m51123.md

title	layout
Buffers	page

By the end of this section, you will be able to: * Describe the composition and function of acid–base buffers * Calculate the pH of a buffer before and after the addition of added acid or base

A mixture of a weak acid and its conjugate base (or a mixture of a weak base and its conjugate acid) is called a buffer solution, or a buffer{: data-type="term"}. Buffer solutions resist a change in pH when small amounts of a strong acid or a strong base are added ([link]). A solution of acetic acid and sodium acetate (CH₃COOH + CH₃COONa) is an example of a buffer that consists of a weak acid and its salt. An example of a buffer that consists of a weak base and its salt is a solution of ammonia and ammonium chloride (NH₃(aq) + NH₄Cl(aq)).

{: #CNX_Chem_14_06_compare}

How Buffers Work

A mixture of acetic acid and sodium acetate is acidic because the K_a of acetic acid is greater than the K_b of its conjugate base acetate. It is a buffer because it contains both the weak acid and its salt. Hence, it acts to keep the hydronium ion concentration (and the pH) almost constant by the addition of either a small amount of a strong acid or a strong base. If we add a base such as sodium hydroxide, the hydroxide ions react with the few hydronium ions present. The decrease in hydronium ion concentration causes the acetic acid hydrolysis equilibrium to shift to the right, restoring the hydronium ion concentration almost to its original value, and yielding a relatively modest increase in pH:

CH3CO2H(aq)+H2O(l)⇌H3O+(aq)+CH3CO2−(aq)

If we add an acid such as hydrochloric acid, the resultant increase in hydronium ion concentration shifts the equilibrium to the left. This effectively converts the added strong acid to a much weaker acid (acetic acid), and the buffer solution thus experiences only a slight decrease in pH.

{: #CNX_Chem_14_06_bufferchrt}

A mixture of ammonia and ammonium chloride is basic because the K_b for ammonia is greater than the K_a for the ammonium ion. It is a buffer because it also contains the salt of the weak base. If we add a base (hydroxide ions), ammonium ions in the buffer react with the hydroxide ions to form ammonia and water and reduce the hydroxide ion concentration almost to its original value:

NH4+(aq)+OH−(aq)⟶NH3(aq)+H2O(l)

If we add an acid (hydronium ions), ammonia molecules in the buffer mixture react with the hydronium ions to form ammonium ions and reduce the hydronium ion concentration almost to its original value:

H3O+(aq)+NH3(aq)⟶NH4+(aq)+H2O(l)

The three parts of the following example illustrate the change in pH that accompanies the addition of base to a buffered solution of a weak acid and to an unbuffered solution of a strong acid.

pH Changes in Buffered and Unbuffered Solutions Acetate buffers are used in biochemical studies of enzymes and other chemical components of cells to prevent pH changes that might change the biochemical activity of these compounds.

(a) Calculate the pH of an acetate buffer that is a mixture with 0.10 M acetic acid and 0.10 M sodium acetate.

Solution To determine the pH of the buffer solution we use a typical equilibrium calculation (as illustrated in earlier Examples):


1. *Determine the direction of change.* The equilibrium in a mixture of H₃O⁺, CH3CO2−,

and CH<sub>3</sub>CO<sub>2</sub>H is:
* * *
{: data-type="newline"}

<div data-type="equation">
<math xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mtext>CH</mtext></mrow><mn>3</mn></msub><msub><mrow><mtext>CO</mtext></mrow><mn>2</mn></msub><mtext>H</mtext><mo stretchy="false">(</mo><mi>a</mi><mi>q</mi><mo stretchy="false">)</mo><mo>+</mo><msub><mtext>H</mtext><mn>2</mn></msub><mtext>O</mtext><mo stretchy="false">(</mo><mi>l</mi><mo stretchy="false">)</mo><mspace width="0.2em" /><mo stretchy="false">⇌</mo><mspace width="0.2em" /><msub><mtext>H</mtext><mn>3</mn></msub><msup><mtext>O</mtext><mtext>+</mtext></msup><mo stretchy="false">(</mo><mi>a</mi><mi>q</mi><mo stretchy="false">)</mo><mo>+</mo><msub><mrow><mtext>CH</mtext></mrow><mn>3</mn></msub><msub><mrow><mtext>CO</mtext></mrow><mn>2</mn></msub><msup><mrow /><mtext>−</mtext></msup><mo stretchy="false">(</mo><mi>a</mi><mi>q</mi><mo stretchy="false">)</mo></mrow></math>
</div>

* * *
{: data-type="newline"}

The equilibrium constant for CH<sub>3</sub>CO<sub>2</sub>H is not given, so we look it up in [Appendix H](/m51225){: .target-chapter}\: *K*<sub>a</sub> = 1.8
<math xmlns="http://www.w3.org/1998/Math/MathML"><mo>×</mo></math>

10<sup>−5</sup>. With \[CH<sub>3</sub>CO<sub>2</sub>H\] =
<math xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo stretchy="false">[</mo><msub><mrow><mtext>CH</mtext></mrow><mn>3</mn></msub><msub><mrow><mtext>CO</mtext></mrow><mn>2</mn></msub><msup><mrow /><mtext>−</mtext></msup><mo stretchy="false">]</mo></mrow></math>

= 0.10 *M* and \[H<sub>3</sub>O<sup>+</sup>\] = ~0 *M*, the reaction shifts to the right to form H<sub>3</sub>O<sup>+</sup>.

2. Determine x and equilibrium concentrations. A table of changes and concentrations follows:

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   ![This table has two main columns and four rows. The first row for the first column does not have a heading and then has the following in the first column: Initial concentration ( M ), Change ( M ), Equilibrium ( M ). The second column has the header of “\[ C H subscript 3 C O subscript 2 H \] \[ H subscript 2 O \] equilibrium arrow H subscript 3 O superscript plus sign \[ C H subscript 3 C O subscript 2 superscript negative sign \].” Under the second column is a subgroup of four columns and three rows. The first column has the following: 0.10, negative x, 0.10 minus sign x. The second column is blank. The third column has the following: approximately 0, x, x. The fourth column has the following: 0.10, x, 0.10 plus sign x.](../resources/CNX_Chem_14_06_ICETable16_img.jpg)

3. Solve for x and the equilibrium concentrations. We find:

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   x=1.8×10−5M

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   and

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   [H3O+]=0+x=1.8×10−5M

   ---

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   Thus:

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   pH=−log[H3O+]=−log(1.8×10−5)

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   =4.74

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4. Check the work. If we calculate all calculated equilibrium concentrations, we find that the equilibrium value of the reaction coefficient, Q = K_a. {: data-number-style="arabic" .stepwise}

(b) Calculate the pH after 1.0 mL of 0.10 M NaOH is added to 100 mL of this buffer, giving a solution with a volume of 101 mL.

First, we calculate the concentrations of an intermediate mixture resulting from the complete reaction between the acid in the buffer and the added base. Then we determine the concentrations of the mixture at the new equilibrium:

![Eight tan rectangles are shown in four columns of two rectangles each that are connected with right pointing arrows. The first rectangle in the upper left is labeled “Volume of N a O H solution.” An arrow points right to a second rectangle labeled “Moles of N a O H added.” A second arrow points right to a third rectangle labeled “Additional moles of N a C H subscript 3 C O subscript 2.” Just beneath the first rectangle in the upper left is a rectangle labeled “Volume of buffer solution.” An arrow points right to another rectangle labeled “Initial moles of C H subscript 3 C O subscript 2 H.” This rectangle points to the same third rectangle, which is labeled “ Additional moles of N a C H subscript 3 C O subscript 2.” An arrow points right to a rectangle labeled “ Unreacted moles of C H subscript 3 C O subscript 2 H.” An arrow points from this rectangle to a rectangle below labeled “\[ C H subscript 3 C O subscript 2 H \].” An arrow extends below the “Additional moles of N a C H subscript 3 C O subscript 2” rectangle to a rectangle labeled “\[ C H subscript 3 C O subscript 2 \].” This rectangle points right to the rectangle labeled “\[ C H subscript 3 C O subscript 2 H \].”](../resources/CNX_Chem_14_06_steps2_img.jpg)

1. *Determine the moles of NaOH.* One milliliter (0.0010 L) of 0.10 *M* NaOH contains: * * * {: data-type="newline"}

<div data-type="equation">
<math xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mn>0.0010</mn><mspace width="0.2em" /><menclose notation="horizontalstrike"><mtext>L</mtext></menclose><mspace width="0.2em" /><mo>×</mo><mspace width="0.2em" /><mrow><mo>(</mo><mrow><mspace width="0.2em" /><mfrac><mrow><mn>0.10</mn><mspace width="0.2em" /><mtext>mol NaOH</mtext></mrow><mrow><mn>1</mn><mspace width="0.2em" /><menclose notation="horizontalstrike"><mtext>L</mtext></menclose></mrow></mfrac></mrow><mo>)</mo><mspace width="0.2em" /></mrow><mspace width="0.2em" /><mo>=</mo><mn>1.0</mn><mspace width="0.2em" /><mo>×</mo><mspace width="0.2em" /><msup><mrow><mn>10</mn></mrow><mrow><mn>−4</mn></mrow></msup><mspace width="0.2em" /><mtext>mol NaOH</mtext></mrow></math>
</div>

2. Determine the moles of CH₂CO₂H. Before reaction, 0.100 L of the buffer solution contains:

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   0.100L×(0.100molCH3CO2H1L)=1.00×10−2molCH3CO2H

3. Solve for the amount of NaCH₃CO₂ produced. The 1.0 ×

   10⁻⁴ mol of NaOH neutralizes 1.0 ×

   10⁻⁴ mol of CH₃CO₂H, leaving:

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   (1.0×10−2)−(0.01×10−2)=0.99×10−2molCH3CO2H

   ---

   {: data-type="newline"}

   and producing 1.0 ×

   10⁻⁴ mol of NaCH₃CO₂. This makes a total of:

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   (1.0×10−2)+(0.01×10−2)=1.01×10−2molNaCH3CO2

4. Find the molarity of the products. After reaction, CH₃CO₂H and NaCH₃CO₂ are contained in 101 mL of the intermediate solution, so:

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   [CH3CO2H]=9.9×10−3mol0.101L=0.098M

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   [NaCH3CO2]=1.01×10−2mol0.101L=0.100M

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   Now we calculate the pH after the intermediate solution, which is 0.098 M in CH₃CO₂H and 0.100 M in NaCH₃CO₂, comes to equilibrium. The calculation is very similar to that in part (a) of this example:

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   This series of calculations gives a pH = 4.75. Thus the addition of the base barely changes the pH of the solution ([link]). {: data-number-style="arabic" .stepwise}

(c) For comparison, calculate the pH after 1.0 mL of 0.10 M NaOH is added to 100 mL of a solution of an unbuffered solution with a pH of 4.74 (a 1.8 ×

10⁻⁵-M solution of HCl). The volume of the final solution is 101 mL.

Solution This 1.8 ×

10⁻⁵-M solution of HCl has the same hydronium ion concentration as the 0.10-M solution of acetic acid-sodium acetate buffer described in part (a) of this example. The solution contains:

0.100L×(1.8×10−5mol HCl1L)=1.8×10−6mol HCl

As shown in part (b), 1 mL of 0.10 *M* NaOH contains 1.0 ×

10⁻⁴ mol of NaOH. When the NaOH and HCl solutions are mixed, the HCl is the limiting reagent in the reaction. All of the HCl reacts, and the amount of NaOH that remains is:

(1.0×10−4)−(1.8×10−6)=9.8×10−5M

The concentration of NaOH is:

9.8×10−5MNaOH0.101L=9.7×10−4M

The pOH of this solution is:

pOH=−log[OH−]=−log(9.7×10−4)=3.01

The pH is:

pH=14.00−pOH=10.99

The pH changes from 4.74 to 10.99 in this unbuffered solution. This compares to the change of 4.74 to 4.75 that occurred when the same amount of NaOH was added to the buffered solution described in part (b).

Check Your Learning Show that adding 1.0 mL of 0.10 M HCl changes the pH of 100 mL of a 1.8 ×

10⁻⁵ M HCl solution from 4.74 to 3.00.

Answer:

Initial pH of 1.8 ×

10⁻⁵ M HCl; pH = −log[H₃O⁺] = −log[1.8 ×

10⁻⁵] = 4.74* * * {: data-type="newline"}

Moles of H₃O⁺ in 100 mL 1.8 ×

10⁻⁵ M HCl; 1.8 ×

10⁻⁵ moles/L ×

0.100 L = 1.8 ×

10⁻⁶* * * {: data-type="newline"}

Moles of H₃O⁺ added by addition of 1.0 mL of 0.10 M HCl: 0.10 moles/L ×

0.0010 L = 1.0 ×

10⁻⁴ moles; final pH after addition of 1.0 mL of 0.10 M HCl:* * * {: data-type="newline"}

pH=−log[H3O+]=−log(total molesH3O+total volume)=−log(1.0×10−4mol+1.8×10−6mol101mL(1L1000mL))=3.00

If we add an acid or a base to a buffer that is a mixture of a weak base and its salt, the calculations of the changes in pH are analogous to those for a buffer mixture of a weak acid and its salt.

Buffer Capacity

Buffer solutions do not have an unlimited capacity to keep the pH relatively constant ([link]). If we add so much base to a buffer that the weak acid is exhausted, no more buffering action toward the base is possible. On the other hand, if we add an excess of acid, the weak base would be exhausted, and no more buffering action toward any additional acid would be possible. In fact, we do not even need to exhaust all of the acid or base in a buffer to overwhelm it; its buffering action will diminish rapidly as a given component nears depletion.

{: #CNX_Chem_14_06_exhaust}

The buffer capacity{: data-type="term"} is the amount of acid or base that can be added to a given volume of a buffer solution before the pH changes significantly, usually by one unit. Buffer capacity depends on the amounts of the weak acid and its conjugate base that are in a buffer mixture. For example, 1 L of a solution that is 1.0 M in acetic acid and 1.0 M in sodium acetate has a greater buffer capacity than 1 L of a solution that is 0.10 M in acetic acid and 0.10 M in sodium acetate even though both solutions have the same pH. The first solution has more buffer capacity because it contains more acetic acid and acetate ion.

Selection of Suitable Buffer Mixtures

There are two useful rules of thumb for selecting buffer mixtures:

1. A good buffer mixture should have about equal concentrations of both of its components. A buffer solution has generally lost its usefulness when one component of the buffer pair is less than about 10% of the other. [link] shows an acetic acid-acetate ion buffer as base is added. The initial pH is 4.74. A change of 1 pH unit occurs when the acetic acid concentration is reduced to 11% of the acetate ion concentration.

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   {: #CNX_Chem_14_06_buffer}

2. Weak acids and their salts are better as buffers for pHs less than 7; weak bases and their salts are better as buffers for pHs greater than 7. {: data-number-style="arabic"}

Blood is an important example of a buffered solution, with the principal acid and ion responsible for the buffering action being carbonic acid, H₂CO₃, and the bicarbonate ion, HCO3−.

When a hydronium ion is introduced to the blood stream, it is removed primarily by the reaction:

H3O+(aq)+HCO3−(aq)⟶H2CO3(aq)+H2O(l)

An added hydroxide ion is removed by the reaction:

OH−(aq)+H2CO3(aq)⟶HCO3−(aq)+H2O(l)

The added strong acid or base is thus effectively converted to the much weaker acid or base of the buffer pair (H₃O⁺ is converted to H₂CO₃ and OH^- is converted to HCO₃^-). The pH of human blood thus remains very near the value determined by the buffer pairs pKa, in this case, 7.35. Normal variations in blood pH are usually less than 0.1, and pH changes of 0.4 or greater are likely to be fatal.

The Henderson-Hasselbalch Equation

The ionization-constant expression for a solution of a weak acid can be written as:

Ka=[H3O+][A−][HA]

Rearranging to solve for [H₃O⁺], we get:

[H3O+]=Ka×[HA][A−]

Taking the negative logarithm of both sides of this equation, we arrive at:

−log[H3O+]=−logKa− log[HA][A−],

which can be written as

pH=pKa+log[A−][HA]

where pK_a is the negative of the common logarithm of the ionization constant of the weak acid (pK_a = −log K_a). This equation relates the pH, the ionization constant of a weak acid, and the concentrations of the weak acid and its salt in a buffered solution. Scientists often use this expression, called the Henderson-Hasselbalch equation{: data-type="term"}, to calculate the pH of buffer solutions. It is important to note that the “x is small” assumption must be valid to use this equation.

Lawrence Joseph Henderson and Karl Albert Hasselbalch

Lawrence Joseph **Henderson**{: data-type="term" .no-emphasis} (1878–1942) was an American physician, biochemist and physiologist, to name only a few of his many pursuits. He obtained a medical degree from Harvard and then spent 2 years studying in Strasbourg, then a part of Germany, before returning to take a lecturer position at Harvard. He eventually became a professor at Harvard and worked there his entire life. He discovered that the acid-base balance in human blood is regulated by a buffer system formed by the dissolved carbon dioxide in blood. He wrote an equation in 1908 to describe the carbonic acid-carbonate buffer system in blood. Henderson was broadly knowledgeable; in addition to his important research on the physiology of blood, he also wrote on the adaptations of organisms and their fit with their environments, on sociology and on university education. He also founded the Fatigue Laboratory, at the Harvard Business School, which examined human physiology with specific focus on work in industry, exercise, and nutrition.

In 1916, Karl Albert Hasselbalch{: data-type="term" .no-emphasis} (1874–1962), a Danish physician and chemist, shared authorship in a paper with Christian Bohr in 1904 that described the Bohr effect, which showed that the ability of hemoglobin in the blood to bind with oxygen was inversely related to the acidity of the blood and the concentration of carbon dioxide. The pH scale was introduced in 1909 by another Dane, Sørensen, and in 1912, Hasselbalch published measurements of the pH of blood. In 1916, Hasselbalch expressed Henderson’s equation in logarithmic terms, consistent with the logarithmic scale of pH, and thus the Henderson-Hasselbalch equation was born.

Medicine: The Buffer System in Blood

The normal pH of human blood is about 7.4. The carbonate buffer system in the blood uses the following equilibrium reaction:

CO2(g)+2H2O(l)⇌H2CO3(aq)⇌HCO3−(aq)+H3O+(aq)

The concentration of carbonic acid, H₂CO₃ is approximately 0.0012 *M*, and the concentration of the hydrogen carbonate ion, HCO3−,

is around 0.024 M. Using the Henderson-Hasselbalch equation and the pK_a of carbonic acid at body temperature, we can calculate the pH of blood:

pH=pKa+log[base][acid]=6.4+log0.0240.0012=7.7

The fact that the H₂CO₃ concentration is significantly lower than that of the HCO3−

ion may seem unusual, but this imbalance is due to the fact that most of the by-products of our metabolism that enter our bloodstream are acidic. Therefore, there must be a larger proportion of base than acid, so that the capacity of the buffer will not be exceeded.

Lactic acid is produced in our muscles when we exercise. As the lactic acid enters the bloodstream, it is neutralized by the HCO3−

ion, producing H₂CO₃. An enzyme then accelerates the breakdown of the excess carbonic acid to carbon dioxide and water, which can be eliminated by breathing. In fact, in addition to the regulating effects of the carbonate buffering system on the pH of blood, the body uses breathing to regulate blood pH. If the pH of the blood decreases too far, an increase in breathing removes CO₂ from the blood through the lungs driving the equilibrium reaction such that [H₃O⁺] is lowered. If the blood is too alkaline, a lower breath rate increases CO₂ concentration in the blood, driving the equilibrium reaction the other way, increasing [H⁺] and restoring an appropriate pH.


View [information][1] on the buffer system encountered in natural waters.

Key Concepts and Summary

A solution containing a mixture of a weak acid and its conjugate base, or of a weak base and its conjugate acid, is called a buffer solution. The presence of a weak conjugate acid-base pair provides reactants that may neutralize small additions of strong acid or base, yielding weaker conjugate partners. The hydronium ion concentration of a buffer solution therefore does not change significantly when a small amount of acid or base is added.

Key Equations

• pK_a = −log K_a
• pK_b = −log K_b
• pH=pKa+log[A−][HA] {: data-bullet-style="bullet"}

Explain why a buffer can be prepared from a mixture of NH₄Cl and NaOH but not from NH₃ and NaOH.

Explain why the pH does not change significantly when a small amount of an acid or a base is added to a solution that contains equal amounts of the acid H₃PO₄ and a salt of its conjugate base NaH₂PO₄.

Excess H₃O⁺ is removed primarily by the reaction:* * * {: data-type="newline"}

H3O+(aq)+H2PO4−(aq)⟶H3PO4(aq)+H2O(l)

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Excess base is removed by the reaction:* * * {: data-type="newline"}

OH−(aq)+H3PO4(aq)⟶H2PO4−(aq)+H2O(l)

Explain why the pH does not change significantly when a small amount of an acid or a base is added to a solution that contains equal amounts of the base NH₃ and a salt of its conjugate acid NH₄Cl.

What is [H₃O⁺] in a solution of 0.25 *M* CH₃CO₂H and 0.030 *M* NaCH₃CO₂?* * * {: data-type="newline"}

CH3CO2H(aq)+H2O(l)⇌H3O+(aq)+CH3CO2−(aq)Ka=1.8×10−5

[H₃O⁺] = 1.5 ×

10⁻⁴ M

What is [H₃O⁺] in a solution of 0.075 *M* HNO₂ and 0.030 *M* NaNO₂?* * * {: data-type="newline"}

HNO2(aq)+H2O(l)⇌H3O+(aq)+NO2−(aq)Ka=4.5×10−5

What is [OH⁻] in a solution of 0.125 *M* CH₃NH₂ and 0.130 *M* CH₃NH₃Cl?* * * {: data-type="newline"}

CH3NH2(aq)+H2O(l)⇌CH3NH3+(aq)+OH−(aq)Kb=4.4×10−4

[OH⁻] = 4.2 ×

10⁻⁴ M

What is [OH⁻] in a solution of 1.25 *M* NH₃ and 0.78 *M* NH₄NO₃?* * * {: data-type="newline"}

NH3(aq)+H2O(l)⇌NH4+(aq)+OH−(aq)Kb=1.8×10−5

What concentration of NH₄NO₃ is required to make [OH⁻] = 1.0 ×

10⁻⁵ in a 0.200-M solution of NH₃?

[NH₄NO₃] = 0.36 *M*

What concentration of NaF is required to make [H₃O⁺] = 2.3 ×

10⁻⁴ in a 0.300-M solution of HF?

What is the effect on the concentration of acetic acid, hydronium ion, and acetate ion when the following are added to an acidic buffer solution of equal concentrations of acetic acid and sodium acetate:

(a) HCl

(b) KCH₃CO₂

(c) NaCl

(d) KOH

(e) CH₃CO₂H

(a) The added HCl will increase the concentration of H₃O⁺ slightly, which will react with CH3CO2−

and produce CH₃CO₂H in the process. Thus, [CH3CO2−]

decreases and [CH₃CO₂H] increases.* * * {: data-type="newline"}

(b) The added KCH₃CO₂ will increase the concentration of [CH3CO2−]

which will react with H₃O⁺ and produce CH₃CO₂ H in the process. Thus, [H₃O⁺] decreases slightly and [CH₃CO₂H] increases.* * * {: data-type="newline"}

(c) The added NaCl will have no effect on the concentration of the ions.* * * {: data-type="newline"}

(d) The added KOH will produce OH⁻ ions, which will react with the H₃O⁺, thus reducing [H₃O⁺]. Some additional CH₃CO₂H will dissociate, producing [CH3CO2−]

ions in the process. Thus, [CH₃CO₂H] decreases slightly and [CH3CO2−]

increases.* * * {: data-type="newline"}

(e) The added CH₃CO₂H will increase its concentration, causing more of it to dissociate and producing more [CH3CO2−]

and H₃O⁺ in the process. Thus, [H₃O⁺] increases slightly and [CH3CO2−]

increases.

What is the effect on the concentration of ammonia, hydroxide ion, and ammonium ion when the following are added to a basic buffer solution of equal concentrations of ammonia and ammonium nitrate:

(a) KI

(b) NH₃

(c) HI

(d) NaOH

(e) NH₄Cl

What will be the pH of a buffer solution prepared from 0.20 mol NH₃, 0.40 mol NH₄NO₃, and just enough water to give 1.00 L of solution?

pH = 8.95

Calculate the pH of a buffer solution prepared from 0.155 mol of phosphoric acid, 0.250 mole of KH₂PO₄, and enough water to make 0.500 L of solution.

How much solid NaCH₃CO₂•3H₂O must be added to 0.300 L of a 0.50-*M* acetic acid solution to give a buffer with a pH of 5.00? (Hint: Assume a negligible change in volume as the solid is added.)

37 g (0.27 mol)

What mass of NH₄Cl must be added to 0.750 L of a 0.100-*M* solution of NH₃ to give a buffer solution with a pH of 9.26? (Hint: Assume a negligible change in volume as the solid is added.)

A buffer solution is prepared from equal volumes of 0.200 *M* acetic acid and 0.600 *M* sodium acetate. Use 1.80 ×

10⁻⁵ as K_a for acetic acid.

(a) What is the pH of the solution?

(b) Is the solution acidic or basic?

(c) What is the pH of a solution that results when 3.00 mL of 0.034 M HCl is added to 0.200 L of the original buffer?

(a) pH = 5.222;* * * {: data-type="newline"}

(b) The solution is acidic.* * * {: data-type="newline"}

(c) pH = 5.221

A 5.36–g sample of NH₄Cl was added to 25.0 mL of 1.00 *M* NaOH and the resulting solution

diluted to 0.100 L.

(a) What is the pH of this buffer solution?

(b) Is the solution acidic or basic?

(c) What is the pH of a solution that results when 3.00 mL of 0.034 M HCl is added to the solution?

Which acid in [[link]](/m51118#fs-idm41095184) is most appropriate for preparation of a buffer solution with a pH of 3.1? Explain your choice.

To prepare the best buffer for a weak acid HA and its salt, the ratio [H3O+]Ka

should be as close to 1 as possible for effective buffer action. The [H₃O⁺] concentration in a buffer of pH 3.1 is [H₃O⁺] = 10^−3.1 = 7.94 ×

10⁻⁴ M* * * {: data-type="newline"}

We can now solve for K_a of the best acid as follows:* * * {: data-type="newline"}

[H3O+]Ka=1Ka=[H3O+]1=7.94×10−4

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In [link], the acid with the closest K_a to 7.94 ×

10⁻⁴ is HF, with a K_a of 7.2 ×

10⁻⁴.

Which acid in [[link]](/m51118#fs-idm41095184) is most appropriate for preparation of a buffer solution with a pH of 3.7? Explain your choice.

Which base in [[link]](/m51118#fs-idm84795184) is most appropriate for preparation of a buffer solution with a pH of 10.65? Explain your choice.

For buffers with pHs > 7, you should use a weak base and its salt. The most effective buffer will have a ratio [OH−]Kb

that is as close to 1 as possible. The pOH of the buffer is 14.00 − 10.65 = 3.35. Therefore, [OH⁻] is [OH⁻] = 10^−pOH = 10^−3.35 = 4.467 ×

10⁻⁴ M.* * * {: data-type="newline"}

We can now solve for K_b of the best base as follows:* * * {: data-type="newline"}

[OH−]Kb=1

---

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K_b = [OH⁻] = 4.47 ×

10⁻⁴* * * {: data-type="newline"}

In [link], the base with the closest K_b to 4.47 ×

10⁻⁴ is CH₃NH₂, with a K_b = 4.4 ×

10⁻⁴.

Which base in [[link]](/m51118#fs-idm84795184) is most appropriate for preparation of a buffer solution with a pH of 9.20? Explain your choice.

Saccharin, C₇H₄NSO₃H, is a weak acid (*K*_a = 2.1 ×

10⁻²). If 0.250 L of diet cola with a buffered pH of 5.48 was prepared from 2.00 ×

10⁻³ g of sodium saccharide, Na(C₇H₄NSO₃), what are the final concentrations of saccharine and sodium saccharide in the solution?

The molar mass of sodium saccharide is 205.169 g/mol. Using the abbreviations HA for saccharin and NaA for sodium saccharide the number of moles of NaA in the solution is:* * * {: data-type="newline"}

9.75 ×

10⁻⁶ mol* * * {: data-type="newline"}

The pKa for [HA] is 1.68, so [HA] = 6.2×19–9

M. Thus, [A–] (saccharin ions) is 3.90×10–5

M.* * * {: data-type="newline"}

Thus, [A⁻](saccarin ions) is 3.90 ×

10⁻⁵ M

What is the pH of 1.000 L of a solution of 100.0 g of glutamic acid (C₅H₉NO₄, a diprotic acid; *K*₁ = 8.5 ×

10⁻⁵, K₂ = 3.39 ×

10⁻¹⁰) to which has been added 20.0 g of NaOH during the preparation of monosodium glutamate, the flavoring agent? What is the pH when exactly 1 mol of NaOH per mole of acid has been added?

### Glossary {: data-type="glossary-title"}

buffer capacity : amount of an acid or base that can be added to a volume of a buffer solution before its pH changes significantly (usually by one pH unit) ^

buffer : mixture of a weak acid or a weak base and the salt of its conjugate; the pH of a buffer resists change when small amounts of acid or base are added ^

Henderson-Hasselbalch equation : equation used to calculate the pH of buffer solutions