Video Transcript : Ka And Ph Shortcut For Weak Acids In Mcat Chemistry
Leah here from leah4sci.com/mcat and in this video we’ll take a look at the calculations for weak acids as it’ll show up on your MCAT. You can find this entire series along with the practice quiz and cheat sheet by visiting my website leah4sci.com/MCATAcidBase.
Now if we’re plugging values into the ka and we have the equation as follows. ka is equal to the H+ concentration of x, the A minus concentration of x, so that’s x times x divided by molarity initial minus x. But we set that x value as negligible, what we’re really looking at is x squared over molarity initial. And this is why you don’t need an ICE chart on the MCAT. Because if you find yourself trying to calculate something like the pH, you’re trying to find the H+ concentration on equilibrium which is X. Simply setup your equation so that ka is equal to x squared over molarity initial of your initial acid, solve for X to get your H+ concentration and then your negative log of the H+ concentration. Estimate your answer and you’ll have your pH. Let’s try a quick example:
Be sure to join me in the next video where we look at how to do calculations for weak bases. And you can find that along with the mentioned links, the rest of this series, practice quiz and cheat sheet by visiting my website leah4sci.com/MCATAcidBase.
How Is A Mole Defined
A mole is defined as 6.02214076 × 1023 of some chemical unit, be it atoms, molecules, ions, or others. The mole is a convenient unit to use because of the great number of atoms, molecules, or others in any substance. The mole was originally defined as the number of atoms in 12 grams of carbon-12, but in 2018 the General Conference on Weights and Measures announced that effective May 20, 2019, the mole would be just 6.02214076 × 1023 of some chemical unit.
Solubility Product Constant Table
Below is a chart showing the $K_s_p$ values for many common substances. The $K_s_p$ values are for when the substances are around 25 degrees Celsius, which is standard. Because the $K_s_p$ values are so small, there may be minor differences in their values depending on which source you use. The data in this chart comes from the University of Rhode Islands Department of Chemistry.
How Is A Mole Calculated
If you want to know how many moles of a material you have, divide the mass of the material by its molar mass. The molar mass of a substance is the mass in grams of one mole of that substance. This mass is given by the atomic weight of the chemical unit that makes up that substance in atomic mass units . For example, silver has an atomic weight of 107.8682 amu, so one mole of silver has a mass of 107.8682 grams.
The mole designates an extremely large number of units, 6.02214076 × 1023. The General Conference on Weights and Measures defined the mole as this number for the International System of Units effective from May 20, 2019. The mole was previously defined as the number of atoms determined experimentally to be found in 12 grams of carbon-12. The number of units in a mole also bears the name Avogadros number, or Avogadros constant, in honour of the Italian physicist Amedeo Avogadro . Avogadro proposed that equal volumes of gases under the same conditions contain the same number of molecules, a hypothesis that proved useful in determining atomic and molecular weights and which led to the concept of the mole.
Acidic Basic And Neutral Solution
In pure water the concentration of hydronium ion and hydroxide ion are same and equal to 10-7 M at 250 C. This types of solution is known as neutral solution. But depending on the difference between their concentration, the solution is named as acidic or basic. Such as
- If = , it is a neutral solution.
- If > , it is an acidic solution.
- If < , it is a basic solution.
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What Is $k: S: P$
$K_s_p$ is known as the solubility constant or solubility product. Its the equilibrium constant used for equations when a solid substance is dissolving in a liquid/aqueous solution. As a reminder, a solute is considered soluble if more than 1 gram of it can be completely dissolved in 100 ml of water.
$K_s_p$ is used for solutes that are only slightly soluble and dont completely dissolve in solution. $K_s_p$ represents how much of the solute will dissolve in solution.
The value of $K_s_p$ varies depending on the solute. The more soluble a substance is, the higher its $K_s_p$ chemistry value. And what are the $K_s_p$ units? Actually, it doesnt have a unit! The $K_s_p$ value does not have any units because the molar concentrations of the reactants and products are different for each equation. This would mean the $K_s_p$ unit would be different for every problem and would be difficult to solve, so in order to make it simpler, chemists generally drop $K_s_p$ units altogether. How nice of them!
A Refresher On The Acid Dissociation Constant Ka
We can take any acid and label it HA where H is the proton associated with the conjugate base, A.
In solution, HA will dissociate into a free proton, H+, and free conjugate base, A. Using the squared bracket notation mentioned already, we can write:
HA -> H+ + A
And given that stoichiometry must be obeyed, using the squared bracket notation described already, we can write:
When the equilibrium is achieved, the concentrations of products and reactants are no longer changing. So the ratio of the products to reactants is also not changing.
The product:reactant ratio is constant. This is the acid dissociation constant, Ka.
Ka = /
The larger the value of Ka, the higher the product:reactant ratio.
Thats to say, the higher the value of Ka, the more free protons there are in the solution.
This, by definition, means the larger the value of Ka, the stronger acid we are dealing with.
Rememberstronger acids have larger Ka values than weaker acids.
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International Union Of Pure And Applied Chemistry
The International Union of Pure and Applied Chemistry is the international body that represents chemistry and related sciences and technologies. Its goal is to unite a fragmented, global chemistry community for the advancement of the chemical sciences via collaboration and the free exchange of scientific information.
The International Union of Pure and Applied Chemistry , established in 1919, was formed by chemists from both industry and academia. These chemists recognized a need for international standardization of weights, measures, names, and symbols in chemistry. Its predecessor body, the International Association of Chemical Societies , had met in Paris in 1911 and established the foundation of the standardization objectives that IUPAC would later center its work around.
Relation Between Pka And Pkb
Let us consider a weak acid HA which ionizes in the aqueous solution as:
HA + H2O H3O+ + A
The dissociation constant of acid is defined by
Ka = / ..
The conjugate base A- behaves as a weak base in water
A + H2O HA + OH
For base Kb = / .
Multiply equation and
Ka x Kb = x = = Kw
Ka x Kb = Kw
On taking negative logarithm on both side
log Ka log Kb = log Kw
pKa + pKb = pKw
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To Find The Solubility Of Solutes
Wondering how to calculate molar solubility from $K_s_p$? Knowing the value of $K_s_p$ allows you to find the solubility of different solutes. Heres an example: The $K_s_p$ value of $Ag_2SO_4$ ,silver sulfate, is 1.4×$10^^5$. Determine the molar solubility.
First, we need to write out the dissociation equation: $K_s_p$=$ ^2$ $$
Next, we plug in the $K_s_p$ value to create an algebraic expression.
1.4×$10^^5$= $^2$ $$
$2x$= =3.0x$10^^2$ M
Pka And Buffer Capacity
Another important point is the relationship between pH and the pKa of an acid. This relationship is described by the following equation.
This equation can be rearranged as follows.
This shows how pKa and pH are equal when exactly half of the acid has dissociated . If the pH changes by 1 near the pKa value, the dissociation status of the acid changes by an extremely large amount.
Fig. Relationship Between pH of Solution and Dissociation Status of Acetic Acid
In the case of acetic acid, for example, if the solution’s pH changes near 4.8, it causes a large change in the presence of acetic acid. When the pH is 3.8, over 90 % exist as acetic acid molecules , but at a pH of 5.8, over 90 % exist as acetate ions .Conversely, to change the pH level near the pKa value of an acid, the dissociation status of the acid must be changed significantly, which requires using an extremely large amount of acid or base. The ability of a substance to maintain the pH of such solutions is referred to as its buffer capacity, where the closer the pKa and pH are, the higher the buffer capacity. Therefore, when selecting a buffer solution, which are used widely in liquid chromatography, an acid or base buffer solution with a pKa value close to the target pH level is selected to maximize this buffer capacity.
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Kw Increases With Increase Of Temperature
Autoionisation of water is an endothermic process. According to Le chateliers principle, if conditions are changed in a equilibrium process, the equilibrium will shift to such a direction where it can minimize the effect of the change of the condition. Thus if water is heated the equilibrium will shift to right to form more ions by absorbing extra heat as this is an endothermic process. According to the equation of Kw, if the concentration of ions increases the Kw increases. So we can say that Kw increases with the increase of temperature.
Understanding Ka And Pka
Ka, pKa, Kb, and pKb are most helpful when predicting whether a species will donate or accept protons at a specific pH value. They describe the degree of ionization of an acid or base and are true indicators of acid or base strength because adding water to a solution will not change the equilibrium constant. Ka and pKa relate to acids, while Kb and pKb deal with bases. Like pH and pOH, these values also account for hydrogen ion or proton concentration or hydroxide ion concentration .
Ka and Kb are related to each other through the ion constant for water, Kw:
- Kw = Ka x Kb
Ka is the acid dissociation constant. pKa is simply the -log of this constant. Similarly, Kb is the base dissociation constant, while pKb is the -log of the constant. The acid and base dissociation constants are usually expressed in terms of moles per liter . Acids and bases dissociate according to general equations:
- HA + H2O â A- + H3O+
In the formulas, A stands for acid and B for base.
- Ka = /
- pKa = – log Ka
- at half the equivalence point, pH = pKa = -log Ka
A large Ka value indicates a strong acid because it means the acid is largely dissociated into its ions. A large Ka value also means the formation of products in the reaction is favored. A small Ka value means little of the acid dissociates, so you have a weak acid. The Ka value for most weak acids ranges from 10-2 to 10-14.
Predicting The Strength Of Bases
So far, we have talked about the strength of the acid. However, it is also very important to understand the strength of the base. Lets look at the dissociation of ethanol and ethyl amine one more time:
We have determined that ethanol is a stronger acid and another way to look at this is to say that the equilibrium is shifted more on the right side compared to the dissociation of the ethyl amine. This, in turn, means that the ethoxide ion is not as reactive as the EtNH ion because otherwise the equilibrium would have been shifted to the left as much as it is for the amine. So, the ethoxide is more stable, less reactive/active than the EtNH ion. All of this, in one word, means that it is a weaker base.
This correlation is true for any acid and conjugate base pair:
The stronger the acid, the weaker its conjugate base.
And vice versa:
The weaker the acid, the stronger its conjugate base.
What Is Ka In Chemistry
The acid dissociation constant is used to differentiate between strong and weak acids. The acid dissociates more as the Ka increases. Strong acids must therefore dissociate more in water. A weak acid, on the other hand, is less likely to ionise and release a hydrogen ion, leading to a less acidic solution.
The acid dissociation constant, denoted by Ka, is the equilibrium constant of an acids dissociation reaction. This equilibrium constant is a numerical representation of an acids strength in a solution. Ka is often stated in mol/L units.
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Pka And Dissociation Equilibrium
Acids include strong acids, which completely dissociate in water, and weak acids, which only partially dissociate. When an acid dissociates, it releases a proton to make the solution acidic, but weak acids have both a dissociated state and undissociated state that coexist according to the following dissociation equilibrium equation.
The concentration ratio of both sides is constant given fixed analytical conditions and is referred to as the acid dissociation constant . Ka is defined by the following equation.
The square brackets indicate the concentration of respective components. Based on this equation, Ka expresses how easily the acid releases a proton . In addition, the equation shows how the dissociation state of weak acids vary according to the level in the solution.Carboxylic acids , such as acetic and lactic acids, normally have a Ka constant of about 10-3 to 10-6. Consequently, expressing acidity in terms of the Ka constant alone can be inconvenient and not very intuitive.Therefore, pKa was introduced as an index to express the acidity of weak acids, where pKa is defined as follows.
For example, the Ka constant for acetic acid is 0.0000158 , but the pKa constant is 4.8, which is a simpler expression. In addition, the smaller the pKa value, the stronger the acid. For example, the pKa value of lactic acid is about 3.8, so that means lactic acid is a stronger acid than acetic acid.
Writing $k: S: P$ Expressions
Below is the solubility product equation which is followed by four $K_s_p$ chemistry problems so you can see how to write out $K_s_p$ expressions.
For the reaction $A_aB_b$ $aA^b^$ + $bB^a^$
The solubility expression is $K_s_p$= $^a$ $^b$
The first equation is known as a dissociation equation, and the second is the balanced $K_s_p$ expression.
For these equations:
- A and B represent different ions and solids. In these equations, they are also referred to as “products”.
- a and b represent coefficients used to balance the equation
- and indicate which state the product is in
- Brackets stand for molar concentration. So represents the molar concentration of AgCl.
In order to write $K_s_p$ expressions correctly, you need to have a good knowledge of chemical names, polyatomic ions, and the charges associated with each ion. Also, the key thing to be aware of with these equations is that each concentration is raised to the power of its coefficient in the balanced $K_s_p$ expression.
Lets look at a few examples.
$PbBr_2$ $Pb^2^$ + $2Br^$
$K_s_p$= $$ $^2$
In this problem, dont forget to square the Br in the $K_s_p$ equation. You do this because of the coefficient 2 in the dissociation equation.
CuS $Cu^$ + S¯
$Ag_2CrO_4$ 2$Ag^$ + $CrO_4^2^$
$Cu_3$ $^2$ $3Cu^2^$ + $2PO_4^3^$
$K_s_p$ = $^3$ $^2$
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How To Calculate Kb From Ka
The Bronsted Lowry definition of an acid and a base is that an acid donates hydrogen ions, whereas a base receives the hydrogen ions. The Kb is the base dissociation constant, or the way in which the ions that compose the base separate into their positive and negative components. The Ka is the acid dissociation constant. The larger the value of Kb, the stronger the base, and the larger the value of Ka, the stronger the acid. By multiplying Ka by Kb, you receive the Kw, or the dissociation constant for water, which is 1.0 x 10^-14. When finding the Kb from the Ka, it is necessary to connect these various parts of the equation.
Read the problem, and write down the information that is given. In a problem that involves calculating the Kb from the Ka, you are usually given the Ka and the Kw. For example, you may be asked to calculate the Kb of the chloride ion. The given Ka of the conjugate acid of the chloride ion, which is hydrogen chloride, is 1.0 x 10^6. The given Kw is 1.0 x 10^-14.
Write down the equation for the the Ka, the Kb, and the Kw, which is Kw = . Solve the equation for Kb by dividing the Kw by the Ka. You then obtain the equation Kb = Kw / Ka.
Put the values from the problem into the equation. For example, for the chloride ion, Kb = 1.0 x 10^-14 / 1.0 x 10^6. The Kb is 1.0×10^-20.
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