKb) 3.24
Smallest Kb: 4.74
*Largest Kb: 3.02
Largest pKa = Smallest pKb = Largest Kb = strongest base
The bigger the pKa, the ____ the base?
Bigger the pKa –>
Smaller the Ka
Smaller the Ka –>
Bigger the Kb
Bigger the Kb –>
Stronger the base
Reacting equivalents
make stoich corrections for how many mol acid/base are produced per mol of reacting source
Ex:
HCl –> H+ + Cl-
(1 mol HCl –> 1 eq acid)
H2SO4 –> 2H+ + SO4^2-
(1 mol HsSO4 –> 2 eq acid)
Ca(OH)2 –> Ca^2+ + 2OH-
(1mol Ca(OH)2 –> 2 eq base)
———————-
(MCAT limits to Acid-Bases… but it applies elsewhere)
Gram-Equivalent Weight
Ex: NaOH and Na2CO3
Stoich correction of molecular weight
g/eq = g/mol X mol/eq
NaOH (1mol:1eq ratio)
40 g/mol x 1 mol/1eq= 40 g/eq
Na2CO3 (1mol:2eq ratio)
106 g/mol x 1 mol/2 eq = 53 g/eq
Normality (N)
Ex: H2SO4
stoich corrections for molality–needed for world of equivalents in world of eq points (as opposed to world of mols)
eq/L = mol/L x eq/mol
1O^-1 mol/L x 2 eq/mol = 2 x 10^-1 eq/L
Titration Curves
Strong Acid-Strong Base: steeper
Weak Acid-Strong Base: less steep
Titrant
solution of known concentration
Equivalence Point Formula
WHEN you are at the equivalence point:
Eq acid = Eq base
- Use this formular to find the M? or vol? or grams?
- Does not care about pH if asking for volume of base added!!!
- If everything is monoprotic/monobasic, you can interchange M and N… but be read JIC
Eq Pt Formula: Both aq
(Normality of Acids)(Volume of Acid) = (Normality of Base)(Volume of Base)
(eq/L)(L) = (eq/L)(L)
Eq Pt Formula: aq/solid
(Normality of Acids)(Volume of Acid) = mass base/gram eq wt = mass/mw x eq/mol
Titration Question Types
pH question families
1) What is the pH at X point in the curve?
2) What is the volume/grams/moles of the solution ?
Eq Pt Formula: if using both solids
mass of acid/gram eq wt = mass of base/ gram eq wt
Set up the formula to find:
How many mg of NaOH (s) is required to neutralize 408 mg of KHP (s), a monoprotic weak acid with a mw of 204 amu?
Monoprotic means Gram Eq Wt (GEW)= mw
Mass of acid/ GEW = mass of base / GEW
408/204 = x/40
2=x/40
x=80
Set up the formula to find:
How many mg of NaOH (s) is required to neutralize 40 mL of a 10^-1 M HCl solution?
(N of acid) * (V of acid) = mass/GEW
(10^-1)(40) = x/40
x=160
In Weak Acid-Strong Base (or SA-WB–same)Titrations, how are the following affected?
- pH initial
- pH Halfway
- pH at eq pt
e.g.
HA + NaOH –> H2O + A-Cl+
Buffer solution: weak acid with its weak CB in solution at the same time (Henderson-Hasselbach time)
Simulation:
- Initial: HCl=30eq, NaCl=0eq
- Halfway: HCl=15eq, NaCl=15eq
- Eq Pt: HCl=0eq, A-Cl+=30eq
- pH initial: must use Ka expression to solve
- pH Halfway: buffer soln (must use H-H eqn to solve)
- pH at eq pt: HA neutralized, A- = wk base, pH > 7
In Weak Acid-Strong Base (or SA-WB–same)Titrations, how are the following affected?
- pH initial
- pH Halfway
- pH at eq pt
e.g.
HA + NaOH –> H2O + A-Cl+
Buffer solution: weak acid with its weak CB in solution at the same time (Henderson-Hasselbach time)
Simulation:
- Initial: HCl=30eq, NaCl=0eq
- Halfway: HCl=15eq, NaCl=15eq
- Eq Pt: HCl=0eq, A-Cl+=30eq
- pH initial: must use Ka expression to solve
- pH Halfway: buffer soln (must use H-H eqn to solve)
- pH at eq pt: HA neutralized, A- = wk base, pH > 7
In Strong Acid-Strong Base Titrations, how are the following affected?
- pH initial
- pH Halfway
- pH at eq pt
e.g.
HCl + NaOH –> H2O + NaCl
*Assume 100% disassociation.
Simulation:
- Initial: HCl=30eq, NaCl=0eq
- Halfway: HCl=15eq, NaCl=15eq
- Eq Pt: HCl=0eq, NaCl=30eq
- pH initial: easy to solve bc [HCl]initial = [H+]
- pH Halfway: NaCl is inert, so [HCl]halfway = [H+]
- pH at eq pt (look at the products): HCl neutralized, NaCl inert, pH = 7
Aside: how do we know NaCl is inert? it is a conjugate of strong species. That is HCl is strong so produces inert Cl. NaOH is strong so produces inert Na.
In Strong Acid-Strong Base Titrations, how are the following affected?
- pH initial
- pH Halfway
- pH at eq pt
e.g.
HCl + NaOH –> H2O + NaCl
*Assume 100% disassociation.
Simulation:
- Initial: HCl=30eq, NaCl=0eq
- Halfway: HCl=15eq, NaCl=15eq
- Eq Pt: HCl=0eq, NaCl=30eq
- pH initial: easy to solve bc [HCl]initial = [H+]
- pH Halfway: NaCl is inert, so [HCl]halfway = [H+]
- pH at eq pt (look at the products): HCl neutralized, NaCl inert, pH = 7
Aside: how do we know NaCl is inert? it is a conjugate of strong species. That is HCl is strong so produces inert Cl. NaOH is strong so produces inert Na.
What is the pH of a 10^-3 HCl soln (pKa = -8)
Strong acid soln: 100% disassociation
[HCl]initial –> [H3O+]
[H+] = 10^-3
pH = 3
What is the pH of a 10^-3 HCl soln (pKa = -8)
Strong acid soln: 100% disassociation
[HCl]initial –> [H3O+]
[H+] = 10^-3
pH = 3
What is the pH of a 10^-3 M CH3COOH soln (pKa = 5)?
-less likely to be on MCAT
Ka=[A-][H3O+]/[HA]
HA + H3O –> A- + H3O+
-x x x
__________________
10^-5 = x^2/10^-3 x^2 = 10^-8 x=10^-4 = H+ pH = 4
What is the pH of a 10^-3 M CH3COOH soln (pKa = 5)?
-less likely to be on MCAT
Ka=[A-][H3O+]/[HA]
HA + H3O –> A- + H3O+
-x x x
__________________
10^-5 = x^2/10^-3 x^2 = 10^-8 x=10^-4 = H+ pH = 4
What is the pH of a 10^-3 M NH3 soln (pKb = 5)
-less likely to be on MCAT bc of math
Kb=[HB+][OH-]/[B] 10^-5 = x^2/10^-3 x=10^-4 = Oh- pOH = 4 pH = 10
What is the pH of a 10^-3 M NH3 soln (pKb = 5)
-less likely to be on MCAT bc of math
Kb=[HB+][OH-]/[B] 10^-5 = x^2/10^-3 x=10^-4 = Oh- pOH = 4 pH = 10
What is the pH of a 10^-3 M HCl at the 1/2 eq pt when titrated with NaOH (s)?
-SA 1/2 eq pt less likely to be asked on MCAT bc pH barely budges from initial
[H+]=(10^-3)/2 = 5 x 10^-4
pH=3.5
What is the pH of a 10^-3 M HCl at the 1/2 eq pt when titrated with NaOH (s)?
-SA 1/2 eq pt less likely to be asked on MCAT bc pH barely budges from initial
[H+]=(10^-3)/2 = 5 x 10^-4
pH=3.5
pH at eq pt
MCAT will ask in terms of acid and basic.
pH at eq pt
MCAT will ask in terms of acid and basic.
Ammonia Buffer Solution
Finish! slide 18
HH in head
Ammonia Buffer Solution
Finish! slide 18
What is the pH of a 10^-3 M CH3COOH (pKa=5) at the 1/2 eq pt when titrated with NaOH (s)?
pH=5
- Weak Acid 1/2 eq pt much more likely to be asked on MCAT bc in a buffer soln
- Buffer soln –> use HH eqn!
pH = pKa + log (A-/HA) –> at 1/2 eq pt, pH = pKa
pH=5
What is the pH of a 10^-3 M CH3COOH (pKa=5) at the 1/2 eq pt when titrated with NaOH (s)?
pH=5
- Weak Acid 1/2 eq pt much more likely to be asked on MCAT bc in a buffer soln
- Buffer soln –> use HH eqn!
pH = pKa + log (A-/HA) –> at 1/2 eq pt, pH = pKa
pH=5
What is the pH of a 10^-3 M NH3 (pKb=5) at the 1/2 eq pt when titrated with NaOH (s)?
pH=9
-Weak Acid 1/2 eq pt much more likely to be asked on MCAT bc in a buffer soln
pOH = pKb + log (HB+/B) –> at 1/2 eq pt, pOH = pKb
pOH=5,
pH = 14-5 = 9
What is the pH of a 10^-3 M NH3 (pKb=5) at the 1/2 eq pt when titrated with NaOH (s)?
pH=9
-Weak Acid 1/2 eq pt much more likely to be asked on MCAT bc in a buffer soln
pOH = pKb + log (HB+/B) –> at 1/2 eq pt, pOH = pKb
pOH=5,
pH = 14-5 = 9
In Weak Acid-Strong Base (or SA-WB–same)Titrations, how are the following affected?
- pH initial
- pH Halfway
- pH at eq pt
e.g.
HA + NaOH –> H2O + A-Cl+
Buffer solution: weak acid with its weak CB in solution at the same time (Henderson-Hasselbach time)
Simulation:
- Initial: HCl=30eq, NaCl=0eq
- Halfway: HCl=15eq, NaCl=15eq
- Eq Pt: HCl=0eq, A-Cl+=30eq
- pH initial: must use Ka expression to solve
- pH Halfway: buffer soln (must use H-H eqn to solve)
- pH at eq pt: HA neutralized, A- = wk base, pH > 7
In Strong Acid-Strong Base Titrations, how are the following affected?
- pH initial
- pH Halfway
- pH at eq pt
e.g.
HCl + NaOH –> H2O + NaCl
*Assume 100% disassociation.
Simulation:
- Initial: HCl=30eq, NaCl=0eq
- Halfway: HCl=15eq, NaCl=15eq
- Eq Pt: HCl=0eq, NaCl=30eq
- pH initial: easy to solve bc [HCl]initial = [H+]
- pH Halfway: NaCl is inert, so [HCl]halfway = [H+]
- pH at eq pt (look at the products): HCl neutralized, NaCl inert, pH = 7
Aside: how do we know NaCl is inert? it is a conjugate of strong species. That is HCl is strong so produces inert Cl. NaOH is strong so produces inert Na.
What is the pH of a 10^-3 HCl soln (pKa = -8)?
Strong acid soln: 100% disassociation
[HCl]initial –> [H3O+]
[H+] = 10^-3
pH = 3
What is the pH of a 10^-3 M CH3COOH soln (pKa = 5)?
-less likely to be on MCAT
Ka=[A-][H3O+]/[HA]
HA + H3O –> A- + H3O+
-x x x
__________________
10^-5 = x^2/10^-3 x^2 = 10^-8 x=10^-4 = H+ pH = 4
What is the pH of a 10^-3 M NH3 soln (pKb = 5)
-less likely to be on MCAT bc of math
Kb=[HB+][OH-]/[B] 10^-5 = x^2/10^-3 x=10^-4 = Oh- pOH = 4 pH = 10
What is the pH of a 10^-3 M HCl at the 1/2 eq pt when titrated with NaOH (s)?
-SA 1/2 eq pt less likely to be asked on MCAT bc pH barely budges from initial
[H+]=(10^-3)/2 = 5 x 10^-4
pH=3.5
What is the pH of a 10^-3 M CH3COOH (pKa=5) at the 1/2 eq pt when titrated with NaOH (s)?
pH=5
- Weak Acid 1/2 eq pt much more likely to be asked on MCAT bc in a buffer soln
- Buffer soln –> use HH eqn!
pH = pKa + log (A-/HA) –> at 1/2 eq pt, pH = pKa
pH=5
What is the pH of a 10^-3 M NH3 (pKb=5) at the 1/2 eq pt when titrated with NaOH (s)?
pH=9
-Weak Acid 1/2 eq pt much more likely to be asked on MCAT bc in a buffer soln
pOH = pKb + log (HB+/B) –> at 1/2 eq pt, pOH = pKb
pOH=5,
pH = 14-5 = 9
pH at eq pt
MCAT will ask in terms of acid and basic.
Ammonia Buffer Solution
Finish! slide 18
HH in head
pH exercises
slide 20
1) Ask: is it a buffer or not?
Yes? HH time
2) No? Is acid or base winning?
And by how much?
If acid is winning by a factor of 10, for example, pH lowers by 1 pH pointBuffer Capacity
1) look at eqs of Acid and CB
* Ability to resist change has everything to do of absolute amt equivalents of acid and CB (not identity of buffer acids/bases or pH value) –> more you have of both–> more capable you are of resisting pH changes due to outside influx of acid/base
if 1:1 ratio, pH = pKa
Buffer Capacity:
How will the following solutions change pH after adding 10 eq H+?
CH3COO- + H3O+ –> CH3COOH + H2O
insert jpeg
Indicators:
pKa of methyl red is 5.2
Describe the relative concentration of the colored form of a 1M solution of methyl red at pH 4.2, 5.2, 6.2
(insert jpeg if desire)
pH = pKa + log(yellow/red)
- 2 = 5.2 + log (1/10) RED
- 2 = 5.2 + log (1/1) ORANGE
- 2 = 5.2 + log (10/1) YELLOW
[Takeaway: color change occurs when pH solution is around pKa of an indicator, thus, choose an indicator whose pKa is NEAR ESTIMATED eq pt pH]
A student was given a flask containing 25.0 mL of an aq soln of ammonia (pKb at 25 deg 4.74) of unknown concentration. The student was asked to determine the concentration of this solution by titrating it against a standard 0.10 M aq HCl soln that was placed in burette. The student added 2 drops of indicator to the soln and titrated it against the HCl solution. The experimental temp was 25 deg C. The student noted that 40.0 mL of the HCl solution was needed to reach the eq pt. Table 1 lists that pH range of commonly used acid-base indicators.
insert jpeg
pOH = pKb,
pH=pKa at top of curve 1/2 eq pt
