In the previous topic (Acid-Base Chemistry Review), you studied the behavior of various acid/base systems, including

strong acids or bases in water
weak acids or bases in water
salts which can undergo hydrolysis (therefore acting as acids or bases), because an ion formed in the dissociation of the salt is a conjugate acid or base.

We will next look at systems that have mixtures containing two of the above (for example, a strong acid/weak acid mixture or a mixture of a weak base and a salt). Some of these mixtures exhibit what is called the common-ion effect.

Objective 13: Define common ion effect and explain how it affects an acid/base or solubility equilibrium.

A common ion appears in an equilibrium reaction but also comes from a source other than that reaction. The common ion effect is the shift in equilibrium (left or right) caused by the addition of a compound having an ion in common with the dissolved substances. In acid-base system, the common ion effect occurs in certain mixtures of

weak acids and salts
weak bases and salts
weak acids and strong acids
weak bases and strong bases

Common examples of the common ion effect include solutions containing both a weak acid HA and a salt providing its conjugate base (NaA, KA, etc.) Examples are:

HF and NaF
HNO₂ and KNO₂
HCN and NaCN

Another way to state this is that the common ion effect occurs in a solution containing a weak acid and its conjugate base (or a weak base and its conjugate acid). For example, in the HCN and NaCN mixture the acid is HCN and the conjugate base is CN^–. Note that it doesn’t matter what the salt’s cation is (Na⁺, K⁺, etc – it will simply act as a spectator ion).

Common ion effect example 1: a mixture of HF (acid) and NaF (salt) in aqueous solution

Before looking at the reactions that will occur, let’s look at a common mistake. Often, students think that when mixed the HF and NaF will react with each other. They will not. Try writing the reaction HF +NaF →, and it becomes apparent no reaction will occur. So what will happen?

The salt will dissociate or ionize: $\text{NaF }\!\!~\!\!\text{ (aq)}\to \text{N}{{\text{a}}^{+}}\text{(aq)+ }\!\!~\!\!\text{ }{{\text{F}}^{-}}\text{(aq)}$

The weak acid will partially ionize: $\text{HF }\!\!~\!\!\text{ (aq)}\rightleftharpoons {{\text{H}}^{+}}\text{(aq)+ }\!\!~\!\!\text{ }{{\text{F}}^{-}}\text{(aq)}$

The common ion is the aqueous F^–. Why? Because it is produced by both the acid and the salt.

The weak acid dissociation is an equilibrium reaction:

$\text{HF }\!\!~\!\!\text{ (aq)}\rightleftharpoons {{\text{H}}^{+}}\text{(aq)+ }\!\!~\!\!\text{ }{{\text{F}}^{-}}\text{(aq)}$

With the common ion effect, the NaF that is present ADDS F^–. Considering LeChatlier’s principle, what effect does this addition of F^– from the salt have on the acid dissociation equilibrium? Answer the three questions.

The presence of the common ion reduces the dissociation of the weak acid HF, making the solution less acidic than it would be were the common ion not present.

This link from LibreTexts explains the common ion effect further. Pay special attention to the section “Common ion effect with weak acids and bases”

Common ion effect example 2: a mixture of a weak base and its salt in aqueous solution

The following questions will let you think about the common ion effect with a weak base (NH₃) and a salt (NH₄Cl) providing its conjugate acid (NH₄⁺):

The weak base NH₃ ionizes as follows: $\text{N}{{\text{H}}_{3}}\text{ }\!\!~\!\!\text{ + }\!\!~\!\!\text{ }{{\text{H}}_{2}}\text{O }\!\!~\!\!\text{ }\rightleftharpoons \text{ }\!\!~\!\!\text{ N}{{\text{H}}_{4}}^{+}\text{+ }\!\!~\!\!\text{ O}{{\text{H}}^{-}}$

The salt (NH₄Cl) ionizes as follows: $\text{N}{{\text{H}}_{4}}\text{Cl }\!\!~\!\!\text{ }\!\!~\!\!\text{ }\to \text{ }\!\!~\!\!\text{ N}{{\text{H}}_{4}}^{+}\text{+ }\!\!~\!\!\text{ C}{{\text{l}}^{-}}$

A numerical example of the common ion effect with pH (and how to calculate pH with common ion)

Before calculating a common ion pH, let’s review a calculation from our acid base review last week

But what happens when, instead of being alone in solution, the 0.10 M acetic acid is accompanied by 0.10 M sodium acetate?

The pH (4.74) is higher than the pH of the 0.10 M acetic acid alone (2.87) due to the common ion effect. The acetic acid dissociates less completely in the presence of the common ion, producing a lower [H⁺] and therefore a higher pH.

Objective 13 Practice

Objective 14: Define, explain, and be able to identify buffer solutions.

A buffer solution resists change in pH when an acid or base is added to it. The common ion effect is the basis for buffer solutions. A solution containing a weak acid and its conjugate base (or a weak base and its conjugate acid) will be a buffer. As an example, a solution containing both HF and NaF is an example of a buffer solution. The HF is the weak acid. The NaF will provide F- ions, which is the conjugate base. Note that this is the same example we first looked at in Objective 13. I just didn’t tell you that it was a buffer solution at that time.

Identifying acid-base buffers – examples

A mixture of 0.1 M HNO₂ and 0.1 M NaNO₂ is a buffer. The HNO₂ is a weak acid and the NaNO₂ supplies the conjugate base NO₂^–.
A mixture of 0.5 M HNO₂ and 0.3 M NaNO₂ is also a buffer, for the same reason as the previous example. Note the conjugate acid and base concentrations do not have to be equal.
A mixture of 0.5 M HNO₃and 0.3 M NaNO₃ is not a buffer. Why not? Because HNO₃is a strong acid.
A mixture of 0.4 mol HF and 0.2 mol NaOH dissolved in a liter of water does not appear at first glance to be a buffer. But it will form a buffer. See the video below for an explanation.

A mixture of 0.4 mol HF and 0.4 mol NaOH dissolved in a liter of water will not make a buffer. Why not? In the previous example, the neutralization reaction between the weak acid HF and the hydroxide ion was this: $\text{HF + O}{{\text{H}}^{-}}\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ }\to \text{ }\!\!~\!\!\text{ }{{\text{F}}^{-}}\text{+ }\!\!~\!\!\text{ }{{\text{H}}_{2}}\text{O}$ . The OH^– converted part, but not all, of the weak acid HF to its conjugate base F^–. Let’s look at what happens in this case, when there are equal moles of HF and OH^–.

	HF	OH^–	→	F^–	H₂O
Starting moles	0.4	0.4		0	n/a
Reacting moles	-0.4	-0.4		+0.4
Moles after neutralization	0	0		0.4

This will not result in a buffer because all of the HF has been converted to conjugate base.

Objective 14 Practice

Objective 17: Calculate the pH of a buffer solution starting with an initial concentration of the buffer components and K_a or using the Henderson-Hasselbalch equation using the pK_aand the initial buffer component concentrations.

You have already seen a pH calculation of a buffer solution ( The first video on this page in objective 13). In that video, the 0.10 M acetic acid/0.10 M sodium acetate mixture is a buffer. Let’s review it with another example:

Calculate the pH of a buffer that contains 0.250 M formic acid (HCHO₂) and 0.100 M NaCHO_2.The K_a of formic acid is 1.8 X 10^-4.

The salt NaCHO₂ will produce 0.100 M CHO₂^– by ionization.

The acid will dissociate as $\text{HCH}{{\text{O}}_{2}}\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ }\rightleftarrows \text{ }\!\!~\!\!\text{ }{{\text{H}}^{+}}\text{+ }\!\!~\!\!\text{ CH}{{\text{O}}_{2}}^{-}$

The ICE table is set up as follows:

	HCHO₂	↔	H⁺	CHO₂^–
I	0.250		0	0.100
C	-x		+x	+x
E	0.250-x		x	0.100+x

The Ka expression is ${{\text{K}}_{a}}=\frac{\left[ {{\text{H}}^{+}} \right]\left[ \text{CH}{{\text{O}}_{2}}^{-} \right]}{[\text{HCH}{{\text{O}}_{2}}]}$

Substituting in the Ka value and the values from the ICE table yields

$1.8\times {{10}^{-4}}=\frac{\left[ x \right]\left[ 0.100+x \right]}{[0.250-x]}$

Neglecting the added and subtracted x yields

$1.8\times {{10}^{-4}}=\frac{\left[ x \right]\left[ 0.100 \right]}{[0.250]}$

Solving for x yields

x = 4.5 X 10^-4M = [H⁺]

and pH = -log (4.5 X 10^-4)= 3.35

Buffer calculations using an ICE table are rather laborious, and the ICE table usually isn’t necessary. Let’s look at why, and some shortcuts for buffer pH calculations next.

Shortcuts for buffer pH calculations (avoiding the ICE table)

In any effective buffer, the molarities of the conjugate acid and base will be sufficiently large that you can neglect the added and subtracted x when solving for x in the Ka expression. If you are going to do that anyway, you can simply plug the initial values for the conjugate acid and base into the Ka expression without doing the ICE table.

Let’s repeat the previous example using this shortcut:

The problem:

Calculate the pH of a buffer that contains 0.250 M formic acid (HCHO₂) and 0.100 M NaCHO_2.The K_a of formic acid is 1.8 X 10^-4.

The Ka expression is ${{\text{K}}_{a}}=\frac{\left[ {{\text{H}}^{+}} \right]\left[ \text{CH}{{\text{O}}_{2}}^{-} \right]}{[\text{HCH}{{\text{O}}_{2}}]}$

Substituting in the Ka value and the given values for the molarities of HCHO₂ and CHO₂^– yields

$1.8\times {{10}^{-4}}=\frac{\left[ {{H}^{+}} \right]\left[ 0.100 \right]}{[0.250]}$

Solving for [H⁺] yields

4.5 X 10^-4M = [H⁺]

and pH = -log (4.5 X 10^-4)= 3.35

In the shortcut just discussed, buffer pH was calculated by plugging the known Ka value and given molarities into the Ka expression, solving for [H⁺], then solving for pH. One could also algebraically manipulate the Ka expression to solve for pH before plugging in any numbers to get:

$\text{pH }\!\!~\!\!\text{ = }\!\!~\!\!\text{ p}{{\text{K}}_{{{a}_{{}}}}}+\log \frac{[\text{conj }\!\!~\!\!\text{ base}]}{[\text{conj }\!\!~\!\!\text{ acid}]}$

This equation is known as the Henderson-Hasselbalch Equation and is commonly seen in textbooks and online resources as an equation for determining buffer pH. Its algebraic derivation is shown in OpenStax Section 14.6. Note that it is no different than the Ka expression (just an algebraic manipulation), so using either method is acceptable for buffer calculations. You can try the previous example using the Henderson-Hasselbalch equation if you wish to confirm that you will get the same result of pH 3.35.

For buffer solutions based on bases and their conjugate acids, the Kb expression can be used in the same fashion. An example follows in the video below:

Calculate the pH of a buffer solution consisting of 0.200 M NH₃ and 0.300 M NH₄Cl

Objective 17 practice

Objective 15: Explain how buffer solutions are prepared and calculate the amount of the components required to prepare a buffer given pH.

The most straightforward way to prepare a buffer solution at a given or desired pH is to mix a weak acid with its conjugate base (or a weak base with its conjugate acid) in solution. In either case, the conjugate will be provided by a salt. Plugging directly into the Ka expression (realizing that you are now given pH and therefore know [H⁺] (if using Ka) or [OH^–] (if using Kb) you can solve for molarity. Try this in the following practice problem:

Another way to prepare a buffer is to start with a weak acid and add strong base to convert part of the weak acid to its conjugate base through neutralization (start with a weak base and add strong acid to convert part of the weak base to its conjugate acid through neutralization). We saw that above in Objective 14 with the example of a mixture of 0.4 mol HF and 0.2 mol NaOH producing a buffer.

Objective 19:Define buffering capacity and given a choice of buffers select which one has a higher buffering capacity.

The function of a buffer solution is to stabilize pH, so it resists change in pH when an acid or base is added to it. The way it works is that added acid is neutralized by the conjugate base in the buffer, and added base is neutralized by the conjugate acid in the buffer.

Buffering Capacity refers to the relative amount of acid (H⁺) or base (OH^–) the buffer can absorb without significantly changing the pH. Solutions with higher concentrations will have higher buffer capacity. You can read more about buffer capacity in the Buffer Capacity section in OpenStax Section 14.6.

A good buffer mixture should have approximately equal concentrations of conjugate acid or base so it has roughly equal capacity against added acid or base. Since ${{K}_{a}}=\frac{\text{ }\!\![\!\!\text{ }{{\text{H}}^{+}}][\text{conjugate base }\!\!]\!\!\text{ }}{[\text{conjugate acid }\!\!]\!\!\text{ }}$ , when [conjugate base]=[conjugate acid], K_a = [H⁺] or pK_a= pH. Therefore, to ensure approximately equal concentrations of conjugate acid and base the pKa of the weak acid should be as close as possible to the desired pH.

Objective 19 Practice

Objective 16: Using chemical equations, describe how a buffer solution can resist a change in pH when a small amount of H⁺ (from strong acid) or OH^– (from strong base) is added.

How buffers resist pH change

A buffer solution will resist change in pH when an acid or base is added to it. Let’s look at chemical equations to see why. Consider a HA/NaA buffer. In this buffer:

The conjugate acid is HA
The conjugate base is: A^–. The sodium is just a spectator ion and should be removed from equations.
NOTE HA/NaA is not a chemical formula and should not a chemical formula. It is simply notation showing the two species making up the buffer solution.

If base (OH^–) is added to the buffer, the conjugate acid HA in the buffer neutralizes the added base and is converted to conjugate base A-. The net ionic equation is: $\text{HA }\!\!~\!\!\text{ + }\!\!~\!\!\text{ O}{{\text{H}}^{-}}\text{ }\!\!~\!\!\text{ }\to \text{ }\!\!~\!\!\text{ }{{\text{A}}^{-}}\text{+ }\!\!~\!\!\text{ }{{\text{H}}_{2}}\text{O}$

Note the added base converts the conjugate acid HA in the buffer to conjugate base A-.

If acid (H⁺) is added to the buffer, the conjugate base A^– in the buffer neutralizes the added acid and is converted to conjugate acid HA. The net ionic equation is: ${{\text{A}}^{-}}\text{ }\!\!~\!\!\text{ + }\!\!~\!\!\text{ }{{\text{H}}^{+}}\text{ }\!\!~\!\!\text{ }\to \text{ }\!\!~\!\!\text{ HA}$

Note the added acid converts the conjugate base A^– in the buffer to conjugate acid HA.

The video below shows how equations would be written for the addition of an acid or a base to a HNO₂/KNO₂ buffer

Further information on this objective in the section “How Buffers Work” at the beginning of OpenStax Section 14.6

Objective 16 example

Objective 18: Calculate the change in pH of a buffer solution of known composition caused by adding a small amount of strong acid (H⁺) or strong base (OH^–).

Calculating the pH change upon addition of addition of acid and base to a buffer involves putting together two concepts you have learned in this unit:

the equations you just saw in Objective 16
the pH calculation for a buffer in objective 17

The procedure that combines these two concepts is this:

Write the neutralization reaction of the added acid or base and the appropriate component of the buffer
Determine the new concentrations of the buffer using stoichiometry of the neutralization reaction (uses moles; looks like an ICE table but it is not!)
Determine pH of the “new” buffer with its new concentrations of conjugate acid and conjugate base. (this will be a buffer pH calculation using Ka, Kb, or Henderson-Hasselbalch). Note when determining pH: K_a or Kb is NOT products/reactants for the neutralization reaction in step 1.

Objective 18 Example

Let’s see how these steps are applied in an example:

Calculate the pH change that takes place when 100.0 mL of 0.0500 M NaOH are added to 400.0 mL of a buffer solution that is 0.200 M in NH₃ and 0.300 M in NH₄Cl. The K_b of NH₃ = 1.8 x 10^-5 .

Step 1 : Write the neutralization reaction of the added acid or base and the appropriate component of the buffer

$\text{N}{{\text{H}}_{4}}^{+}\text{ }\!\!~\!\!\text{ + }\!\!~\!\!\text{ O}{{\text{H}}^{-}}\text{ }\!\!~\!\!\text{ }\to \text{ }\!\!~\!\!\text{ N}{{\text{H}}_{3}}\text{+ }\!\!~\!\!\text{ }{{\text{H}}_{2}}\text{O}$

Step 2: Determine the new concentrations of the buffer using stoichiometry of the neutralization reaction (uses moles; looks like an ICE table but it is not!)

Initial moles of NH₄⁺: (0.400 L)(0.300 M) = 0.120 mol

Initial moles of NH3: (0.400 L)(0.300 M) = 0.080 mol

Initial moles of OH^-: (0.100 L)(0.0500 M) = 0.005 mol

Stoichiometry table:

	NH₄⁺	OH^–	→	NH3	H₂O
Starting moles	0.120	0.005		0.080	n/a
Reacting moles	-0.005	-0.005		+0.005
Moles after neutralization	0.115	0		0.085

Step 3: Determine pH of the “new” buffer with its new concentrations of conjugate acid and conjugate base. (this will be a buffer pH calculation using Ka, Kb, or Henderson-Hasselbalch).

After neutralization, we have a “new buffer” that is made up of 0.115 mol NH₄⁺and 0.085 mol NH₃ in 500.0 mL of solution (100 mL were added to the original 400). So there are new molarities of the conjugate acids and bases.

$[\text{N}{{\text{H}}_{4}}^{+}]\text{ }\!\!~\!\!\text{ = }\frac{0.115mol}{0.500L}$

$[\text{N}{{\text{H}}_{3}}^{{}}]\text{ }\!\!~\!\!\text{ = }\frac{0.085mol}{0.500L}$

For the pH calculation, the Kb expression of NH₃ can be used:

${{K}_{b}}=\frac{[\text{conjugate acid }\!\!]\!\!\text{ }\!\![\!\!\text{ O}{{\text{H}}^{-}}]}{[\text{conjugate base }\!\!]\!\!\text{ }}=\frac{[N{{H}_{4}}^{+}\text{ }\!\!]\!\!\text{ }\!\![\!\!\text{ O}{{\text{H}}^{-}}]}{[N{{H}_{3}}\text{ }\!\!]\!\!\text{ }}$

Note when determining pH: K_a or Kb is NOT products/reactants for the neutralization reaction in step 1.

Substituting in for Kb,

$1.8\times {{10}^{-5}}=\frac{\left( \frac{0.115mol}{0.500L} \right)\text{ }\!\![\!\!\text{ O}{{\text{H}}^{-}}]}{\left( \frac{0.085mol}{0.500L} \right)}$

Note cancelling the 0.500 L volumes makes the math easier.

Solving for [OH^–] yields [OH^–] =1.33 X 10^-5M. pOH = -log 1.33 X 10^-5= 4.88 and pH = 9.12.

Therefore, the pH after the NaOH was added to the buffer was 9.12.

The initial pH of the buffer was 9.08 (solved in the video in objective 17). Therefore the pH was increased by 0.04, which is the small increase expected as buffers minimize pH changes.

Another example is shown at the bottom if Example 14.20 in OpenStax section 14.6

Bruce’s notes – Common Ion Effect and Buffers

Objective 13: Define common ion effect and explain how it affects an acid/base or solubility equilibrium.

Common ion effect example 1: a mixture of HF (acid) and NaF (salt) in aqueous solution

Common ion effect example 2: a mixture of a weak base and its salt in aqueous solution

A numerical example of the common ion effect with pH (and how to calculate pH with common ion)

Objective 13 Practice

Objective 14: Define, explain, and be able to identify buffer solutions.

Identifying acid-base buffers – examples

Objective 14 Practice

Objective 17: Calculate the pH of a buffer solution starting with an initial concentration of the buffer components and K_a or using the Henderson-Hasselbalch equation using the pK_aand the initial buffer component concentrations.

Shortcuts for buffer pH calculations (avoiding the ICE table)

Objective 15: Explain how buffer solutions are prepared and calculate the amount of the components required to prepare a buffer given pH.

Objective 19:Define buffering capacity and given a choice of buffers select which one has a higher buffering capacity.

Objective 19 Practice

Objective 16: Using chemical equations, describe how a buffer solution can resist a change in pH when a small amount of H⁺ (from strong acid) or OH^– (from strong base) is added.

How buffers resist pH change

Objective 16 example

Objective 18: Calculate the change in pH of a buffer solution of known composition caused by adding a small amount of strong acid (H⁺) or strong base (OH^–).

Objective 18 Example

Objective 18 Practice

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Objective 13: Define common ion effect and explain how it affects an acid/base or solubility equilibrium.

Common ion effect example 1: a mixture of HF (acid) and NaF (salt) in aqueous solution

Common ion effect example 2: a mixture of a weak base and its salt in aqueous solution

A numerical example of the common ion effect with pH (and how to calculate pH with common ion)

Objective 13 Practice

Objective 14: Define, explain, and be able to identify buffer solutions.

Identifying acid-base buffers – examples

Objective 14 Practice

Objective 17: Calculate the pH of a buffer solution starting with an initial concentration of the buffer components and Ka or using the Henderson-Hasselbalch equation using the pKa and the initial buffer component concentrations.

Shortcuts for buffer pH calculations (avoiding the ICE table)

Objective 15: Explain how buffer solutions are prepared and calculate the amount of the components required to prepare a buffer given pH.

Objective 19:Define buffering capacity and given a choice of buffers select which one has a higher buffering capacity.

Objective 19 Practice

Objective 16: Using chemical equations, describe how a buffer solution can resist a change in pH when a small amount of H+ (from strong acid) or OH– (from strong base) is added.

How buffers resist pH change

Objective 16 example

Objective 18: Calculate the change in pH of a buffer solution of known composition caused by adding a small amount of strong acid (H+) or strong base (OH–).

Objective 18 Example

Objective 18 Practice

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Objective 17: Calculate the pH of a buffer solution starting with an initial concentration of the buffer components and K_a or using the Henderson-Hasselbalch equation using the pK_aand the initial buffer component concentrations.

Objective 16: Using chemical equations, describe how a buffer solution can resist a change in pH when a small amount of H⁺ (from strong acid) or OH^– (from strong base) is added.

Objective 18: Calculate the change in pH of a buffer solution of known composition caused by adding a small amount of strong acid (H⁺) or strong base (OH^–).