Chemistry Review - Acids Bases & Ph Buffers Mastering Biology
Acids and Bases Definitions
This AP® Chemistry crash course review will go over the fundamentals of acids and bases in chemistry and their applications to the AP® Chemistry exam. There are three definitions of what constitutes an acid and a base. Each of these definitions is useful for different purposes.
1. Arrhenius Definition: Acids are compounds that increase the concentration of protons \left( { H }^{ + } \correct) in an aqueous solution. Bases are compounds that increase the concentration of hydroxide ions \left( { OH }^{ - } \right) in an aqueous solution. When acids and bases react, they undergo a neutralization reaction, and the products are a salt and h2o.
2. Brønsted-Lowry Definition: Acids are compounds that donate hydrogen atoms \left( { H }^{ + } \right), and bases are compounds that accept hydrogen atoms. Acids donate their hydrogen and get conjugate bases, and bases react with protons to form conjugate acids. Acids and bases that dissociate completely are chosen strong acids and bases, and those that simply partially dissociate are called weak acids and bases and grade equilibrium mixtures in solution.
iii. Lewis Definition: Acids are compounds that can have an incoming electron pair, and bases are compounds that donate electron pairs to other compounds.
The three definitions of acids or bases are more generalized as we go from top to bottom. The Arrhenius definition only encompasses compounds that increment the concentration of hydrogen ion and hydroxide moieties in an aqueous solution. The Brønsted-Lowry definition encompasses all compounds that are capable of donating hydrogen atoms to 1 another, regardless of the solvent they reside in; this definition encompasses those compounds that are classified as acids or bases in the Arrhenius definition, such as acerb acid. The Lewis takes away the necessity for hydrogen ions and instead defines acids in bases in terms of electron pairs; this applies to whatsoever compounds that are acids and bases under either the Arrhenius or the Brønsted-Lowry definition. The primary signal to remember is that all three definitions of acids and bases are internally consistent with each other.
The get-go topic that must be covered in any discussion about acids and bases is how we measure a solutions' acerbity. The acidity of a solution is most commonly measured on the pH scale. The pH of a solution is a measure of the concentrated of hydrogen ions \left( { H }^{ + } \correct), or protons) in solution. To be exact, the pH of a solution is the negative log of the concentration of protons in solution.
pH=-log { { [H }^{ + }] }
The pH scale ranges from 0 to 14 for reasons we will discuss below. Earlier nosotros get there, some other useful scale to know is the pOH scale. The pOH of a solution is a measure of the concentration of hydroxide ions \left( { OH }^{ - } \right) in solution, too in the negative log form.
pOH=-log { { [OH }^{ - }] }
The pH and pOH scales are related to each other due to the equilibrium that exists between hydrogen ions and hydroxide in solution. For a given solution, the sum of the pH and pOH of the solution must always equal 14.
pH+pOH=fourteen
To understand why this human relationship must be true, we have to review in more detail the term acid and base as defined in the Brønsted-Lowry definition. The Brønsted-Lowry definition considers acids to exist any protonated compound capable of donating its protons; these species are represented by the figureAH, where "A" is the central compound and "H" is the donated proton. A base is defined as whatsoever species that can accept a donated proton. The deprotonated base is represented by the symbol "B". When a protonated acrid comes in contact with a base, the acid donates the proton to the base, thus protonating the base and itself become deprotonated. This tin be represented by the chemical expression below:
AH + B \leftrightarrow A^-+ BH^+
And the equilibrium expression for the reaction would exist:
G = \dfrac{ { [AH] }{ [B] } }{ { [A] }^{ - }{ [BH] }^{ + } }
1 of the nigh important consequences of the Brønsted-Lowry definition is that a deprotonated acid can regain its donated proton, and a protonated base tin can become deprotonated again. We call the deprotonated acid species the cohabit base considering of its ability to function as a proton acceptor (Brønsted-Lowry base). The protonated base then becomes a conjugate acid because of its power to donate protons. The 2 species be in an equilibrium mixture represented by the expression above. The only exceptions to this rule are the seven so-chosen "strong" acids and 8 strong bases whose conjugate acids and bases are so weak that their dissociation in h2o is not considered to exist reversible.
How does this tie into the pH of a solution? Well, in aqueous media, the hydrogen ion is also known as hydronium since in reality protons dissolved in water form weak bonds to the solvent water molecules. This relationship is represented by the molecule known as the hydronium ion, { H }_{ 3 }{ O }^{ + }. The Brønsted-Lowry definition enumerates that water molecules are anamphoteric species, meaning that they can act both as an acid or a base. The reaction that illustrates this is the grabbing of a proton from one water molecule by some other water molecule, resulting in the formation of a hydronium { H }_{ 3 }{ O }^{ + } and hydroxide\left( { OH }^{ - } \right) ion:
{ 2H }_{ 2 }O\leftrightarrow { OH }^{ - }+{ H }_{ 3 }{ O }^{ + }
And the reaction is represented by the equilibrium expression beneath, where w stands for water.
{ Thou }_{ due west }=\left[ { OH }^{ - } \right] \left[ { H }_{ 3 }{ O }^{ + } \right] ={ 1.0 \times 10 }^{ -14 } at 25 ^{\circ} \text{C}
Hydronium is, therefore, the conjugate acid of water, and hydroxide the conjugate base, and both species are in equilibrium with each other. The concentration of water in water does not appear as function of the equilibrium expression, and then what nosotros are left with are the concentrations of the two product molecules { OH }^{ - } and { H }_{ 3 }{ O }^{ ii }. Considering the dissociation constant (K_{ westward }) of h2o is always equal to { one.0 \times x }^{ -fourteen }, the addition of the pH and pOH of a solution must always equal the negative log of that value, which is 14. That is how we explicate the relationship between pH and pOH in solution.
Example one
What is the concentration of hydronium ion in a 0.500 \text{ 50} aqueous solution of 1 \text{ M} sodium hydroxide (NaOH)?
Sodium hydroxide is one of the potent bases and therefore dissociated fully in aqueous solution.
NaOH\rightarrow { Na }^{ + }+{ OH }^{ - }
To find the concentration of hydronium ion, we will need to take advantage of the relationship between pH and pOH of a solution. We brainstorm by computing the molar amount of sodium hydroxide in the aqueous solution using the unit cancellation method:
\dfrac{1 \text{ mol } OH^-}{1 \text{ L}} \times 0.500 \text{ Fifty} = 0.500 \text{ mol } OH^-
We will and so calculate the pOH of the solution.
pOH=-log\left[ { OH }^{ - } \right] = 0.3010
Note that the concluding number of meaning figures has changed. This is because when nosotros take the log of a number, the number of significant figures represents the number of decimal places that must be shown in the logarithm. The contrary is true when going the other direction: the number of decimal places in the log equals the number of significant figures in the number.
We then rearrange the human relationship between pH and pOH so solve for the pH of the solution.
pH+pOH=fourteen
pH=14-pOH=14-(0.301)=13.699
Finally, we rearrange the equation for pH to solve for the hydronium ion concentration.
pH=-log\left[ { H }_{ iii }{ O }^{ + } \right]
\left[ { H }_{ three }{ O }^{ + } \right] ={ 10 }^{ -pH }={ 10 }^{ -(xiii.699) }={ ii.00 \times 10 }^{ -14 } \text{ mol }{ H }_{ three }{ O }^{ + }
Qualitatively, this answer makes sense because nosotros expect to have an extremely low concentration of hydronium ions in a basic solution of sodium hydroxide.
One feature of conjugate acids and bases is that they are always weaker than their original counterparts. What does this mean? An acid has a certain power to donate its proton based on the stability of the fundamental acidic species A, and a base has a certain ability to have protons based on its ability to stabilize the incoming positive charge. Compounds with pi-systems, such every bit carboxylic acids and aromatic systems, are generally stronger acids and bases due to their ability to delocalize the electric charge on the molecule.
A Brønsted acrid forms an equilibrium mixture when dissolved in h2o. The water molecule acts as a base, accepting the donated proton from the acrid, and becomes the conjugate acid hydronium ion. The acid turns into its deprotonated conjugate base.
HA+{ H }_{ 2 }O\leftrightarrow { H }_{ 3 }{ O }^{ + }+{ A }^{ - }
In this style, Brønsted-Lowry acids are consistent with the Arrhenius definition of an acrid considering they increase the concentration of { H }_{ 3 }{ O }^{ + } in an aqueous solution. The equilibrium expression for this reaction is called the acid dissociation constant and is designated \left( { K }_{ a } \right). The larger the value of { 1000 }_{ a }, the stronger the acid is. Most weak acids have a dissociation constant in the range of ten^{-2} to 10^{-14}.
{ K }_{ a }=\dfrac { { [{ H }_{ iii }O^{ + }] }{ [{ A }^{ - }] } }{ { [HA] } }
The acid dissociation constant is a measure of how fully an acid dissociated upon solvation in water. It is a useful measure for the "forcefulness" of an acid, i.due east., its ability to stabilize its conjugate base. The acid dissociation constant is often reported as in the negative logarithmic form \left( { pK }_{ a } \right), similar to how pH is reported. The smaller the value of \left( { pK }_{ a } \right), the stronger the acid is. Most weak acids have a \left( { pK }_{ a } \correct) betwixt 2 and 14.
{ pK }_{ a } =-log\left( { K }_{ a } \correct)
A Brønsted base forms the aforementioned equilibrium mixture when dissolved in water, except that hither water acts as an acid, donating a proton to becomes the conjugate base hydroxide. This is consistent with the Arrhenius definition of a base of operations, i.e., increasing the concentration of hydroxide ions in aqueous solution. The chemic reaction is expressed as:
B+{ H }_{ ii }O\leftrightarrow { OH }^{ - }+{ BH }^{ + }
The equilibrium expression for this reaction is chosen the Base Dissociation Abiding and is designated { M }_{ b } . It is a measure of how strongly a base of operations grabs upwardly protons in solution and a good judge of the strength of a base of operations.The larger the value of { K }_{ b } is, the stronger the base is. Base dissociation constants are also reported as the negative log of { K }_{ b }, and this value is called the { pK }_{ b }. The smaller the value of { pK }_{ b } is, the stronger the base of operations is.
{ K }_{ b }=\dfrac { [{ OH }^{ - }][{ BH }^{ + }] }{ [B] }
Example 2
A given monoprotic acrid is dissolved in one liter of water. The acrid has a dissociation abiding equal to two \times 10^{-five}. Calculate the pH of the solution at equilibrium if 0.100 \text{ moles} of the acid are dissolved.
Solving this problem requires us to utilise Ice tables. For those of you who don't remember, Water ice stands for Initial [Status], Change, and End[ing Status]. ICE tables are a common style to solve for equilibrium problems. We write out the balanced chemical equation at the elevation of the table, then write the initial concentrations of all reagents, the change in their concentration, and the final concentration. The alter is usually represented by the variable ten. For the dissociation of an acid in water:
| Equation | HA + | H_2O \rightleftharpoons | H_3O^+ + | A^{-} |
| Initial | 0.100 \text{ mol} | 1 \times 10^{-7} | 0 | |
| Change | -10 | +ten | +ten | |
| Terminate | 0.100 - x | one \times 10^{-7} + ten | x |
Delight annotation that the equilibrium constant of pure h2o with hydroxide and hydronium is equal to 1 \times x^{-14}, and so the starting concentration of hydronium (and hydroxide) ions in a solution of pure h2o is ever equal to 1 \times 10^{-7}.
At this indicate, we substitute the terminate values into the equilibrium expression for acrid dissociation:
{ K }_{ a }=\dfrac { { (1cdot 10^{ -7 }-x) }{ (10) } }{ { (0.100-x) } }
There is a simplifying assumption that we can make if nosotros satisfy sure criteria. In order to avoid calculations involving the quadratic equation, which are both time consuming and slow, we tin make every bit elementary calculation to decide if the value of x in the denominator can exist removed. For the simplifying assumption to be true, the initial concentration of the acid divided past the dissociation constant of the acid must exist greater than 500. In this example:
\dfrac{0.100}{{ 2 \times x }^{ -five }}=2{,}500
2{,}500 > > 500
We can, therefore, make the simplifying assumption that x in the denominator is negligible. The expression tin can be further simplified past bold that the value of x is much greater than the starting concentration of hydronium ion. The terminal concentration of hydronium in solution can and then be simplified simply to x. The equilibrium expression then becomes:
{ Thou }_{ a }=\dfrac { { ({ x }^{ ii }) } }{ { (0.100) } }
Substituting in the value for { K }_{ a } allows us to solve for x.
{ K }_{ a }={ 2 \times 10 }^{ -5 }=\dfrac{{ x }^{ 2 }}{0.100}
x={ \left( { 2 \times ten }^{ -half dozen } \right) }^{ \dfrac { 1 }{ ii } }={ ane.41 \times 10 }^{ -3 } \text{ mol}
At this signal, we go back and check that our supposition that the value of ten is much greater than the starting concentration of hydronium ion by dividing our value past the starting concentration. If the number is greater than 500, then our assumption was practiced; if not, the assumption is incorrect, and we must go back and consider the hydronium ions initially present in the solution. In this example:
\dfrac{1.4{ ane \times 10 }^{ -3 }}{{ 1 \times x }^{ -7 }} \approx xiv{,}000
14{,}000 > > 500
So our assumption stands. Now to solve the final slice of the puzzle, we substitute the value of ten for the concentration of hydronium ions in the pH equation, and make it at the pH of the acidic solution:
pH=-log({ 1.41 \times 10 }^{ -3 })=two.851
The final equation we will learn relating to acids and bases is the Henderson-Hasselbalch equation, which directly relates the pH of a solution to the { pK }_{ a } of an acid. This equation is useful for computing the fraction of protonated and deprotonated species of a given acid in a solution of a given pH.
pH=p{ Thou }_{ a }+log { \dfrac { [{ H }^{ - }] }{ [HA] } }
It makes sense as the concentration of hydronium ions rise in a solution, the acid is less and less probable to become deprotonated by the solvent, and therefore a greater fraction of the acid will exist in the protonated form. The same is true for the reverse. This adding tin also be performed using the base dissociation constant together with the pOH of a solution.
Finally, permit u.s.a. note that the production of the base and acid dissociation constants of a given chemical compound and its conjugate pair is ever equal to { K }_{ due west } due to the equilibrium that exists between hydronium and hydroxide in aqueous solution.
{ One thousand }_{ w }={ Thou }_{ a } \times { One thousand }_{ b }={ 1 \times 10 }^{ -xiv }
And considering of this human relationship, the sum of the { pK }_{ a } and { pK }_{ b } of an acid or base of operations and its conjugate pair is always equal to 14.
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