Classes 10 and 11: Salt Hydrolysis and Buffer Systems

Question 1

Predict whether a salt dissolved in water will be acidic, basic, or neutral. Be able to recognize spectator ions that have no effect on pH.

Answer 1

• Determine if the salt is formed from a strong acid and strong base (neutralization reaction)
  
  • If so, it will be neutral and contain spectator ions
  • Spectator ions have no effect on pH
• If formed from a strong acid and weak base
  
  • It will be acidic due to leftover H+ ions from the strong acid
• If formed from a weak acid and strong base
  
  • It will be basic due to leftover OH- ions from the strong base
• Memorize strong acids (HCl, HNO3, H2SO4, HClO4, etc.)
• Memorize strong bases (NaOH, KOH, Ca(OH)2, etc.)
• Anything else is a weak acid/base
• Spectator ions are the leftovers that do not participate in acid-base reactions, e.g. Na+, Cl-, NO3-

Question 2

Write out any appropriate reactions with curved arrow notation to support your prediction of the acidity or basicity of various salts.

Answer 2

1. Salt formed from strong acid and strong base (neutral):
   HCl + NaOH → NaCl + H2O
   No curved arrows needed as it’s a simple neutralization
2. Salt formed from strong acid and weak base (acidic):
   HCl + NH3 → NH4Cl
   Base accepts proton with curved arrow:
   H-Cl + :NH3 → NH4+ Cl-
3. Salt formed from weak acid and strong base (basic):
   CH3COOH + NaOH → CH3COONa + H2O
   Acid donates proton with curved arrow:
   CH3COOH + :OH- → CH3COO- + H2O
4. Identifying spectator ions:
   In NaNO3 formed from HNO3 + NaOH
   Na+ and NO3- are spectator ions
   No curved arrows as they don’t participate

The curved arrows show the movement of protons and electron pairs, allowing you to predict if excess H+ (acidic) or OH- (basic) remains after the acid-base reaction forming the salt.

Question 3

Calculate the pH of a solution containing a non-neutral salt.

Answer 3

For an Acidic Salt:
1. Identify the salt as being formed from a strong acid and weak base
2. Write out the dissociation reaction of the salt in water
3. Calculate the [H+] from the Ka of the conjugate acid of the weak base
4. Use the equation: pH = -log[H+]

For a Basic Salt:
1. Identify the salt as being formed from a weak acid and strong base
2. Write out the dissociation reaction of the salt in water
3. Calculate the [OH-] from the Kb of the conjugate base of the weak acid
4. Calculate [H+] = Kw/[OH-]
5. Use the equation: pH = -log[H+]

Key Points:
- Identify if it’s an acidic or basic salt
- Write out the full dissociation reaction
- Use Ka, Kb, or Kw to calculate [H+] or [OH-]
- Substitute into pH = -log[H+] equation

Include any relevant equilibrium constant values (Ka, Kb, Kw) and show the calculated pH value from start to finish.

Question 4

Predict shifts in equilibrium due to the “common ion effect” and calculate the resulting changes in pH.

Answer 4

Common Ion Effect:
- Adding a compound that contains an ion already present in the equilibrium shifts the equilibrium in the direction that reduces the concentration of that ion

To Predict the Shift:
1. Identify the ions present in the added compound
2. Determine which of those ions are common to the equilibrium reaction
3. Equilibrium will shift away from the side that produces that common ion

To Calculate pH Change:
1. Write out the full equilibrium reaction
2. Identify the equilibrium constant (Ka, Kb, or Kw) involved
3. Calculate initial concentrations after adding the common ion
4. Use the IC*Q equation to calculate the new equilibrium concentrations
5. Substitute [H+] or [OH-] into pH = -log[H+] to find new pH

Key Points:
- Common ions cause equilibrium to shift away from producing more of that ion
- Use IC*Q equation with new initial concentrations accounting for common ion
- Calculate new [H+] or [OH-] and determine new pH

Include a full example calculation showing the equilibrium reaction, added common ion, IC*Q setup, and final pH.

Question 5

Review the identification of weak conjugate acid/base pairs in order to identify a buffer solution.

Answer 5

• A buffer solution contains a weak acid and its conjugate base, or a weak base and its conjugate acid
• It resists large changes in pH when small amounts of an acid or base are added
• To identify a buffer pair:
  
  • Weak acid (HA) and its conjugate base (A-)
  • Weak base (B) and its conjugate acid (BH+)
• Common weak acid examples: acetic acid, carbonic acid
• Common weak base examples: ammonia, amine
• The conjugate pairs must come from the same reaction (acid dissociation or base protonation)
• The relative concentrations of the conjugate pairs determine the buffer pH

Question 6

Identify the “major species” in solution – the ones in the highest concentration that will affect the pH.

Answer 6

• Major species are the ions/molecules present in the highest concentrations
• They determine the overall properties and behavior of the solution
• For strong acids/bases:
  
  • Major species are the ions from complete dissociation
• For weak acids/bases:
  
  • Major species are the undissociated acid/base and its conjugate
• To identify major species:
  
  • Consider initial concentrations
  • Consider Ka/Kb values to determine extent of dissociation
  • Higher Ka/Kb means more dissociation
• The major species control the pH through equilibrium concentrations

Question 7

Write the uni-directional reaction or equilibrium reactions needed to calculate the final pH of a buffer or the pH after acid or base has been added and perform these calculations.

Answer 7

• For strong acid/base additions:
  
  • Use strong ionization reactions (complete dissociation)
  • Calculate new concentrations of H3O+/OH- from added amounts
• For weak acid/base buffers:
  
  • Write dissociation equilibrium reaction
  • Use Ka or Kb expression
  • Set up ICE table for equilibrium concentrations
  • Substitute concentrations into Ka/Kb equation and solve for [H3O+] or [OH-]
• For weak acid/base + strong acid/base additions:
  
  • First calculate pH from initial buffer
  • Then use strong ionization to calculate pH change from acid/base addition
  • Final pH = initial buffer pH +/- pH change
• Use pH = -log[H3O+] or pOH = -log[OH-] as needed

Question 8

Predict acid-base reactions and determine whether or not a buffer has been formed.

Answer 8

• Identify acid-base conjugate pairs
  
  • Weak acid + conjugate base
  • Weak base + conjugate acid
• If both members of a conjugate pair are present, a buffer is formed
• No buffer if:
  
  • Only strong acid/base present
  • Only one member of conjugate pair present
• To predict reactions:
  
  • Strong acid reacts with weak base to form conjugate acid
  • Strong base reacts with weak acid to form conjugate base
  • Weak acid/base essentially does not react with itself
• Buffer capacity depends on:
  
  • Concentrations of acid and base
  • Acid dissociation constant (Ka)
  • Ratio of acid to base concentrations

Question 9

Predict whether a salt dissolved in water will be acidic, basic, or neutral. Be able to recognize spectator ions that have no effect on pH.

Answer 9

If a weak acid is neutral ⇒ charge on conjugate base is negative
If a weak base is neutral ⇒ charge on conjugate acid is positive
An ion cannot exist without a counterion in water
Salt = charged conjugate base/acid with a spectator ion
Common spectator ions:
Li+, Na+, K+
Cl-, Br-, NO3-, ClO4-
STRATEGY FOR SALTS:
Identify spectator ions to ignore
Identify weak acids and bases
Check to see if you have Ka or Kb
Set up an ICE table calculation for any weak acid/base problem

Question 10

Predict shifts in equilibrium due to the “common ion effect” and calculate the resulting changes in pH.

Answer 10

The common-ion effect is used to describe the effect on an equilibrium when one or more species in the reaction is shared with another reaction. This results in a shifting of the equilibrium properties.
Le Chatelier’s Principle states that if an equilibrium becomes unbalanced, the reaction will shift to restore the balance. If a common ion is added to a weak acid or weak base equilibrium, then the equilibrium will shift towards the reactants, in this case the weak acid or base.

Question 11

Buffers—general info

Answer 11

A buffer: a solution containing significant, comparable concentrations of both a weak acid and its conjugate base
Comparable: meaning no more than a factor f 10 difference between [HA] and [A-]
Significant: meaning both weak acid and base must be significantly greater than [H3O+] and [OH-]

Question 12

Identify the “major species” in solution – the ones in the highest concentration that will affect the pH.

Answer 12

Major species are the ones that will impact the reaction—e.g acids and bases
A weak base is GREAT at neutralizing a strong acid
A weak acid is GREAT at neutralizing a strong base
Reaction of a weak acid with a strong base favors the products strongly
A buffer has both weak acid and weak base and so is good at neutralizing a strong acid or a strong base.
Matters: proteins have links through charged molecules—they could be protonated.

Question 13

Write the unidirectional reaction or equilibrium reactions needed to calculate the final pH of a buffer or the pH after acid or base has been added and perform these calculations.

Answer 13

Use an ICE table
Determine the limiting reactant to determine how much is actually neutralized
A weak base + a strong acid ⇒ goes to completion
A weak acid + a strong base ⇒ goes to completion
Limiting reactant is consumed completely
If the result is a buffer, then limiting reactant
If not, then do normal ICE table

Question 14

Predict acid-base reactions and determine whether or not a buffer has been formed.

Answer 14

Henderson-Hasselbalch equation: pH=pKa + log([A-]/[HA])

Weak acid + OH- ⇒ Weak base

Weak base + H3O+ ⇒ Weak acid

How to make a buffer:
Weak acid + conjugate weak base
Weak acid + strong base
Weak base + strong acid

Question 15

Extra

Answer 15

IF BUFFER:
WEAK ACID+WEAK BASE → ICE table
WEAK ACID/BASE + STRONG ACID/BASE → mole table then ICE table or use HH
Use original concentrations, not in equilibrium
NOTE: PAY ATTENTION TO YOUR HYBRID ORBITALS!!!! Are the electrons fixed? What orbital are they in? Are other electrons already in the P orbital?
Charged Acids and Bases cannot exist in solution without counter-ions (e.g.
Na+, K+, Mg2+ and Cl-, Br-, I-, NO3-)
RULES OF RESONANCE
Atomic connectivity is conserved—only electron positions vary
Each resonance structure must be a valid lewis structure
Contributor with less charge separation is the greatest contributor
The resonance structure with a formal charge on the more electronegative atom is the greatest contributor
No loss or gain of NET formal charge
All resonance structures have the same number of unpaired electrons
RESONANCE DELOCALIZATION STABILIZES A MOLECULE
NOTE: the lone pair must be a part of resonance/impact resonance
NOTE: look at the impact of those resonance forms when both atoms have resonance structures
TIP: when it comes to base strength, whichever base is the least stable will be the most likely to attract an electron

Question 16

How to figure out whether a salt when dissolved in water will affect the pH of the solution

Answer 16

Determining if a Salt Affects pH:
- Consider if the salt is formed from a strong acid and strong base (neutral solution), weak acid and strong base (basic solution), or strong acid and weak base (acidic solution).

Strong Acid + Strong Base Salts:
- Salts like NaCl, KNO3, etc. are neutral and do not affect pH significantly
- The ions present (Na+, Cl-, K+, NO3-) do not hydrolyze in water
- The resulting solution has a pH ≈ 7

Weak Acid + Strong Base Salts:
- The conjugate base of the weak acid can undergo hydrolysis, producing OH- and causing a basic solution
- Example: Na2CO3, NaC2H3O2 - the CO3(2-) and C2H3O2(-) ions hydrolyze, forming OH-
- Calculate pH using Kb of conjugate base

Strong Acid + Weak Base Salts:
- The conjugate acid of the weak base can undergo hydrolysis, producing H3O+ and causing an acidic solution
- Example: NH4Cl, CH3NH3Cl - the NH4+ and CH3NH3+ ions hydrolyze, forming H3O+
- Calculate pH using Ka of conjugate acid

Key Points:
- Identify if acid/base components are strong or weak
- Conjugate of weak acid makes basic solution; conjugate of weak base makes acidic
- For basic solution, calculate pH from Kb of conjugate base
- For acidic solution, calculate pH from Ka of conjugate acid

So in summary, determine the acid/base strengths, identify the ion that can hydrolyze, and calculate the resulting pH based on the equilibrium involving that ion.

Question 17

Calculate the pH of a 0.10 M solution of ammonium chloride.

Answer 17

NH4+ + H2O <-> NH3. + H3O+
Ka=5.6x10^-10

I. 0.1. 0. 0
C. -x. +x. +x
E. 0.1-x. x. x

Ka= [NH3][H3O+]/[NH4+] = x^2/0.1-x. = 5.6x10^-10

x= 7.5 x 10^-6

pH=-log(7.5x10^-6) =5.12

Question 18

To identify which species are acids, bases, and spectator ions in a chemical solution

Answer 18

1. Identify the solute and solvent:
   
   • In aqueous solutions (solutions involving water), water is the solvent.
   • The solute is the substance dissolved in water.
2. Identify potential acids and bases:
   
   • Acids are substances that release hydrogen ions (H+) when dissolved in water.
   • Bases are substances that release hydroxide ions (OH-) when dissolved in water.
3. Identify the dissociation of the solute:
   
   • Strong acids and bases dissociate completely in water, producing their respective ions.
   • Weak acids and bases dissociate partially, with an equilibrium between the undissociated and dissociated forms.
4. Classify the ions:
   a. Acids:
   
   • The hydrogen ions (H+) released by acids are considered acid ions.
   • The remaining portion of the acid is typically a negatively charged ion (anion).
   b. Bases:
   - The hydroxide ions (OH-) released by bases are considered base ions.
   - The remaining portion of the base is typically a positively charged ion (cation).c. Spectator ions:
   - Ions present in the solution that do not participate in the acid-base reaction are called spectator ions.
   - These ions do not change the overall pH of the solution.
5. Identify the conjugate acid-base pairs:
   
   • In an acid-base reaction, the acid donates a hydrogen ion (H+) to the base.
   • The remaining portion of the acid after losing H+ is called the conjugate base.
   • The remaining portion of the base after accepting H+ is called the conjugate acid.

By following these steps, you can identify the acids, bases, and spectator ions present in a chemical solution, as well as their conjugate acid-base pairs. This understanding is crucial for analyzing acid-base reactions, calculating pH, and predicting the behavior of chemical systems involving acids and bases.

Question 19

The Henderson-Hasselbalch equation and the relationship between pKa and Ka

Answer 19

1. Henderson-Hasselbalch Equation:
   The Henderson-Hasselbalch equation relates the pH of a solution to the ratio of the concentrations of the conjugate acid-base pair. It is given by the following equation:

pH = pKa + log([base] / [acid])

Where:
- pH is the negative logarithm of the hydrogen ion concentration
- pKa is the negative logarithm of the acid dissociation constant (Ka)
- [base] is the molar concentration of the conjugate base
- [acid] is the molar concentration of the conjugate acid

This equation is particularly useful for calculating the pH of buffer solutions, which are solutions composed of a weak acid and its conjugate base (or a weak base and its conjugate acid).

2. pKa and Ka Relationship:
   The pKa (negative logarithm of the acid dissociation constant) is related to the acid dissociation constant (Ka) by the following equation:

pKa = -log(Ka)

Where:
- pKa is the negative logarithm of the acid dissociation constant
- Ka is the acid dissociation constant

The acid dissociation constant (Ka) is a measure of the strength of an acid in aqueous solution. It is the equilibrium constant for the dissociation reaction of the acid in water.

The pKa value is widely used in acid-base chemistry because it provides a convenient way to compare the strengths of different acids. A smaller pKa value indicates a stronger acid, while a larger pKa value indicates a weaker acid.

Buffers and pKa:
The pKa value is crucial for understanding and designing buffer solutions. A buffer solution is most effective when the pH is close to the pKa value of the acid or base used in the buffer. The Henderson-Hasselbalch equation can be rearranged to calculate the pH of a buffer solution:

pH = pKa + log([salt] / [acid])

Where:
- [salt] is the molar concentration of the conjugate base (or the salt formed by the acid and base)
- [acid] is the molar concentration of the acid

By adjusting the ratio of the concentrations of the acid and its conjugate base, the pH of the buffer solution can be controlled and maintained within a desired range, even upon the addition of small amounts of strong acids or bases.

In summary, the Henderson-Hasselbalch equation and the relationship between pKa and Ka are fundamental concepts in acid-base chemistry, enabling the calculation of pH, understanding of acid-base equilibria, and the design and analysis of buffer solutions.

Question 20

Making a buffer #1 - 100.0 mL of 0.350 M HNO, is combined with 250.0 mL of 0.280 M NaNO and diluted to 500.0 mL. What is the pH of the buffer solution? (Hint: You must always write out the relevant chemistry first before beginning your calculations)

Answer 20

[HNO2] = (0.100L)(0.350M)/(0.500L) = 0.0700 M

[NO2-] = (0.250L)(0.280M)/(0.500L) = 0.140 M

HNO2 +H2O <-> NO2- + H3O+

I 0.0700 0.140 0
C -x +x +x
E 0.0700-x 0.140+x x

Ka= 4.0x10^-4 = (x)(0.140)/(0.0700)

x=2.0x10^-4 M
pH=3.70

pH=pKa+log([NO2-]/[HNO2])

=-log(4.0x10^-4) + log(0.140/0.0700)
=3.4-+0.301. = 3.70 =pH

Question 21

Making a buffer #2 - 100.0 g NH.CI (MW 53.49 g/mol) and 65.0 g NaOH (40.00 g/mol) are diluted to 1.00 L with water. What is the pH of the solution? (Hint: You must always write out the relevant chemistry first before beginning your calculations)

Answer 21

NH4+ + OH- -> NH3 + H2O

pH = pKa NH4+. + log([NH3]/[NH4+]
= -log(5.6x10^-10) +log ((1.625/1.00L)/(0.245/1.00 L))
=9.25 + 0.821 = 10.07

To find the pH of the solution, we need to follow these steps:

Step 1: Write the balanced chemical equation for the reaction between ammonium chloride (NH4Cl) and sodium hydroxide (NaOH).

NH4Cl + NaOH ⇌ NH3 + NaCl + H2O

Step 2: Calculate the moles of NH4Cl and NaOH present in the solution.

Moles of NH4Cl = 100.0 g / 53.49 g/mol = 1.87 mol
Moles of NaOH = 65.0 g / 40.00 g/mol = 1.625 mol

Step 3: Identify the limiting reactant and determine the moles of NH3 and NH4+ in the solution at equilibrium.

Since NaOH has fewer moles (1.625 mol) than NH4Cl (1.87 mol), NaOH is the limiting reactant.
Moles of NH3 formed = 1.625 mol
Moles of NH4+ remaining = 1.87 mol - 1.625 mol = 0.245 mol

Step 4: Calculate the concentrations of NH3 and NH4+ in the 1.00 L solution.

[NH3] = 1.625 mol / 1.00 L = 1.625 mol/L
[NH4+] = 0.245 mol / 1.00 L = 0.245 mol/L

Step 5: Use the Henderson-Hasselbalch equation to calculate the pH of the solution.

pH = pKa(NH4+) + log([NH3] / [NH4+])
pKa(NH4+) = -log(5.6 × 10^-10) = 9.25 (at 25°C)
pH = 9.25 + log(1.625 / 0.245)
pH = 9.25 + 0.821
pH = 10.07

Therefore, the pH of the solution obtained by diluting 100.0 g of NH4Cl and 65.0 g of NaOH to 1.00 L with water is approximately 10.07.

Question 22

How it is possible that the calculation from “Making a buffer #2” led to a buffer solution when a weak acid was not mixed with its weak conjugate acid-base salt?

Answer 22

After the strong base completely reacts with the weak acid NH4+, the OH- is completely used up because it’s the limiting reagent. At the end of the reaction, the major species left are a weak conjugate acid-base pair.

Question 23

In your biology class, you are required to monitor the reproductive processes of a rare colony of bacteria. In their metabolic pathways, the bacteria produce OH at the rate of 2.00x10^-4 mol/hour. You plan to monitor the bacteria over a period of 12 hours in a small 100.0 mL culture dish that also contains a buffer that is 0.500 M in CHNH;CI and 0.500 M in CH;NH. (The Kb of CH;NH is 4.3x10^-4)

14. What is the initial pH of the buffered solution in the dish before any bacteria metabolism has started?

Ka CH3NH3+. = (1.0x10^-14/4.3x10-4) = 2.33 x 10^-11

pKa = -log Ka
pH = pKa + log [CH3NH2]/[CH3NH3+]
= -log(2.33x10^-11)+log(0.500/0.500) =10.63

Answer 23

15. What is the pH of the dish at the end of the 12-hour experiment?

mol OH- =12 hrs [2.00x10^-4 mol /hr]= 2.40 x 10 ^-3 mol

Question 24

Consider 2 unknown salts, NaX and NaY in equal concentrations. What would you need
to know to determine which solution has the higher pH? Explain how you would decide
using some chemical equations and a few brief words. Then do sample calculations to
back up your reasoning using 0.10 M aqueous solutions of NaNO2 and NaClO2.

Answer 24

We know that Na+ is the corresponding cation to a strong base, NaOH. This means that Na+ is
unlikely to react with water, which makes it a spectator ion. What we’re looking at here is the
anions, NO2
– vs. ClO2
- . When we react these anions with water, OH- will be produced. To
calculate the pH of each solution, you would need the Ka value in order to solve for the Kb and
then solve for x.
Weaker acids have stronger conjugate bases. Qualitatively, we can infer that the anion with the
weaker acid conjugate (smaller Ka) will be more basic, or have a higher pH:
Ka of HNO2 = 4.0 x 10^-4
Ka of HClO2 = 1.1 x 10^-2
Let’s test our idea out numerically using some ICE tables:
1. NO2-
NO2- + H2O ↔ HNO2 + OH-

NO2- - H2O HNO2 OH-
Initial 0.10 0 0
Change -x + x + x
E. 0.10 - x x x
Kb = Kw/Ka -> Kb = (1 x 10^-14)/(4.0 x 10^-4) -> Kb = 2.5 x 10^-11
Kb = 2.5 x 10-11 = (x^2)/(0.10 – x) -> x = [OH-] = 1.58 x 10^-6
pOH = -log(1.58 x 10^-6) = 5.80
pH = 14 – pOH = 8.2
pH = 8.2

2. ClO2-
   ClO2- + H2O ↔ HClO2 + OH-

       ClO2- H2O HClO2 OH Initial 0.10                0        0 Change -x                 + x + x Equil 0.10 - x             x       x

Kb = Kw/Ka -> Kb = (1 x 10^-14)/(1.1 x10^-2) -> Kb = 9.09 x 10-13

Kb = 9.09 x 10^-13 = (x^2)/(0.10 – x) -> x = [OH-] = 3.01 x 10^-7

pOH = -log(3.01 x 10^-7) = 6.52
pH = 14 – pOH = 7.5
pH = 7.5

Question 25

How do we use this information to rank the pH values?

Answer 25

Salts will dissociate completely, releasing more ions into solution that a weak acid or
base (ex. NaF dissociates completely into F- ions, HF only dissociates partly into Fions)
- For comparing salts, look at the Ka or Kb values of their constituent ions.
o Even for two neutral salts, one cation may for example have a higher Ka than
the other, which will affect the acidity of the salt solution overall. A solution
made from a salt composed of a weak acid AND a weak base will have a pH
that is approximately the average of the pKa’s of the acid and the conjugate
acid of the base.

Question 26

What is the pH of a 0.567 M solution of CH3NH3Cl? Show the Lewis structure of the relevant species affecting the pH with water and used the curved arrow formalism to
show the transfer of the proton, as well as the mathematical reasoning

Answer 26

CH3NH3 + H2O <-> CH2NH2 + H30+

I 0.567 M. 0. 0
C. -x. +x. +x
E. 0.567-x. x. x

2.51 x 10^-11 8 (x^2/(0.567-x))
x=3.77*10^-6

pH * -log(x) =5.42

Question 27

Several years ago on an exam, a student argued (incorrectly) that if you dissolved any
weak acid in water, you would obtain a buffer solution. The reasoning was that anytime
you dissolve a weak acid in water, some of the conjugate base will also form when the
weak acid dissociates. Explain why this thinking is incorrect

Answer 27

Many weak acids have a Ka value in the order of magnitude of 10-5 or 10-6
. This means
that only a very small amount of acid will dissociate into its conjugate base. The most
effective buffer solutions have an equal concentration of acid and conjugate base. The
acceptable range of concentration ratios of weak acid to conjugate base is between 1:10
and 10:1. In other words, their concentrations need to be within one order of
magnitude of each other. Many weak acids have an acid:conjugate base ratio on the
order of 10,000:1, which is pretty far off.

Question 28

Could a buffered solution by made by mixing aqueous solutions of HCl and NaOH?
Explain. Why isn’t a mixture of a strong acid and its conjugate base considered a buffer
solution?

Answer 28

This solution is NOT a buffer system. When you mix strong acids and bases, they will
fully react – leaving nothing to buffer the system. You will end up with water and NaCl
(salt). For a buffer you want a weak acid and its conjugate base or a weak base and its
conjugate acid so they can react weakly with other species and establish an equilibrium.

Question 29

Phenolphthalein often used as an indicator in acid-base reactions – you will use it in a
lab experiment after spring break. It’s pKa is 9.24. An aqueous solution containing
phenolphthalein may change color depending on the pH of the solution.
a) A reaction that occurs with this molecule is shown below. Draw the curved arrows in
Step 1 to show the proton transfer. Then draw the curved arrows on HIn- (A) to show
how the ring structure opens to form HIn- (B). All arrows should be drawn on the
structures provided (ie. do NOT draw additional structures or change formal charges!).
Based on the chemistry above, when will a solution of phenolphthalein be pink? When
the solution is acidic or basic? Explain

Answer 29

The solution will be pink when the solution is basic. LeChatelier states that increasing the
amount of base will shift the equilibrium to the right. You can also consider the pKa (9.24)
which is higher than 7 meaning that the color will change when the solution is basic (above
pH of 7). Think of the Henderson Hasselbach equation – as you add more A- (base) the log
value will increase. This added to the pKa yields the pH which will rise as the amount of base
in solution rises. See class 12 for more about the ratio of base/acid as it relates to pH.

Question 30

A buffer was prepared by adding 4.95 g of sodium acetate to 250.0 mL of 0.150 M
CH3COOH. Assume no volume change when the sodium acetate is added.
a) What is the pH of the buffer?

Answer 30

4.95g CH3COONa x (1 mol/82.0343h) = (0.060341 mol CH3COONa)/(0.250 L) = 0.24136

acetic acid = CH3COOH

((0.150 mol CH3COOH)/ 1L) * 0.250 L = 0.0375 mol CH3COOH

pH = pKa + log ([CH3COO-]/[CH3COOH])
= 4.74 + log (0.241362/0.150 M) = 4.95

Question 31

Take the 250.0 mL buffer solution that described above and divide it into 2 beakers:
Beaker A contains 100.0 mL and Beaker B contains 150.0 mL.
(i) To beaker A, you add 1.00 mL of 12.0 M HCl. What is the new pH? Does a
buffered solution remain after the addition? How do you know?

Answer 31

[CH3C00-] =0.241362 M
[CH3COOH] = 0.150 M
V =100.0 ml
Add 100 ml of 12.0 M HO

CH3COO- (aq) + H3O+(aq) -> CH3COOH (ag) + H2O(l)

I 0.24136. 0.012. 0.0150
C -0.012. -0.012. +0.012
E 0.012136 0. 0.027

pH = 4.74 + log (0.012136/0.027) =4.39

Question 32

Phosphoric acid is one of the acids in Coca-Cola. If you have 335 mL of 0.0800 M H3PO4,
what is the final pH if you add 0.035 moles of NaOH to it?

Answer 32

Phosphoric acid = H3PO4
M=0.0800 M
V=335 mL

(0.0800mol/1L)* (0.335L) = 0.0268 mol

pH = 6.84