# Solution 9: Acids/Bases, Common Ion Effect, and Buffers


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Work in groups on these problems. You should try to answer the questions without referring to your textbook. If you get stuck, try asking another group for help.

## Unit I: Solubility (A Heterogeneous Equilbrium)

Both ionic salts and covalent molecules dissolve in solutions to varying degrees. The limit to how much salt can be dissolved in a given volume of water is given by the solubility product, $$K_{sp}$$, which is the relevant equilibrium constant for the process (just with a special name) and naturally depends on the type of salt and temperature or other species in solution (the common ion effect). Before beginning with calculations, let's calibrate things a bit:

### Q9.1: Definition of $$K_{sp}$$

Write the relevant solubility product for dissolving $$AgCl$$ (with a molar mass of 143.3 g/mol) in water:

$AgCl (s) \rightarrow Ag^+(aq) + Cl^−(aq) \label{W1}$

And equilibrium constant is an equilibrium constant is an equilibrium constant....

$K = \dfrac{[\ce{Ag^{+} (aq)}][\ce{Cl^{-} (aq) }]}{[AgCl (s)]} \nonumber$

but should be written in activities. We call this equilibrium constant the solubility product $$K_{sp}$$ (just to be annoying) and we can rewrite it to look like this

$K_{sp} = [\ce{Ag^{+} (aq)}][\ce{Cl^{-} (aq) }] \nonumber$

This specific equilibrium constant is $$1.8 \times 10^{−10}$$ 25 °C (Reference Table E3 on the Libretexts).

### Q9.2: Thermodynamic and Intermolecular Forces

How to you expect $$K_{sp}$$ to change under the following conditions:

1. If $$\Delta G_{solution}$$ for the dissolving process were made more negative?
2. If $$\Delta H_{solution}$$ for the dissolving process were made more negative?
3. If $$\Delta S_{solution}$$ for the dissolving process were made more negative?
4. If the lattice energy for $$AgCl$$ were increased and all other factors maintained the same?
5. If the solvation energies for $$Ag^+$$ and $$Cl^-$$ were increased and all other factors maintained the same?
6. If the intermolecular forces between the solute and solvent molecules were increased and all other factors maintained the same?
7. If the entropy of dissolving were increased and and all other factors maintained the same?

Intuitively, decreasing the $$\Delta G^o_{sol}$$ will increased the "driving force" for the reaction and make it more spontaneous. We can do it numerically also,

$K_{sp} = -exp (\Delta G^o_{sol}/RT) \nonumber$

So as $$\Delta G^o_{sol}$$ decreases, $$K_{sp}$$ increases. That then favors the products of the Equation \ref{W1} and makes the salt more soluble in agreement with our expectations.

We know that $\Delta G^o_{sol} = \Delta H^o_{sol} - T \Delta S^o_{sol}\nonumber$ Presuming everything is the same except for more negative $$\Delta H^o$$, then that makes a more negative $$\Delta G^o$$. From the answer to part a, this makes a greater $$K_{sp}$$ and favors the products (i.e., makes the salt more soluble). This also make intuitive sense, since this means the interactions (i.e., hydration) between the water and the salt ions are stronger.

We know that $\Delta G^o_{sol} = \Delta H^o_{sol} - T \Delta S^o_{sol}\nonumber$ Presuming everything is the same except for more negative $$\Delta S^o_{sol}$$, then that makes a more positive $$\Delta G^o_{sol}$$. From the answer to part a, this makes a smaller $$K_{sp}$$ and favors the reactants (i.e., makes the salt less soluble). This also make intuitive sense, since this means that entropy does not favor the dissolving process and the salt would be less soluble.

The lattice energy is directly related to the $$\Delta H_{sol}^o$$ component of the reaction, which can be writen as $\Delta H_{sol}^o = -\Delta H_{hydration} (Na^{+}) -\Delta H_{hydration} (Cl^{-}) + \Delta H_{\text{lattice energy}}\nonumber$ Be careful with signs here... the heats of hydration are the energy given off when hydrating the ion (always exothermic since interactions are made and while lattice energy is typically positive, it relates to an endothermic process (negative $$\Delta H$$). This is normally not a problem as long as you are careful when you define the process and accompanied energy.

From the above equation, increasing the lattice energy will increase $$\Delta H_{sol}^o$$ and make $$K_{sp}$$ smaller. The salt will be less soluble. This makes intutive sense since it will be "harder" (require more thermal energy) to break the stronger lattice.

Solvation energies and hydration energies are the same thing, except the latter is for water and the former is for any solvent (they can be used interchangeably for water as I often do). So we use the equation for $$\Delta H_{sol}^o$$ in part d and not as the hydration energy increase, $$\Delta H_{sol}^o$$ becomes more negative. This means (from part b) that the dissolving process if more favored (since $$\Delta G_{sol}^o$$ is more negative). The salt becomes more soluble.

Intuitively, this makes sense since there is a driving energy from the stronger interactions between the ions and the water molecules.

Increasing the IMF between solution and solvent in effect increases the solvation energies. So all of part e applies.

This is asking if $$\Delta S_{solution}^o$$ were increased, which would decrease $$\Delta G_{solution}^o$$ and then increase $$K_{sp}$$. The reaction is driven harder and the solubility is increased.

### Q9.3: A Le Chatelier

If a system is under equilibrium, then the solubility constant expression you derived for $$AgCl$$ involves the activities of both undissolved solute and dissolve solute ions (i.e., both exist). How would this equilibrium shift for Reaction $$\ref{W1}$$, if the amount of dissolved species were to be removed? This is essentially saying that $$Q < K_{sp}$$.

Key to the start of this problem is that the system has established a dynamic equilibrium with some solid salt and some some salt in solution (as ions). If the amound of dissolved species were removed, then the solid $$AgCl$$ salt will dissolve. This is since $$Q_{sp}$$ would be zero (and hence $$<K_{sp}$$).

### Q9.4: Solubility

If $$Q$$ can never equal $$K_{sp}$$ for a dissolving process, that means that all of the solute dissolves into solution (i.e., an equilibrium can never be established between solid $$AgCl$$ and dissolved $$Ag^+ (aq)$$ and $$Cl^- (aq)$$ ions) and no $$AgCl (s)$$ will exist. Adding more $$AgCl(s)$$ will result in increasing $$Ag^+ (aq)$$ and $$Cl^- (aq)$$ ions until that equilibrium is established - the system is then saturated.

This is the trick with solubility problems. You can have an equilibrium situation in which a dynamic equilibrium CANNOT be established. How? Well, the $$K_{sp}$$ wants a specific concentration of ions in solution. What if you add less $$AgCl$$ that the amount $$K_{sp}$$ wants? All of it goes into solution and none stays as a solid. The dynamics equilibrium established in Question 9.1 is not established? The system is frustrated, but deals the best it can. This means the $$Q_{sp} < K_{sp}$$ and the system has adjusted as much as it can.

Once enough salt is added to the solution and $$Q_{sp} = K_{sp}$$, the system is happy. Adding more salt will not mean more will go into solution since $$K_{sp}$$ is already established. That solid salt just sits around (but participates in a dynamic equilibrium naturally with the ions in solution).

### Q9.5: ICE Table

Construct an ICE table to demonstrate what will happen when one adds $$AgCl(s)$$ to a saturated solution of $$Ag^+ (aq)$$ and $$Cl^- (aq)$$ ions. What is $$Q$$ before adding $$AgCl(s)$$ and afterward?

From the dynamic equilibrium (Equation \red{W1}):

$AgCl (s) \rightarrow Ag^+(aq) + Cl^−(aq)$

ICE Table $$AgCl (s)$$ $$Ag^+(aq)$$ $$Cl^−(aq)$$
Initial - $$\sqrt{1.8 \times 10^{-10}}$$ $$\sqrt{1.8 \times 10^{-10}}$$
Change - +x +x
Equilibrium - $$\sqrt{1.8 \times 10^{-10}} +x$$ $$\sqrt{1.8 \times 10^{-10}} +x$$

Where $$AgCl(s)$$ is unity since we are talking activities.

The relevant equilibrium condition is (from part 9.1)

$K_{sp} = [\ce{Ag^{+} (aq)}][\ce{Cl^{-} (aq) }] = 1.8 \times 10^{-10} \nonumber$

which was established since were we under saturated conditions (hence initial conditions in ICE table). So what x, the perturbation by adding more $$AgCl(s)$$ to the equilibrium?

$K_{sp} = (\sqrt{1.8 \times 10^{10}} +x) (\sqrt{1.8 \times 10^{10}} +x) = 1.8 \times 10^{10}\nonumber$

Once can do a quadratic equation on this, but it is clear that $$x=0$$ is the only solution possible. What does this mean? Adding salt to a saturated solution does not increase the concentration of the salt ions in solution.

The $$Q_{sp}$$ before was $$K_{sp} = 1.8 \times 10^{-10}$$ since the system has established a proper equilibrium (= saturated). And after adding salt, the $$Q_{sp}$$ was still $$K_{sp} = 1.8 \times 10^{-10}$$.

### Q9.5: Solubility Again

Based off of the above discussions, what is the maximum concentration of $$Ag^+$$ that can be generated by adding $$AgCl(s)$$ to water?

That is established by the $$K_{sp}$$ value at $$\sqrt{1.8 \times 10^{-10}}$$.

What is the maximum concentration of $$Cl^-$$ that can be generated by adding $$AgCl(s)$$ to water?

That is established by the $$K_{sp}$$ value at $$\sqrt{1.8 \times 10^{-10}}$$ since there is a 1:1 stoichiometry.

What is maximum mass of $$AgCl$$ that can be dissolved by adding $$AgCl(s)$$ to 1 L of water?

$\dfrac{\sqrt{1.8 \times 10^{-10}} \,M}{1\,L} = \sqrt{1.8 \times 10^{-10}} \,\text{moles of AgCl will dissolve in 1 L}$

Molar mass of $$AgCl$$ is 143.32 g/mol so

$\sqrt{1.8 \times 10^{-10}} \,\cancel{mol} \times (143.32\, g/\cancel{mol})=1.92 \times 10^{-3} g (\text{per L})$

That is a very small number. This is why chemists commonly argue that $$AgCl$$ is insoluble (or super weakly soluble). You would have guessed that though by the magnitude of $$K_{sp}$$ already.

The amount above is generally referred to as the solubility of a substance ($$AgCl$$. However, you can also express this as a molar solubility instead, which is the number of moles of the solute that can be dissolved per liter. What is the molar solubility of $$AgCl$$?

That is just $$\sqrt{1.8 \times 10^{-10}} = 1.34 \times 10^{-5} \,\text{moles of AgCl per L}$$ from above.

## Unit II: Common Ion Effects (Sharing Species)

### Q9.6: Intuition

Sodium chloride $$NaCl$$ is appreciably more soluble than $$AgCl$$ in water ($$K_{sp} = 36$$). From a simple Le Chatelier argument, how would you expect the concentration of $$Ag^-$$ to shift if some $$NaCl$$ were added to a saturated solution of $$AgCl$$?

Here is the equilibrium we are discussing

$AgCl (s) \rightarrow Ag^+(aq) + Cl^−(aq)$

adding $$NaCl$$ to solution will add $$Na^+$$ and $$Cl^-$$ to solution. The sodium cations do nothing, but adding the chloride anions will push the reaction to the reactants.

### Q9.7: Spectator Ions

Does the $$Na^+$$ affect any equilibrium shift (why or why not)?

The sodium cations do nothing since they are not involved directly into the equilibrium (at least at this level of consideration).

### Q9.8: ICE

Construct an ICE table to describe the effect of adding $$1 \times 10^{-3} M$$ of $$Na^+$$ and $$Cl^-$$ ions to a saturated solution of $$AgCl$$. Remember that the first row in the table is the saturated solution condition + the added solute - you can call this a IACE table, but that is a silly acronym).

From the dynamic equilibrium (Equation \ref{W1}):

$AgCl (s) \rightarrow Ag^+(aq) + Cl^−(aq)$

ICE Table $$AgCl (s)$$ $$Ag^+(aq)$$ $$Cl^−(aq)$$
Initial - $$\sqrt{1.8 \times 10^{-10}}$$ $$\sqrt{1.8 \times 10^{-10}}$$
Added $$1 \times 10^{-3}$$
Change - -x -x
Equilibrium - $$\sqrt{1.8 \times 10^{-10}} -x$$ $$\sqrt{1.8 \times 10^{-10}} + 1 \times 10^{-3} -x$$

We can ignore $$Na^{+}$$ since it just "spectates" and only the chloride ions matter.

### Q9.9: Equilibrated System

Using the ICE table you constructed in Q9.8 and the fact that the equilibrium constant (i.e., $$K_{sp}$$ for this system does not change), what is the final concentration of $$Ag^+$$ in solution after adding the salt to the original solution?

As long as the equilibrium is established (i.e., the solid $$AgCl (s)$$ does not all go into solution), then this relationship must hold

\begin{align*} K_{sp} &= [\ce{Ag^{+} (aq)}][\ce{Cl^{-} (aq) }] \\[5pt] &= 1.8 \times 10^{-10} \\[5pt] (\sqrt{1.8 \times 10^{-10}} -x)(\sqrt{1.8 \times 10^{-10}} + 1 \times 10^{-3} -x) \\[5pt] &= 1.8 \times 10^{-10} \\[5pt] (1.34 \times 10^{-5} -x)( 1.0134 \times 10^{-3} -x) \\[5pt] &= 1.8 \times 10^{-10} \\[5pt] (1.35 \times 10^{-8} - (1.34 \times 10^{-5})x - (1.0134 \times 10^{-3})x +x^2 \\[5pt] &= 1.8 \times 10^{-10} \\[5pt] (1.35 \times 10^{-8} - (1.03 \times 10^{-3})x +x^2 \\[5pt] &= 1.8 \times 10^{-10} \end{align*}

There are two solution $$x=0.00101672$$ or $$x=1.327 \times 10^{-5}$$. The former would make a negative $$Ag^+(aq)$$ concentration, so the answer is $$x=1.327 \times 10^{-5}$$.

Substitute this into the expression for the equilibrium value of $$Ag^+(aq)$$ and $$Cl^−(aq)$$:

\begin{align*}[Ag^+(aq)] &= \sqrt{1.8 \times 10^{-10}} -x \\[5pt] &= 1.34 \times 10^{-5} -x \\[5pt] &= 1.34 \times 10^{-5} -1.327 \times 10^{-5} \\[5pt] &= 1.3 \times 10^{-7}\,M \end{align*}

\begin{align*}[Cl^−(aq)] &= \sqrt{1.8 \times 10^{-10}} + 1 \times 10^{-3} -x \\[5pt] &= 1.0134 \times 10^{-3} -x \\[5pt] &= 1.0134 \times 10^{-3} - 1.327 \times 10^{-5} \\[5pt] &= 1.0 \times 10^{-3}\, \end{align*}

### Q9.10:

In general, how would you describe the change of solubility of a ionic compound upon adding an additional solute that shares a common ion?

Adding the $$NaCl$$ salt pushed out (actually precipitates) close to ~99% of the $$AgCl$$ salt that was in solution. This is a general feature of common ions and solubility. Add common ions and decrease the solubility of the salt. Naturally, we can have common ions as reactants for other types of equilibria and they can push the equation forward. As long as you formulate the ICE table properly, you are golden!

## Unit III: Acids and Base Equilibria

### Q9.11: Acetic Acid is a Weak Acid

Acetic acid ( $$HC_2H_3O_2$$) is a weak acid that will dissociate in aqueous solutions to generate hydrated protons and acetate ions:

$HC_2H_3O_2 (aq) + H_2O (l) \rightleftharpoons H_3O^+ (aq) + C_2H_3O_2^- (aq) \label{W3}$

The equilibrium constant for this reaction is an acid dissociation constant ($$K_a$$, but is just an equilibrium constant with equilibrium constant needs. Use the Law of Mass Action to construction the equation of the acid dissociation constant in terms of concentrations of relevant species in this reaction (pay attention to phase).

$K_a = \dfrac{[H_3O^+ (aq)][C_2H_3O_2^- (aq)]}{[HC_2H_3O_2 (aq)]} \nonumber$

### Q9.12: Logarithms are the Way of the Future

Another way to tabulate $$K_a$$ values is as $$pK_a$$ values which are defined as

$pK_A = -\log_{10} K_a$

Why would be do such a thing (hint: it has nothing to do with scientists loving logarithms)? If $$K_a$$ defined in Q9.11 is $$1.8 \times 10^{-5}$$, what is the corresponding $$pK_a$$ for acetic acid?

We use negative logarithms since we have such a great dynamics range in equilibrium constants (from very big to very small).

$pK_a = -\log_{10} K_a = -\log (1.8 \times 10^{-5}) = 4.7$

### Q9.13: ICE Table

Construct an ICE Table to describe the reaction in Equation $$\ref{W3}$$ when 0.010 mol acetic acid is added to 100.0 mL of water.

From the dynamic equilibrium in section 9.11

$AgCl (s) \rightarrow Ag^+(aq) + Cl^−(aq)$

ICE Table $$HC_2H_3O_2 (aq)$$ $$H_2O (l)$$ $$H_3O^+ (aq)$$ $$C_2H_3O_2^- (aq)$$
Initial (mol) 0.10 - 0 0
Change (mol) -x - x x
Equilibrium (mol) 0.10 -x - x x

Notice that the dissociation of water was ignored since we are dumping so much acetic acid into the liquid.

### Q9.14: Solve the ICE Table

Given the information above about the value of $$K_a$$ (or $$pKa$$), identify the concentration of hydronium in solution upon equilibration. As with $$pK_a$$, we can construct a new value $$pH$$ that describes this

$pH \approx -\log_{10} [H_3O^+] \label{W4}$

Why is Equation $$\ref{W4}$$ written with a $$\approx$$ instead of a $$=$$ sign? What is the pH of the solution described in Q9.13. You can assume all $$H_3O^+$$ originate from the added acetic acid, however, there is another source of $$H_3O^+$$ ions; what do you think it is?

From Q9.11

\begin{align*} K_a = \dfrac{[H_3O^+ (aq)][C_2H_3O_2^- (aq)]}{[HC_2H_3O_2 (aq)]} &= 1.8 \times 10^{-5} \\[5pt] \dfrac{x^2}{0.10 -x} &= 1.8 \times 10^{-5} \\[5pt] x^2 &= (1.8 \times 10^{-5})(0.10 -x) \\[5pt] &= 1.8 \times 10^{-6} - (1.8 \times 10^{-5})x \\[5pt] x^2 + (1.8 \times 10^{-5})x - 1.8 \times 10^{-6} &= 0 \end{align*}

There are two solutions to this quadratic equation $$x=0.00133267$$ and $$x=−0.00135067$$. We can drop the latter since it generates a negative concentration. Therefore the equilibrium concentrations for this system are

• $$[HC_2H_3O_2 (aq)]_{equil}=0.1 - 0.00133 = 0.0986\,M$$
• $$[H_3O^+ (aq)]_{equil}= 0.00133\,M$$
• $$[C_2H_3O_2^- (aq)]_{equil}=0.00133\,M$$

Using Eqution \ref{W4}, we can approximate the pH as

$pH \approx -\log_{10} [H_3O^+] = -\log_{10} 0.00133 = 2.87$

Makes sense that it would be acidic since we dumped a bunch of acid into the solution. The approximation originates from using concentration instead of activity for calculating pH.

### Q9.15: Unbuffered Solution

Consider what will happen if we add 0.005 mol of HCl to this solution (fully dissociates). The strong acid will react with the acetate ion. What is the pH of this new solution?

We construct an ICE table with an "Added" term like in the common ion effect above.

$AgCl (s) \rightarrow Ag^+(aq) + Cl^−(aq) \nonumber$

ICE Table $$HC_2H_3O_2 (aq)$$ $$H_2O (l)$$ $$H_3O^+ (aq)$$ $$C_2H_3O_2^- (aq)$$
Initial (mol) 0.0986 - 0.00133 0.00133
Change (mol) +x - -x -x
Equilibrium (mol) 0.0986 +x - 0.05133 -x 0.00133 - x

We ignore the $$Cl^-$$ since it is a spectator ion for this system. We also converted moles to molarity (divide by 10) and assumed no volume change occurs when adding the $$HCl$$.

Intuitively, we would expect the added $$HCl$$ to push the reaction to the right (reactants) since this is a common ion, but how much requires calculating the ICE table properly.

\begin{align*} K_a = \dfrac{[H_3O^+ (aq)][C_2H_3O_2^- (aq)]}{[HC_2H_3O_2 (aq)]} &= 1.8 \times 10^{-5} \\[5pt] \dfrac{(0.05133 -x)(0.00133 - x)}{0.0986 +x} &= 1.8 \times 10^{-5} \\[5pt] (0.05133 -x)(0.00133 - x) &= (1.8 \times 10^{-5})(0.0986 +x) \\[5pt] 6.8 \times 10^{-5} - (0.05133)x - (0.00133)x + x^2 &= 1.77 \times 10^{-6} - (1.8 \times 10^{-5})x \\[5pt] x^2 -0.0526x + 6.62 \times 10^{-5} &= 0 \end{align*}

This quadratic equation has two roots: $$x=0.0513098$$ and $$x=0.0012902$$. The latter is the correct solution so $$x=1.3 \times 10^{-3}$$. The addition of so much hydronium pushed the reaction essentailly all the way to the reactants so there is no acetate conjugate acid.

The pH is

$-\log_{10} 0.05004 = 1.3 \nonumber$

No big surprise that adding more acid increases the pH of the system.

### Q9.16

However, we often find it useful and accurate to make the approximation that the weak acid is only slightly dissociated. Thus the equilibrium concentration of $$HC_2H_3O_2$$ is approximately equal to the initial concentration:

$[ HC_2H_3O_2 ]_{eq} \approx [ HC_2H_3O_2 ]_o$

Is this a justified assumption for the solution you described in Q9.13 and Q9.14 (why or why not)?

Since only 1.4% of the undissociated acid has dissociated (section 9.13), this is a decent approximation.

### Q9.17: Common Ion Effect in Acids and Bases Equilibria

Construct an ICE table for adding 0.010 mol sodium acetate, $$NaC_2H_3O_2$$ into a 100.0 mL solution of 0.010 mol acetic acid (assume approximation above). What is the pH of this new solution? Does this result make sense within a Le Chatelier picture?

We construct an ICE table with an "Added" term like in the common ion effect above.

$AgCl (s) \rightarrow Ag^+(aq) + Cl^−(aq) \nonumber$

ICE Table $$HC_2H_3O_2 (aq)$$ $$H_2O (l)$$ $$H_3O^+ (aq)$$ $$C_2H_3O_2^- (aq)$$
Initial (mol) 0.0986 - 0.00133 0.00133
Change (mol) +x - -x -x
Equilibrium (mol) 0.0986 +x - 0.00133 -x 0.10133 - x

Ignore the sodium cations again since the acetate ion is the common ion in the system.

\begin{align*} K_a = \dfrac{[H_3O^+ (aq)][C_2H_3O_2^- (aq)]}{[HC_2H_3O_2 (aq)]} &= 1.8 \times 10^{-5} \\[5pt] \dfrac{(0.00133 -x)(0.10133 - x)}{0.0986 +x} &= 1.8 \times 10^{-5} \\[5pt] (0.00133 -x)(0.10133 - x) &= (1.8 \times 10^{-5})(0.0986 +x) \\[5pt] 1.34 \times 10^{-4} - (0.00133)x - (0.10133)x + x^2 &= 8.2 \times 10^{-6} - (1.8 \times 10^{-5})x \\[5pt] x^2 -0.102x + 1.26 \times 10^{-4} &= 0 \end{align*}

The solutions to this quadratic equation are $$x=0.100749$$ and $$x=0.00125063$$. The latter is the only valid one so the concentration of hydronium is

$[H_3O^+ (aq)] = 0.00133 -x = 0.00133 - 0.001 = 3.3 \times 10^{-4}\,M$

$[C_2H_3O_2^- (aq)] = 0.10133 -x = 0.10133 - 0.001 = 0.10033 \,M$

$HC_2H_3O_2 (aq)] = 0.0986 +x = 0.0986 + 0.001 = 0.0996\, M$

and

$pH = -\log_{10} 3.3 \times 10^{-4} = 3.48$

And it is more basic. That makes sense since the acetate will push the reaction toward the reactants and that lowers the hydronium concentration.

### Q9.18: Buffer

Consider what will happen if we add 0.005 mol of HCl to this solution (fully dissociates). The strong acid will react with the acetate ion. What is the pH of this new solution?

ICE Table $$HC_2H_3O_2 (aq)$$ $$H_2O (l)$$ $$H_3O^+ (aq)$$ $$C_2H_3O_2^- (aq)$$
Initial (mol) 0.0996 - $$3.3 \times 10^{-4}$$ 0.10033
Change (mol) +x - -x -x
Equilibrium (mol) 0.0996 +x - 0.05033 -x 0.10033 - x

Ignore the sodium cations again since the acetate ion is the common ion in the system.

\begin{align*} K_a = \dfrac{[H_3O^+ (aq)][C_2H_3O_2^- (aq)]}{[HC_2H_3O_2 (aq)]} &= 1.8 \times 10^{-5} \\[5pt] \dfrac{(0.05033 -x)(0.10033 - x)}{0.0996 +x} &= 1.8 \times 10^{-5} \\[5pt] (0.05033 -x)(0.10033 - x) &= (1.8 \times 10^{-5})(0.0996 +x) \\[5pt] 5.04 \times 10^{-3} - (0.005033)x - (0.10033)x + x^2 &= 1.8 \times 10^{-6} - (1.8 \times 10^{-5})x \\[5pt] x^2 -0.153x + 5.05 \times 10^{-3} &= 0 \end{align*}

The solutions to this quadratic equation are $$x=0.104824$$ and $$x=0.048176$$. The latter is the one that makes sense and $$x=0.048$$. So

• $[H_3O^+ (aq)] = 0.05033 -x = 0.05033 - 0.048176 = 2.3 \times 10^{-3}\,M$
• $[C_2H_3O_2^- (aq)] = 0.10033 - x = 0.10033 - 0.048176 = 0.0523 \,M$
• $HC_2H_3O_2 (aq)] = 0.0996 +x = 0.0996 + 0.001 = 0.147\, M$

The pH after adding this extra acid is

$pH = -\log_{10} 2.3 \times 10^{-3} = 2.2$

### Q9.19

Explain the origin of the difference between the final pH calculated in Q9.18 and Q9.15.

You will notice that the pH went up when adding the acid (compared to 3.48 from Q9.17), but the pH is lower than that of the system without the conjugate base (pH=1.3) from question 9.15. This is the basis for a buffer whereby adding the salt of the conjugate base hinders the pH range. This may not appear to be a great acid buffer, but a lot of acid as added and close to a order of magnitude of hydronium concentration was "absorbed" by the buffer.

Solution 9: Acids/Bases, Common Ion Effect, and Buffers is shared under a not declared license and was authored, remixed, and/or curated by Delmar Larsen.