All High School Chemistry Resources
Example Questions
Example Question #22 : Acid Base Chemistry
What volume of a 1.2M solution of hydrochloric acid is needed to neutralize 50mL of a 3M sodium hydroxide solution?
The equation to use here is:
Here, is the molarity of the acid, is the volume of the acid, is the molarity of the base, and is the volume of the base. Don't forget to convert the volume to liters!
Example Question #2 : Help With Acid Base Reactions
What kind of reaction is an acid-base neutralization reaction?
Decomposition
Double-replacement
Single-replacement
Addition (synthesis)
Oxidation-reduction
Double-replacement
Below is a generic acid-base neutralization reaction:
The products are always water, and a salt. This salt is produced from the resulting ions and . The from the acid replaces the from the base, and the from the acid replaces the from the base. Since there are two replacements, acid-base neutralizations are classified as double-replacement reactions.
Example Question #2 : Help With Acid Base Reactions
Which of the following is the definition for an Arrhenius acid?
A proton donor
A substance that increases the concentration when added to water
A substance that increases the concentration when added to water.
A proton acceptor
None of these
A substance that increases the concentration when added to water
Arrhenius acid/base theory was created by Swedish chemist Svante Arrhenius, and is the oldest acid/base classification. According to his classification, acids are compounds that increase the concentration of ions in a solution, while bases are compounds or elements that either decrease the concentration of ions in solution or increase the concentration of ions in a solution. The other two answers describe the Brønsted–Lowry theory of acids and bases.
Example Question #1 : Using Acid Dissociation Constant (Ka)
What is the concentration of hydronium ions in a solution if the hydroxide ion concentration is ?
For every acidic or basic solution, the product of the hydroxide ion concentration and the hydronium ion concentration will be equal to , the dissociation constant for water. In other words:
Since we are given the hydroxide ion concentration, we can determine the hydronium ion concentration using this equation.
Example Question #1 : Using Acid Dissociation Constant (Ka)
A student has a 0.50M solution of acetic acid. She adds solid sodium acetate until the concentration of sodium acetate is 0.050M. What is the final pH of the solution?
The for acetic acid is . Assume the volume remains constant.
Acetic acid is a weak acid, and its dissociation reaction is written as:
Using an ICE table, we can find the pH of the solution when the sodium acetate is added:
I: There is initially a 0.50M concentration of acetic acid. Because sodium acetate will dissolve completely, there will be a 0.050M concentration of acetate ions in the solution.
C: As acetic acid dissociates, the concentrations of acetate ions and hydronium ions will increase by an unknown concentration . Conversely, the acetic acid concentration will decrease by the same amount.
E: By setting the equilibrium expression equal to the acid equilibrium constant, we can solve for the value of .
*Because the value for will be so much lower than the initial concentration, we can omit it from the acetate expression as well as the acetic acid concentration:
Keep in mind that is equal to the concentration of hydronium ions now in the solution.
Use this value in the equation for pH:
Example Question #2 : Using Acid Dissociation Constant (Ka)
What property distinguishes a strong acid from a weak acid?
A weak acid will fail to dissociate completely in water and have a high Ka value
There is no difference between a strong acid and a weak acid based on their properties in water
A strong acid will dissociate completely in water and have a high Ka value
A strong acid will dissociate completely in water and have a low Ka value
A weak acid will dissociate completely in water and have a low Ka value
A strong acid will dissociate completely in water and have a high Ka value
The acid dissociation constant (Ka) is used to distinguish strong acids from weak acids. Strong acids have exceptionally high Ka values.
The Ka value is found by looking at the equilibrium constant for the dissociation of the acid. The higher the Ka, the more the acid dissociates. Thus, strong acids must dissociate more in water. In contrast, a weak acid is less likely to ionize and release a hydrogen ion, thus resulting in a less acidic solution.
This equation makes it clear that the more the acid converts from its original form to its ionzied, dissociated form, the higher the Ka value will be.
Example Question #3 : Using Acid Dissociation Constant (Ka)
Which of the following is what determines the strength of an acid?
The Ka
How many bonds the central atom makes
The Kb
Electronegativity values
Its physical state
The Ka
The Ka is the acid dissociation constant, and thus it is what determines how strong the acid is. Stronger acids dissociate to a greater extent and produce lower pH values.
Example Question #2 : Using Acid Dissociation Constant (Ka)
Determine the equilibrium concentration of ions in a solution.
The definition of for is:
Set up an "ICE" table:
Plug in values:
Use the quadratic formula to solve:
Example Question #33 : Acid Base Chemistry
Determine the equilibrium concentration of ions in a solution.
Set up an "ICE" table
Plug in values:
Use the quadratic formula:
Example Question #31 : Acids And Bases
A 1M solution of a monoprotic acid has a pH of 4.6. What is the value for the conjugate base of the acid?
In order to find the base dissociation constant for the conjugate base, we can start by finding the acid dissociation constant for the acid. Since a 1M solution of the acid has a pH of 4.6, we can find the proton concentration of the solution.
Since the acid is monoprotic, we can set the following equilibrium expression equal to its acid dissociation constant.
We can see that, since the acid is monoprotic, the concntration of protons will be equal to the concentration of the acid anion. The final concentration of the acid molecule will be equal to the initial concentration, minus the amount of protons formed. Using these values, we can solve for the equilibrium constant for the acid.
Now that we have the acid dissociation constant, we can find the conjugate base's dissociation constant by setting the product of the two values equal to the autoionization of water.
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