Can a Buffer Solution Have a Strong Acid and Strong Base

Preparing a Buffer Solution with a Specific pH

A buffer is a solution of weak acrid and cohabit base or weak base and conjugate acid used to resist pH modify with added solute.

Learning Objectives

Describe the properties of a buffer solution.

Cardinal Takeaways

Key Points

• Buffer solutions are resistant to pH change because of the presence of an equilibrium betwixt the acid (HA) and its conjugate base (A-).
• When some stiff acrid is added to a buffer, the equilibrium is shifted to the left, and the hydrogen ion concentration increases by less than expected for the amount of stiff acid added.
• Buffer solutions are necessary in biology for keeping the correct pH for proteins to piece of work.
• Buffers can be prepared in multiple ways by creating a solution of an acrid and its cohabit base.

Fundamental Terms

• aqueous: Consisting mostly of water.
• equilibrium: The state of a reaction in which the rates of the forward (reactant to product) and opposite (production to reactant) reactions are the same.
• pKa: A quantitative measure of the forcefulness of an acid in solution; a weak acid has a pKa value in the gauge range −2 to 12 in water and a strong acid has a pKa value of less than nearly −2.

Buffers

A buffer is an aqueous solution containing a weak acrid and its conjugate base or a weak base of operations and its cohabit acid. A buffer's pH changes very little when a pocket-size amount of stiff acrid or base is added to it. It is used to forbid whatever modify in the pH of a solution, regardless of solute. Buffer solutions are used as a means of keeping pH at a nearly constant value in a wide diverseness of chemical applications. For case, blood in the human body is a buffer solution.

Buffer solutions are resistant to pH modify because of the presence of an equilibrium between the acid (HA) and its conjugate base (A^–). The balanced equation for this reaction is:

[latex]{ \text{HA}\rightleftharpoons \text{H} }^{ + }+{ \text{A} }^{ - }[/latex]

When some strong acrid (more than H⁺) is added to an equilibrium mixture of the weak acid and its conjugate base, the equilibrium is shifted to the left, in accord with Le Chatelier's principle. This causes the hydrogen ion (H⁺) concentration to increase by less than the amount expected for the quantity of strong acid added. Similarly, if a potent base of operations is added to the mixture, the hydrogen ion concentration decreases by less than the amount expected for the quantity of base added. This is considering the reaction shifts to the right to accommodate for the loss of H⁺ in the reaction with the base.

Buffer solutions are necessary in a wide range of applications. In biology, they are necessary for keeping the correct pH for proteins to work; if the pH moves outside of a narrow range, the proteins terminate working and can fall apart. A buffer of carbonic acid (H₂CO_three) and bicarbonate (HCO₃ ⁻) is needed in claret plasma to maintain a pH between 7.35 and 7.45. Industrially, buffer solutions are used in fermentation processes and in setting the correct conditions for dyes used in coloring fabrics.

Preparing a Buffer Solution

There are a couple of means to prepare a buffer solution of a specific pH. In the first method, prepare a solution with an acrid and its conjugate base by dissolving the acid form of the buffer in well-nigh lx% of the volume of water required to obtain the final solution volume. Then, measure the pH of the solution using a pH probe. The pH can be adjusted upwards to the desired value using a strong base like NaOH. If the buffer is made with a base of operations and its conjugate acid, the pH tin be adjusted using a potent acid like HCl. One time the pH is correct, dilute the solution to the concluding desired volume.

pH probe: The probe can be inserted into a solution to measure the pH (reading viii.61 in this case). Probes need to be regularly calibrated with solutions of known pH to be accurate.

Alternatively, you tin ready solutions of both the acid form and base of operations form of the solution. Both solutions must incorporate the same buffer concentration as the concentration of the buffer in the final solution. To go the final buffer, add one solution to the other while monitoring the pH.

In a third method, yous can make up one's mind the exact amount of acid and cohabit base needed to make a buffer of a certain pH, using the Henderson-Hasselbach equation:

[latex]\text{pH}=\text{p}{ \text{K} }_{ \text{a} }+\text{log}(\frac { { [\text{A} }^{ - }] }{ [\text{HA}] } )[/latex]

where pH is the concentration of [H+], pK_a is the acid dissociation constant, and [\text{A}-] and [\text{HA}] are concentrations of the cohabit base and starting acid.

Computing the pH of a Buffer Solution

The pH of a buffer solution can be calculated from the equilibrium constant and the initial concentration of the acid.

Learning Objectives

Calculate the pH of a buffer made only from a weak acid.

Key Takeaways

Key Points

• The strength of a weak acid ( buffer ) is usually represented as an equilibrium constant.
• The acid-dissociation equilibrium constant, which measures the propensity of an acid to dissociate, is described using the equation: [latex]{ \text{K}}_{\text{a} }=\frac { { [\text{H} }^{ + }][{ \text{A} }^{ - }] }{ [\text{HA}] }[/latex].
• Using K_a and the equilibrium equation, yous tin can solve for the concentration of [H⁺].
• The concentration of [H⁺] tin can and so be used to calculate the pH of a solution, as part of the equation: pH = -log([H⁺]).

Key Terms

• equilibrium: The state of a reaction in which the rates of the frontwards (reactant to product) and reverse (product to reactant) reactions are the same.

What Does pH Mean in a Buffer?

In chemistry, pH is a measure of the hydrogen ion (H⁺) concentration in a solution. The pH of a buffer can be calculated from the concentrations of the various components of the reaction. The balanced equation for a buffer is:

[latex]\text{HA} \rightleftharpoons \text{H}^+ + \text{A}^-[/latex]

The force of a weak acid is usually represented as an equilibrium constant. The acid-dissociation equilibrium constant (K_a), which measures the propensity of an acrid to dissociate, for the reaction is:

[latex]\text{G}_{\text{a}} = \frac{\left [\text{H}^{+} \right ]\left [\text{A}^{-} \right ]}{\left [\text{HA} \correct ]}[/latex]

The greater [H⁺] x [A^–] is than [HA], the greater the value of K_a, the more the formation of H⁺ is favored, and the lower the pH of the solution.

Ice Tables: A Useful Tool For Solving Equilibrium Problems

ICE (Initial, Change, Equilibrium) tables are very helpful tools for understanding equilibrium and for computing the pH of a buffer solution. They consist of using the initial concentrations of reactants and products, the change they undergo during the reaction, and their equilibrium concentrations. Consider, for example, the following trouble:

Calculate the pH of a buffer solution that initially consists of 0.0500 G NH₃ and 0.0350 M NH₄ ⁺. (Notation: K_a for NH_four ⁺ is v.half dozen 10 10^-10). The equation for the reaction is every bit follows:

[latex]\text{NH}_4^+ \rightleftharpoons \text{H}^+ + \text{NH}_3[/latex]

We know that initially there is 0.0350 M NH₄ ⁺ and 0.0500 M NH₃. Earlier the reaction occurs, no H⁺ is present so it starts at 0.

ICE table – initial: Water ice table for the buffer solution of NH4+ and NH3 with the starting concentrations.

During the reaction, the NH_iv ⁺ volition dissociate into H⁺ and NH₃. Because the reaction has a 1:i stoichiometry, the amount that NH₄ ⁺ loses is equal to the amounts that H⁺ and NH₃ will gain. This change is represented past the alphabetic character x in the following table.

Ice table – modify: Describes the change in concentration that occurs during the reaction.

Therefore the equilibrium concentrations will look similar this:

ICE table – equilibrium: Describes the concluding concentration of the reactants and products at equilibrium.

Utilize the equilibrium values to the expression for K_a.

[latex]{ 5.6 \times 10^{-10}} = \frac { { [\text{H} }^{ + }][{ \text{NH}_3 }] }{ [\text{NH}_4^+] } = \frac { x (0.0500+\text{ten})}{ 0.0350-\text{10} }[/latex]

Assuming x is negligible compared to 0.0500 and 0.0350 the equation is reduced to:

[latex]{ five.half dozen \times 10^{-10}} = \frac { { [\text{H} }^{ + }][{ \text{NH}_3 }] }{ [\text{NH}_4^+] } = \frac { \text{x} (0.0500)}{ 0.0350 }[/latex]

Solving for x (H⁺):

10 = [H⁺] = 3.92 x 10^-10

pH = -log(3.92 ten 10^-10)

pH = 9.41

The Henderson-Hasselbalch Equation

The Henderson–Hasselbalch equation connects the measurable value of the pH of a solution with the theoretical value pKa.

Learning Objectives

Calculate the pH of a buffer system using the Henderson-Hasselbalch equation.

Central Takeaways

Central Points

• The Henderson-Hasselbalch equation is useful for estimating the pH of a buffer solution and finding the equilibrium pH in an acid – base reaction.
• The formula for the Henderson–Hasselbalch equation is: [latex]\text{pH}=\text{p}{ \text{Thou} }_{ \text{a} }+\text{log}(\frac { { [\text{A} }^{ - }] }{ [\text{HA}] } )[/latex], where pH is the concentration of [H+], pK_a is the acid dissociation constant, and [A^–] and [HA] are concentrations of the cohabit base and starting acid.
• The equation can exist used to determine the amount of acid and cohabit base needed to make a buffer solution of a sure pH.

Key Terms

• pKa: A quantitative mensurate of the strength of an acid in solution; a weak acid has a pKa value in the approximate range -two to 12 in h2o and a strong acid has a pKa value of less than most -ii.

The Henderson–Hasselbalch equation mathematically connects the measurable pH of a solution with the pK_a (which is equal to -log One thousand_a) of the acrid. The equation is also useful for estimating the pH of a buffer solution and finding the equilibrium pH in an acid-base reaction. The equation can exist derived from the formula of pK_a for a weak acid or buffer. The balanced equation for an acid dissociation is:

[latex]\text{HA}\rightleftharpoons { \text{H} }^{ + }+{ \text{A} }^{ - }[/latex]

The acid dissociation constant is:

[latex]{ \text{M} }_{ \text{a} }=\frac { [{ \text{H} }^{ + }][\text{A}^{ - }] }{ [\text{HA}] }[/latex]

Afterward taking the log of the unabridged equation and rearranging it, the result is:

[latex]\text{log}({ \text{K} }_{ \text{a} })=\text{log}[{ \text{H} }^{ + }]+\text{log}(\frac { { [\text{A} }^{ - }] }{ [\text{HA}] } )[/latex]

This equation can be rewritten equally:

[latex]-\text{p}{ \text{G} }_{ \text{a} }=-\text{pH}+\text{log}(\frac { [\text{A}^{ - }] }{ [\text{HA}] } )[/latex]

Distributing the negative sign gives the last version of the Henderson-Hasselbalch equation:

[latex]\text{pH}=\text{p}{ \text{Thou} }_{ \text{a} }+\text{log}(\frac { { [\text{A} }^{ - }] }{ [\text{HA}] } )[/latex]

In an alternate application, the equation tin be used to decide the amount of acid and conjugate base of operations needed to make a buffer of a certain pH. With a given pH and known pK_a, the solution of the Henderson-Hasselbalch equation gives the logarithm of a ratio which can be solved by performing the antilogarithm of pH/pK­_a:

[latex]{ 10 }^{ \text{pH}-\text{p}{ \text{K} }_{ \text{a} } }=\frac { [\text{base}] }{ [\text{acrid}] }[/latex]

An instance of how to use the Henderson-Hasselbalch equation to solve for the pH of a buffer solution is equally follows:

What is the pH of a buffer solution consisting of 0.0350 K NH₃ and 0.0500 M NH₄ ⁺ (K_a for NH₄ ⁺ is 5.6 10 10^-10)? The equation for the reaction is:

[latex]{\text{NH}_4^+}\rightleftharpoons { \text{H} }^{ + }+{ \text{NH}_3}[/latex]

Assuming that the alter in concentrations is negligible in order for the arrangement to reach equilibrium, the Henderson-Hasselbalch equation will exist:

[latex]\text{pH}=\text{p}{ \text{K} }_{ \text{a} }+\text{log}(\frac { { [\text{NH}_3}] }{ [\text{NH}_4^+] } )[/latex]

[latex]\text{pH}=9.25+\text{log}(\frac{0.0350}{0.0500} )[/latex]

pH = nine.095

Calculating Changes in a Buffer Solution

The changed pH of a buffer solution in response to the add-on of an acid or a base can be calculated.

Learning Objectives

Summate the final pH of a solution generated by the addition of a strong acid or base of operations to a buffer.

Key Takeaways

Key Points

• If the concentrations of the weak acid and its conjugate base in a buffer solution are reasonably loftier, then the solution is resistant to changes in hydrogen ion concentration, or pH.
• The change in pH of a buffer solution with an added acrid or base tin be calculated by combining the balanced equation for the reaction and the equilibrium acid dissociation constant (K_a).
• Comparing the last pH of a solution with and without the buffer components shows the effectiveness of the buffer in resisting a modify in pH.

Key Terms

• pH: The negative of the logarithm to base 10 of the concentration of hydrogen ions, measured in moles per liter; a measure of acerbity or alkalinity of a substance, which takes numerical values from 0 (maximum acidity) through vii (neutral) to 14 (maximum alkalinity).
• acid dissociation constant: Quantitative measure of the strength of an acid in solution; typically written as a ratio of the equilibrium concentrations of products to reactants.

If the concentrations of a solution of a weak acrid and its cohabit base are reasonably loftier, then the solution is resistant to changes in hydrogen ion concentration. These solutions are known as buffers. It is possible to calculate how the pH of the solution will change in response to the improver of an acid or a base of operations to a buffer solution.

Calculating Changes in a Buffer Solution, Instance 1:

A solution is 0.050 One thousand in acerb acid (HC₂H₃O_two) and 0.050 One thousand NaC_iiH₃O₂. Calculate the alter in pH when 0.001 mole of hydrochloric acid (HCl) is added to a liter of solution, assuming that the volume increment upon adding the HCl is negligible. Compare this to the pH if the same corporeality of HCl is added to a liter of pure h2o.

Pace 1:

[latex]{\text{HC}_2\text{H}_3\text{O}_2}(\text{aq})\leftrightharpoons {\text{H}^+}(\text{aq})+{\text{C}_2\text{H}_3\text{O}_2^-}(\text{aq}) [/latex]

Retrieve that sodium acetate,
NaC_twoH₃O₂, dissociates into its component ions, Na⁺ and C₂H_threeO_two ^– (the acetate ion) upon dissolution in water. Therefore, the solution will contain both acetic acid and acetate ions.

Before adding HCl, the acetic acrid equilibrium abiding is:

[latex]{ \text{K} }_{ \text{a} }=\frac { { [\text{H} }^{ + }]{ [\text{C}_2{\text{H}}_3{\text{O}}_2 }^{ - }] }{ [\text{H}{\text{C}}_2{\text{H}}_3{\text{O}}_2] } =\frac { \text{x}(0.050) }{ (0.050) }[/latex]

(bold that x is pocket-sized compared to 0.050 M in the equilibrium concentrations)

Therefore:

[latex]\text{x}=[\text{H}^+]={ \text{K} }_{ \text{a} }=one.76\times one{ 0 }^{ -5 }\text{M}[/latex]

[latex]\text{pH}={ \text{pK} }_{ \text{a} }=4.75[/latex]

In this example, ignoring the 10 in the [C_iiH₃O_ii ^–] and [HC_iiH_threeO₂] terms was justified because the value is pocket-sized compared to 0.050.

Step ii:

The added protons from HCl combine with the acetate ions to form more than acetic acid:

[latex]\text{C}_2\text{H}_3\text{O}_2^{ - }+{ \text{H} }^{ + }(\text{from HCl})\rightarrow \text{HC}_2\text{H}_3\text{O}_2[/latex]

Since all of the H⁺ volition be consumed, the new concentrations will be [latex][\text{HC}_2\text{H}_3\text{O}_2]=0.051 \text{M}[/latex] and [latex][\text{C}_2\text{H}_3\text{O}_2^-]=0.049 \text{Grand}[/latex] before the new equilibrium is to be established. Then, we consider the equilibrium conentrations for the dissociation of acerb acid, equally in Pace ane:

[latex]{\text{HC}_2\text{H}_3\text{O}_2}(\text{aq})\leftrightharpoons {\text{H}^+}(\text{aq})+{\text{C}_2\text{H}_3\text{O}_2^-}(\text{aq}) [/latex]

we have,

[latex]{ \text{Chiliad} }_{ \text{a} }=\frac { \text{x}(0.049) }{ (0.051) }[/latex]

[latex]x=[\text{H}^+]=(1.76\times i{ 0 }^{ -5 })\frac { 0.051 }{ 0.049 } =1.83\times 1{ 0 }^{ -5 }\text{M}[/latex]

[latex]\text{pH}=-\text{log}([{ \text{H} }^{ + }])=4.74[/latex]

In the presence of the acerb acid-acetate buffer system, the pH only drops from 4.75 to 4.74 upon improver of 0.001 mol of strong acid HCl, a difference of only 0.01 pH unit of measurement.

Pace 3:

Adding 0.001 G HCl to pure water, the pH is:

[latex]\text{pH}=-\text{log}([{ \text{H} }^{ + }])=three.00[/latex]

In the absence of HC_twoH_iiiO₂ and C₂H₃O_two ^– _, the same concentration of HCl would produce a pH of 3.00.

Calculating Changes in a Buffer Solution, Example 2:

A formic acid buffer is prepared with 0.010 M each of formic acid (HCOOH) and sodium formate (NaCOOH). The K_a for formic acrid is one.viii x x^-4. What is the pH of the solution? What is the pH if 0.0020 M of solid sodium hydroxide (NaOH) is added to a liter of buffer? What would be the pH of the sodium hydroxide solution without the buffer? What would the pH accept been afterward adding sodium hydroxide if the buffer concentrations had been 0.10 M instead of 0.010 M?

Step ane:

Solving for the buffer pH:

[latex]\text{HCOOH} \leftrightharpoons {\text{H}^+} + {\text{HCOO}^-}[/latex]

Assuming x is negligible, the K_a expression looks like:

[latex]{ \text{K} }_{ \text{a} }=\frac { \text{10}(0.010) }{ (0.010) }[/latex]

ane.viii x 10^-4 = 10 = [H⁺]

pH = -log [H⁺] = 3.74

Buffer: pH = 3.74

Footstep 2:

Solving for the buffer pH after 0.0020 Grand NaOH has been added:

[latex]\text{OH}^- + \text{HCOOH} \rightarrow {\text{H}_2O} + {\text{HCOO}^-}[/latex]

The concentration of HCOOH would modify from 0.010 M to 0.0080 M and the concentration of HCOO^– would change from 0.010 M to 0.0120 M.

[latex]{ \text{M} }_{ \text{a} }=\frac { \text{x}(0.0120) }{ (0.0080) }[/latex]

After calculation NaOH, solving for [latex]\text{x}=[\text{H}^+][/latex] and then calculating the pH = 3.92. The pH went up from iii.74 to three.92 upon addition of 0.002 Yard of NaOH.

Step 3:

Solving for the pH of a 0.0020 M solution of NaOH:

pOH = -log (0.0020)

pOH = ii.70

pH = xiv – pOH

pH = 11.30

Without buffer: pH = 11.30

Step 4:

Solving for the pH of the buffer solution if 0.1000 G solutions of the weak acid and its conjugate base had been used and the same amount of NaOH had been added:

The concentration of HCOOH would change from 0.yard Yard to 0.0980 Grand and the concentration of HCOO^– would change from 0.k M to 0.1020 M.

[latex]{ \text{One thousand} }_{ \text{a} }=\frac { \text{10}(0.1020) }{ (0.0980) }[/latex]

pH if 0.one thousand M concentrations had been used = 3.77

This shows the dramatic effect of the formic acid-formate buffer in keeping the solution acidic in spite of the added base. It as well shows the importance of using loftier buffer component concentrations so that the buffering capacity of the solution is non exceeded.

Buffers Containing a Base of operations and Cohabit Acrid

An alkali metal buffer can be made from a mixture of the base and its conjugate acrid, but the formulas for determining pH accept a different class.

Learning Objectives

Calculate the pH of an alkaline buffer system consisting of a weak base and its cohabit acid.

Key Takeaways

Primal Points

• The pH of bases is usually calculated using the hydroxide ion (OH^–) concentration to find the pOH first.
• The formula for pOH is pOH=-log[OH-]. A base of operations dissociation constant (Kb) indicates the strength of the base.
• The pH of a basic solution can be calculated by using the equation: pH = 14.00 – pOH.

Key Terms

• alkaline: Having a pH greater than seven.
• buffers: A weak acid or base used to maintain the acidity (pH) of a solution well-nigh a called value and which prevent a rapid modify in pH when acids or bases are added to the solution.

A base is a substance that decreases the hydrogen ion (H⁺) concentration of a solution. In the more generalized Brønsted-Lowry definition, the hydroxide ion (OH^–) is the base considering it is the substance that combines with the proton. Ammonia and some organic nitrogen compounds tin can combine with protons in solution and human action as Brønsted-Lowry bases. These compounds are generally weaker bases than the hydroxide ion because they accept less attraction for protons. For case, when ammonia competes with OH^– for protons in an aqueous solution, it is only partially successful. It can combine with merely a portion of the H⁺ ions, so it will have a measurable equilibrium abiding. Reactions with weak bases consequence in a relatively low pH compared to strong bases. Bases range from a pH of greater than 7 (7 is neutral like pure water) to 14 (though some bases are greater than fourteen).

An alkaline buffer can be made from a mixture of a base and its conjugate acid, similar to the way in which weak acids and their conjugate bases tin can be used to make a buffer.

Ammonia to ammonium ion: 2-dimensional epitome depicting the association of proton (H⁺) with the weak base ammonia (NH_three) to form its cohabit acrid, ammonium ion (NH₄ ⁺).

Computing the pH of a Base

The pH of bases is normally calculated using the OH^– concentration to find the pOH outset. This is done because the H⁺ concentration is not a part of the reaction, while the OH^– concentration is. The formula for pOH is:

[latex]\text{pOH}=-\text{log}([\text{O}{ \text{H} }^{ - }])[/latex]

By multiplying a cohabit acid (such as NH_four ⁺) and a conjugate base (such as NH₃) the following is given:

[latex]{ \text{Yard} }_{ \text{a} }\times { \text{One thousand} }_{ \text{b} }=\frac { [{ \text{H} }_{ iii }{ \text{O} }^{ + }][\text{N}{ \text{H} }_{ three }] }{ [\text{N}{ \text{H} }_{ 4 }^{ + }] } \times \frac { [\text{North}{ \text{H} }_{ 4 }^{ + }][{ \text{OH} }^{ - }] }{ [\text{Due north}{ \text{H} }_{ 3 }] }[/latex]

[latex]{ \text{Chiliad} }_{ \text{a} }\times { \text{K} }_{ \text{b} }=[{ \text{H} }_{ iii }{ \text{O} }^{ + }][{ \text{OH} }^{ - }]={ \text{One thousand} }_{ \text{w} }[/latex]

[latex]{ \text{log}(\text{K} }_{ \text{a} })+{ \text{log}(\text{K} }_{ \text{b} })=\text{log}({ \text{K} }_{ \text{west} })[/latex]

[latex]{ \text{pK} }_{ \text{a} }+{ \text{pK} }_{ \text{b} }=\text{p}{ \text{K} }_{ \text{w} }=fourteen.00[/latex]

The pH tin exist calculated using the formula:

[latex]{ \text{pH}={xiv}-\text{pOH} }[/latex]

Weak bases exist in chemical equilibrium much in the aforementioned style equally weak acids do. A base dissociation abiding (Thou_b) indicates the force of the base. For example, when ammonia is put in water, the following equilibrium is set up:

[latex]\text{N}{ \text{H} }_{ 3 }+{ \text{H}_2\text{O}}\rightleftharpoons \text{N}{ \text{H} }_{ 4 }^{ + } +{\text{OH}^-}[/latex]

[latex]{ \text{K} }_{ \text{b} }=\frac { [\text{Northward}{ \text{H} }_{ iv }^{ + }][{ \text{OH} }^{ - }] }{ [\text{Northward}{ \text{H} }_{ 3 }] }[/latex]

Bases that take a large K_b will ionize more completely, pregnant they are stronger bases. NaOH (sodium hydroxide) is a stronger base of operations than (CH_threeCH₂)₂NH (diethylamine) which is a stronger base than NH_iii (ammonia). As the bases get weaker, the K_b values get smaller.

Example:

Calculate the pH of a buffer solution consisting of 0.051 M NH_iii and 0.037 1000 NH₄ ⁺. The K_b for NH_three = 1.eight x ten^-5.

[latex]\text{N}{ \text{H} }_{ iii }+{ \text{H}_2\text{O}}\rightleftharpoons \text{N}{ \text{H} }_{ 4 }^{ + } +{\text{OH}^-}[/latex]

[latex]{ \text{K} }_{ \text{b} }=\frac { [\text{N}{ \text{H} }_{ 4 }^{ + }][{ \text{OH} }^{ - }] }{ [\text{North}{ \text{H} }_{ 3 }] }[/latex]

Assuming the change (x) is negligible to 0.051 M and 0.037 M solutions:

[latex]{ \text{K} }_{ \text{b} }=\frac { [0.037][\text{x}] }{ [0.051] }[/latex]

1.eight 10 10^-five[latex]=\frac { [0.037][\text{x}] }{ [0.051] }[/latex]

x = [OH^–] = two.48 x 10^-five

pOH = 4.61

pH = xiv – four.61 = 9.39

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Source: https://courses.lumenlearning.com/boundless-chemistry/chapter/buffer-solutions/

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