How Do The Results In Data Tables 1 And 2 Support The Role Of A Buffer?

Preparing a Buffer Solution with a Specific pH

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

Learning Objectives

Depict the backdrop of a buffer solution.

Central Takeaways

Key Points

Buffer solutions are resistant to pH change because of the presence of an equilibrium betwixt the acid (HA) and its cohabit base (A-).
When some strong acid 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 strong 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 acid and its conjugate base.

Key Terms

aqueous: Consisting mostly of water.
equilibrium: The state of a reaction in which the rates of the forward (reactant to product) and reverse (product to reactant) reactions are the same.
pKa: A quantitative measure of the force of an acid in solution; a weak acrid has a pKa value in the approximate range −ii to 12 in water and a strong acid has a pKa value of less than nigh −2.

Buffers

A buffer is an aqueous solution containing a weak acid and its conjugate base of operations or a weak base and its conjugate acid. A buffer'south pH changes very little when a minor amount of strong acid or base is added to information technology. Information technology is used to preclude whatever change in the pH of a solution, regardless of solute. Buffer solutions are used as a means of keeping pH at a nearly abiding value in a wide diverseness of chemical applications. For example, blood in the human body is a buffer solution.

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

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

When some strong acid (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 accordance with Le Chatelier'due south principle. This causes the hydrogen ion (H⁺) concentration to increment by less than the amount expected for the quantity of strong acrid added. Similarly, if a stiff base is added to the mixture, the hydrogen ion concentration decreases past less than the amount expected for the quantity of base of operations added. This is because 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 stop working and tin can fall apart. A buffer of carbonic acid (H₂CO₃) and bicarbonate (HCO₃ ⁻) is needed in blood plasma to maintain a pH between vii.35 and 7.45. Industrially, buffer solutions are used in fermentation processes and in setting the correct weather for dyes used in coloring fabrics.

Preparing a Buffer Solution

There are a couple of ways to prepare a buffer solution of a specific pH. In the first method, set up a solution with an acid and its cohabit base past dissolving the acid form of the buffer in about threescore% of the volume of h2o required to obtain the last solution volume. Then, measure the pH of the solution using a pH probe. The pH can be adjusted upward to the desired value using a strong base like NaOH. If the buffer is fabricated with a base of operations and its conjugate acid, the pH tin can be adapted using a stiff acid like HCl. Once the pH is correct, dilute the solution to the terminal desired volume.

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

Alternatively, you can set up solutions of both the acid form and base class of the solution. Both solutions must comprise the same buffer concentration as the concentration of the buffer in the final solution. To get the final buffer, add 1 solution to the other while monitoring the pH.

In a third method, you tin determine the exact amount of acid and conjugate base needed to make a buffer of a certain pH, using the Henderson-Hasselbach equation:

[latex]\text{pH}=\text{p}{ \text{Thousand} }_{ \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 conjugate base and starting acid.

Calculating 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 normally represented as an equilibrium constant.
The acrid-dissociation equilibrium constant, which measures the propensity of an acid to dissociate, is described using the equation: [latex]{ \text{Yard}}_{\text{a} }=\frac { { [\text{H} }^{ + }][{ \text{A} }^{ - }] }{ [\text{HA}] }[/latex].
Using K_a and the equilibrium equation, you can solve for the concentration of [H⁺].
The concentration of [H⁺] tin can then 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 forrad (reactant to product) and opposite (product to reactant) reactions are the same.

What Does pH Mean in a Buffer?

In chemistry, pH is a mensurate of the hydrogen ion (H⁺) concentration in a solution. The pH of a buffer can exist 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 strength of a weak acrid is usually represented as an equilibrium abiding. The acid-dissociation equilibrium constant (G_a), which measures the propensity of an acid to dissociate, for the reaction is:

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

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

ICE Tables: A Useful Tool For Solving Equilibrium Problems

Water 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 case, the following problem:

Calculate the pH of a buffer solution that initially consists of 0.0500 G NH_iii and 0.0350 M NH₄ ⁺. (Note: K_a for NH₄ ⁺ is 5.6 10 ten^-x). The equation for the reaction is as follows:

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

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

Water 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. Considering the reaction has a 1:ane stoichiometry, the amount that NH₄ ⁺ loses is equal to the amounts that H⁺ and NH_iii will gain. This change is represented by the letter x in the following table.

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

Therefore the equilibrium concentrations volition look like this:

Ice tabular array – equilibrium: Describes the final concentration of the reactants and products at equilibrium.

Apply the equilibrium values to the expression for K_a.

[latex]{ 5.half dozen \times 10^{-10}} = \frac { { [\text{H} }^{ + }][{ \text{NH}_3 }] }{ [\text{NH}_4^+] } = \frac { x (0.0500+\text{x})}{ 0.0350-\text{x} }[/latex]

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

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

Solving for x (H⁺):

x = [H⁺] = 3.92 x 10^-10

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

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

Summate the pH of a buffer organization using the Henderson-Hasselbalch equation.

Central Takeaways

Key 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{K} }_{ \text{a} }+\text{log}(\frac { { [\text{A} }^{ - }] }{ [\text{HA}] } )[/latex], where pH is the concentration of [H+], pK_a is the acrid dissociation constant, and [A^–] and [HA] are concentrations of the conjugate base and starting acid.
The equation tin be used to determine the corporeality of acid and cohabit base needed to make a buffer solution of a sure pH.

Fundamental Terms

pKa: A quantitative measure of the forcefulness of an acid in solution; a weak acrid has a pKa value in the estimate range -two to 12 in water and a strong acrid has a pKa value of less than about -ii.

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

After taking the log of the entire equation and rearranging information technology, the result is:

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

This equation tin can exist rewritten as:

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

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

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

In an alternate awarding, the equation can be used to determine the amount of acid and conjugate base 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{Thousand} }_{ \text{a} } }=\frac { [\text{base}] }{ [\text{acrid}] }[/latex]

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

What is the pH of a buffer solution consisting of 0.0350 M NH₃ and 0.0500 G NH₄ ⁺(K_a for NH₄ ⁺ is 5.six x 10^-10)? The equation for the reaction is:

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

Assuming that the modify in concentrations is negligible in order for the system to reach equilibrium, the Henderson-Hasselbalch equation will be:

[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 = 9.095

Calculating Changes in a Buffer Solution

The inverse pH of a buffer solution in response to the addition of an acid or a base of operations can exist calculated.

Learning Objectives

Calculate the final pH of a solution generated by the improver of a strong acid or base to a buffer.

Key Takeaways

Key Points

If the concentrations of the weak acrid and its conjugate base in a buffer solution are reasonably high, so the solution is resistant to changes in hydrogen ion concentration, or pH.
The change in pH of a buffer solution with an added acid or base tin can be calculated past combining the balanced equation for the reaction and the equilibrium acid dissociation abiding (1000_a).
Comparing the final 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 acidity or alkalinity of a substance, which takes numerical values from 0 (maximum acidity) through 7 (neutral) to fourteen (maximum alkalinity).
acid dissociation constant: Quantitative measure out of the strength of an acrid 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 conjugate base of operations are reasonably high, then the solution is resistant to changes in hydrogen ion concentration. These solutions are known as buffers. Information technology is possible to calculate how the pH of the solution will change in response to the improver of an acid or a base to a buffer solution.

Calculating Changes in a Buffer Solution, Example i:

A solution is 0.050 One thousand in acetic acid (HC₂H₃O_ii) and 0.050 M NaC₂H₃O₂. Summate the change in pH when 0.001 mole of muriatic acid (HCl) is added to a liter of solution, assuming that the volume increase upon adding the HCl is negligible. Compare this to the pH if the aforementioned amount of HCl is added to a liter of pure water.

Footstep i:

[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]

Remember that sodium acetate,
NaC₂H₃O_two, dissociates into its component ions, Na⁺ and C₂H₃O_ii ^– (the acetate ion) upon dissolution in h2o. Therefore, the solution will comprise both acerb acrid and acetate ions.

Earlier adding HCl, the acerb 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{10}(0.050) }{ (0.050) }[/latex]

(assuming that x is small compared to 0.050 G in the equilibrium concentrations)

Therefore:

[latex]\text{ten}=[\text{H}^+]={ \text{K} }_{ \text{a} }=1.76\times 1{ 0 }^{ -5 }\text{M}[/latex]

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

In this instance, ignoring the x in the [C_twoH₃O₂ ^–] and [HC₂H_iiiO₂] terms was justified because the value is small-scale compared to 0.050.

Step two:

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

[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{M}[/latex] before the new equilibrium is to be established. Then, we consider the equilibrium conentrations for the dissociation of acerb acid, as in 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]

we have,

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

[latex]10=[\text{H}^+]=(1.76\times one{ 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 acetic acid-acetate buffer arrangement, the pH only drops from 4.75 to 4.74 upon addition of 0.001 mol of strong acrid HCl, a departure of only 0.01 pH unit.

Step three:

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

[latex]\text{pH}=-\text{log}([{ \text{H} }^{ + }])=3.00[/latex]

In the absence of HC₂H_threeO_iiand C_iiH₃O₂ ^– _, the aforementioned concentration of HCl would produce a pH of 3.00.

Calculating Changes in a Buffer Solution, Example ii:

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 1.8 x ten^-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 take been after calculation sodium hydroxide if the buffer concentrations had been 0.10 M instead of 0.010 Grand?

Step i:

Solving for the buffer pH:

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

Bold x is negligible, the K_a expression looks like:

[latex]{ \text{Yard} }_{ \text{a} }=\frac { \text{x}(0.010) }{ (0.010) }[/latex]

1.8 x 10^-iv = 10 = [H⁺]

pH = -log [H⁺] = three.74

Buffer: pH = 3.74

Pace 2:

Solving for the buffer pH afterwards 0.0020 G NaOH has been added:

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

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

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

Afterward calculation NaOH, solving for [latex]\text{x}=[\text{H}^+][/latex] and and so calculating the pH = iii.92. The pH went up from 3.74 to 3.92 upon improver of 0.002 Grand of NaOH.

Step 3:

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

pOH = -log (0.0020)

pOH = 2.lxx

pH = 14 – pOH

pH = xi.30

Without buffer: pH = 11.30

Step four:

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

The concentration of HCOOH would change from 0.1000 Thousand to 0.0980 M and the concentration of HCOO^– would change from 0.1000 M to 0.1020 Chiliad.

[latex]{ \text{1000} }_{ \text{a} }=\frac { \text{10}(0.1020) }{ (0.0980) }[/latex]

pH if 0.g 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 likewise shows the importance of using high buffer component concentrations so that the buffering chapters of the solution is not exceeded.

Buffers Containing a Base and Cohabit Acrid

An alkaline buffer tin be made from a mixture of the base of operations and its cohabit acid, but the formulas for determining pH take a dissimilar grade.

Learning Objectives

Calculate the pH of an alkaline buffer system consisting of a weak base and its conjugate acrid.

Key Takeaways

Key Points

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

Key Terms

alkaline: Having a pH greater than 7.
buffers: A weak acid or base used to maintain the acidity (pH) of a solution almost a chosen 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 because it is the substance that combines with the proton. Ammonia and some organic nitrogen compounds can combine with protons in solution and act equally Brønsted-Lowry bases. These compounds are mostly weaker bases than the hydroxide ion because they take less allure for protons. For instance, when ammonia competes with OH^– for protons in an aqueous solution, it is only partially successful. It can combine with only a portion of the H⁺ ions, and so information technology will have a measurable equilibrium abiding. Reactions with weak bases outcome in a relatively low pH compared to potent bases. Bases range from a pH of greater than 7 (7 is neutral similar pure water) to 14 (though some bases are greater than 14).

An alkaline buffer tin can be fabricated from a mixture of a base of operations and its conjugate acrid, similar to the manner in which weak acids and their conjugate bases can be used to brand a buffer.

Ammonia to ammonium ion: Two-dimensional epitome depicting the association of proton (H⁺) with the weak base ammonia (NH₃) to form its conjugate acid, ammonium ion (NH₄ ⁺).

Calculating the pH of a Base

The pH of bases is unremarkably calculated using the OH^– concentration to observe the pOH commencement. 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 conjugate acid (such as NH_iv ⁺) and a conjugate base (such as NH₃) the following is given:

[latex]{ \text{K} }_{ \text{a} }\times { \text{One thousand} }_{ \text{b} }=\frac { [{ \text{H} }_{ 3 }{ \text{O} }^{ + }][\text{N}{ \text{H} }_{ 3 }] }{ [\text{N}{ \text{H} }_{ 4 }^{ + }] } \times \frac { [\text{Northward}{ \text{H} }_{ four }^{ + }][{ \text{OH} }^{ - }] }{ [\text{North}{ \text{H} }_{ 3 }] }[/latex]

[latex]{ \text{K} }_{ \text{a} }\times { \text{K} }_{ \text{b} }=[{ \text{H} }_{ 3 }{ \text{O} }^{ + }][{ \text{OH} }^{ - }]={ \text{K} }_{ \text{west} }[/latex]

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

[latex]{ \text{pK} }_{ \text{a} }+{ \text{pK} }_{ \text{b} }=\text{p}{ \text{Thousand} }_{ \text{w} }=14.00[/latex]

The pH tin can exist calculated using the formula:

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

Weak bases exist in chemical equilibrium much in the same fashion as weak acids do. A base dissociation constant (K_b) indicates the strength of the base of operations. For example, when ammonia is put in water, the post-obit equilibrium is set upwards:

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

Bases that have a large G_b will ionize more completely, meaning they are stronger bases. NaOH (sodium hydroxide) is a stronger base than (CH_threeCH₂)₂NH (diethylamine) which is a stronger base than NH₃ (ammonia). As the bases get weaker, the K_b values get smaller.

Example:

Calculate the pH of a buffer solution consisting of 0.051 K NH_iii and 0.037 Chiliad NH₄ ⁺. The One thousand_b for NH₃ = 1.8 10 x^-five.

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

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

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

[latex]{ \text{Yard} }_{ \text{b} }=\frac { [0.037][\text{ten}] }{ [0.051] }[/latex]

1.8 x x^-5[latex]=\frac { [0.037][\text{x}] }{ [0.051] }[/latex]

10 = [OH^–] = 2.48 ten 10^-five

pOH = 4.61

pH = xiv – 4.61 = 9.39