Objective 1: Write the equation for an acid/base reaction (either ionization or neutralization) identifying the Brønsted – Lowry acids or bases present.

Bronsted Lowry acids and bases

Bronsted-Lowry acids are proton (H⁺) donors. They donate or give up H⁺ in a chemical reaction. Bronsted-Lowry bases are proton (H⁺) acceptors. They accept or gain H⁺ in a chemical reaction.

Objective 1 Example 1

Hydrochloric acid (HCl) reacts as an Bronsted-Lowry acid with water.

$\text{HCl(}aq\text{) + }{{\text{H}}_{2}}\text{O (}l\text{) }\to {{\text{H}}_{3}}{{\text{O}}^{+}}(aq)+\text{C}{{\text{l}}^{-}}(aq)$

HCl donates H⁺to water, so Cl^– is left behind (HCl loses one hydrogen and lowers by one unit of charge). Since water accepts H⁺from HCl, becoming H₃O⁺ (water gains one hydrogen and increases by one unit of charge), water is a Bronsted-Lowry base.

Objective 1 Example 2

Ammonia (NH₃) reacts as an Bronsted-Lowry base with water.

$\text{N}{{\text{H}}_{3}}(aq\text{)}+{{\text{H}}_{2}}\text{O}(l\text{)}\rightleftarrows \text{N}{{\text{H}}_{4}}^{+}(aq\text{)}+\text{O}{{\text{H}}^{-}}(aq\text{)}$

NH₃ accepts H⁺from water, becoming NH₄⁺(ammonia gains one hydrogen and increases by one unit of charge). Since water donates H⁺ to NH₃ , becoming OH- (water loses one hydrogen and lowers by one unit of charge), water is a Bronsted-Lowry acid.

If you want more information and examples, you can read about it in Section 14.1 of OpenStax (this reading will also address Objective 2 conjugate acids and bases).

Objective 1 Practice

Identifying Acids and Bases by Chemical Formula

So far, I have told you whether a compount is a Bronsted-Lowry acid or base. I would expect you to be able to identify them by formula, though.

Bronsted-Lowry acids are often ionic compounds with H⁺ as the cation and a nonmetal or polyatomic anion. Their formulas typically start with H. Examples of acids are listed on this periodic table handout made available to all CHEM 152 students and this table of Ka values in OpenStax.

Bronsted-Lowry bases are often ammonia (NH₃) or amines (organic compounds related to ammonia). Examples of bases are listed in the periodic table in the previous link and This link of Kb values in OpenStax.

Objective 4: Recognize the common strong acids

Objective 5: Recognize the common strong bases.

Strong acids

Strong acids dissociate completely in water. Their reaction can be written in one of two ways:

as a simple dissociation: $\text{HI }\!\!~\!\!\text{ (}aq\text{) }\to \text{ }\!\!~\!\!\text{ }{{\text{H}}^{\text{+}}}(aq)\text{+ }\!\!~\!\!\text{ }{{\text{I}}^{\text{-}}}(aq)$
as an ionization reaction with water: $\text{HI }\!\!~\!\!\text{ (}aq\text{) + }\!\!~\!\!\text{ }{{\text{H}}_{\text{2}}}\text{O }\!\!~\!\!\text{ (}l\text{)}\to \text{ }\!\!~\!\!\text{ }{{\text{H}}_{\text{3}}}{{\text{O}}^{\text{+}}}(aq)\text{+ }\!\!~\!\!\text{ }{{\text{I}}^{\text{-}}}(aq)$

In the first form, the water is understood to be there (note the HI is in aqueous solution). In the second form, the HI is shown donating a proton to water in a Bronsted-Lowry equation. These forms are two different ways of writing the same thing, and we can think of the H⁺(aq) (hydrogen ion) and the H₃O ⁺(aq) (hydronium ion) as being two different ways of writing the same thing.

You are expected to recognize the following acids are strong acids: HCl(aq), HBr(aq), HI(aq), HNO₃, H₂SO₄, HClO₃, and HClO₄.

Weak acids

Weak acids dissociate partially (usually only slightly) in water. Their reactions can be written in the same two ways as strong acids. The main difference is that they are equilibrium reactions rather than one way or complete reactions.

$\text{HN}{{\text{O}}_{2}}(aq)\text{ }\!\!~\!\!\text{ }\rightleftarrows \text{ }\!\!~\!\!\text{ }{{\text{H}}^{\text{+}}}(aq)\text{+ }\!\!~\!\!\text{ N}{{\text{O}}_{2}}^{\text{-}}(aq)$

$\text{HN}{{\text{O}}_{2}}(aq)\text{ }\!\!~\!\!\text{ + }\!\!~\!\!\text{ }{{\text{H}}_{\text{2}}}\text{O (}l\text{) }\!\!~\!\!\text{ }\rightleftarrows \text{ }\!\!~\!\!\text{ }{{\text{H}}_{\text{3}}}{{\text{O}}^{\text{+}}}(aq)\text{+ }\!\!~\!\!\text{ N}{{\text{O}}_{2}}^{\text{-}}(aq)$

K_a (acid dissociation constant)

The K_a (acid dissociation constant) is the equilibrium constant for the acid dissociation or ionization. Since it is an equilibrium constant, K_a only applies to weak acids.

An expression for K_a is written as the product concentrations over the reactant concentrations (all raised to the power of the coefficients) just as you wrote all equilibrium constant expressions in CHEM 151. For the weak acid HNO₂, it would be:

${{K}_{a}}=\frac{\left[ {{H}^{+}} \right]\left[ N{{O}_{2}}^{-} \right]}{\left[ HN{{O}_{2}} \right]}$ for the simple dissociation form

${{K}_{a}}=\frac{\left[ {{H}_{3}}{{O}^{+}} \right]\left[ N{{O}_{2}}^{-} \right]}{\left[ HN{{O}_{2}} \right]}$ for the Bronsted – Lowry reaction with water form of the equation.

Note the two expressions are actually the came as H⁺ and H₃O⁺ are equivalent.

Given any weak acid, you should be able to write its ionization reaction (either form is OK) and Ka expression.

Weak acids and their numerical Ka values are listed on this periodic table handout made available to all CHEM 152 students and this table of Ka values in OpenStax.

Strong bases

Strong bases ionize completely in water to produce hydroxide ions. They are best written as simple dissociations rather than reactions with water.

You are expected to recognize these Group I and Group II hydroxides as strong bases:

LiOH (lithium hydroxide)

NaOH (sodium hydroxide)

KOH (potassium hydroxide)

RbOH (rubidium hydroxide)

CsOH (cesium hydroxide)

Ca(OH)₂(calcium hydroxide)

Sr(OH)₂ (strontium hydroxide)

Ba(OH)₂ (barium hydroxide)

Weak bases

Weak bases ionize partially (usually only slightly) in water. Like weak acids, they are equilibrium reactions rather than one way or complete. Their equilibrium constants are called K_bor base dissociation constants.. Unlike weak acids, their ionization reactions must be written as ionizations reactions with water – NOT as dissociations. Note in the reactions they accept a proton from water (act as Bronsted-Lowry bases).

$\text{N}{{\text{H}}_{3}}(aq\text{)}+{{\text{H}}_{2}}\text{O}(l\text{)}\rightleftarrows \text{N}{{\text{H}}_{4}}^{+}(aq\text{)}+\text{O}{{\text{H}}^{-}}(aq\text{)}$

${{K}_{b}}=\frac{\left[ N{{H}_{4}}^{+} \right]\left[ O{{H}^{-}} \right]}{\left[ N{{H}_{3}} \right]}$

Weak bases and their numerical Kb values are listed on this periodic table handout made available to all CHEM 152 students and this link of Kb values in OpenStax.

More reading on these objectives is available here.

Objective 4 and 5 Practice

Objective 2: Given an acid or base, write the conjugate base or acid.

Acid-base reactions, according to the Bronsted-Lowry definition, involve proton (H⁺) transfer. Each acid has a conjugate base, which is what is left after a proton (H⁺) is removed from the acid. Each base has a conjugate acid, which is formed when the base accepts or adds a proton (H⁺).

Earlier, we examined the reaction of the weak acid nitrous acid and water:

$\text{HN}{{\text{O}}_{2}}(aq)\text{ }~\text{ + }~\text{ }{{\text{H}}_{\text{2}}}\text{O (}l\text{) }~\text{ }\rightleftarrows \text{ }~\text{ }{{\text{H}}_{\text{3}}}{{\text{O}}^{\text{+}}}(aq)\text{+ }~\text{ N}{{\text{O}}_{2}}^{-}(aq)$

When nitrous acid donates a proton to water, the nitrite ion NO₂^– is left behind. NO₂^– is the conjugate base of HNO_2.When water (acting as a base), accepts a proton, it becomes H₃O⁺. H₃O⁺ is the conjugate acid of water.

If we look at the reverse reaction in the same way, we see that HNO₂is the conjugate acid of NO₂^–, and that water is the conjugate base of H₃O⁺.

HNO₂and NO₂^– are a conjugate acid-base pair. If HNO₂is the conjugate acid of NO₂^–, then NO₂^– must be the conjugate base of HNO_2. It is similar to the instance in which a man and woman are a married couple: if the man is the woman’s husband, then the woman is the man’s wife.

What is the other conjugate acid/base pair in the reaction?

This video explains the concept of conjugate acids and bases in more detail: click here for the video. You can read more about them in the OpenStax text in this section.

Objective 2 Practice

Consider the reaction: $\text{HC}{{\text{O}}_{3}}^{-}(\text{aq)}+{{\text{H}}_{2}}\text{O}(\text{l)}\rightleftarrows {{\text{H}}_{2}}\text{C}{{\text{O}}_{3}}(\text{aq)}+\text{O}{{\text{H}}^{-}}(\text{aq)}$

For this reaction, fill in the blanks correctly by dragging the words:

Autoionization of water and K_w

An amphoteric substance an act as either an acid or a base. Amphoteric substances typically have both transferable H and an atom with lone pair. Water is an example of an amphoteric substance. Recall we have seen it act as an acid (donating a proton) in some reactions and a base (accepting a proton) in others.

Other amphoteric substances include NH₃ and anions that include a removable hydrogen like HCO₃^–, HSO₄^–, and H₂PO₄^–.

In autoionization, an amphoteric substance undergoes an acid/base reaction with itself. Water undergoes autoionization in this reaction:

${{\text{H}}_{\text{2}}}\text{O (}l\text{) }~\text{ + }~\text{ }{{\text{H}}_{\text{2}}}\text{O (}l\text{) }~\text{ }\rightleftarrows \text{ }~\text{ }{{\text{H}}_{\text{3}}}{{\text{O}}^{\text{+}}}(aq)\text{+ }~\text{ O}{{\text{H}}^{-}}(aq)$ (written as a reaction with water)

or ${{\text{H}}_{\text{2}}}\text{O (}l\text{) }~\text{ }~\text{ }\rightleftarrows \text{ }~\text{ }{{\text{H}}^{\text{+}}}(aq)\text{+ }~\text{ O}{{\text{H}}^{-}}(aq)$ (written as a simple dissociation)

The equilibrium expression for the autoionization of water is ${{\text{K}}_{w}}=[{{\text{H}}_{3}}{{\text{O}}^{_{+}}}][\text{O}{{\text{H}}^{-}}]$

or ${{\text{K}}_{w}}=[{{\text{H}}^{+}}][\text{O}{{\text{H}}^{-}}]$

This special equilibrium constant is referred to as the ion-product constant for water, K_w. At 25°C, K_w = 1.0 × 10^-14. As you may remember (or will see in a bit, this is the basis for the pH scale.

Autoionization is discussed in OpenStax in Section 14.1.

Objective 3: Explain the relationship between the strength of an acid and the strength of its conjugate base.

Objective 10: Interconvert between the K_a of an acid and the K_b of its conjugate base.

Let’s look at the conjugate acid base pair HF and F^–.

The K_a expression for HF is: ${{K}_{a}}=\frac{\left[ {{H}^{+}} \right]\left[ {{F}^{-}} \right]}{\left[ HF \right]}$

The K_b expression for F^– is: ${{K}_{b}}=\frac{\left[ HF \right]\left[ O{{H}^{-}} \right]}{\left[ {{F}^{-}} \right]}$

Multiplying this K_a and K_b,

${{K}_{a}}{{K}_{b}}=\frac{\left[ {{H}^{+}} \right]\left[ {{F}^{-}} \right]}{\left[ HF \right]}\frac{\left[ HF \right]\left[ O{{H}^{-}} \right]}{\left[ {{F}^{-}} \right]}={{K}_{w}}={{10}^{-14}}$

This will be true for all conjugate acid base pairs in water. Let’s look at that a little more closely. For a conjugate acid base pair,

K_aK_b=K_w=10^-14

This is useful because

If we know (or can look up) the Ka of a weak acid, we can calculate the Kb of its conjugate base (or vice versa). The conjugate values are usually not tabulated, so this is useful
The larger the Ka of an acid, the smaller the Kb of its conjugate base (and vice versa). This means the stronger a weak acid is, the weaker its conjugate base is (and vice versa). They are still both weak, though. The conjugate base of a weak acid is still a weak base.

Objective 3 and 10 example

This example uses data from this periodic table handout made available to all CHEM 152 students.

Calculate the K_b of the acetate ion, C₂H₃O₂^–.

${{K}_{a}}{{K}_{b}}={{K}_{w}}={{10}^{-14}}$

${{K}_{b}}=\frac{{{K}_{w}}}{{{K}_{a}}}=\frac{{{10}^{-14}}}{{{K}_{a}}}=\frac{{{10}^{-14}}}{1.8\times {{10}^{-5}}}=5.6\times {{10}^{-10}}$

Similarly, the Kb of the chlorite ion, ClO₂^– can be calculated to be 9.1 x 10^-13. Since the Kb of acetate is larger than the Kb of chlorate, acetate ion is the stronger base (note both conjugate bases are still weak, though).

Objective 3 and 10 practice

Objective 6: Calculate [H+], pH, [OH-], and/or pOH given the value of any one of the variables or the concentration of a strong acid or base.

In work in the field of chemistry, and in everyday life, we normally deal with dilute concentrations of acids and bases (in the hundredths, thousandths, millionths, or even smaller molarities). Because the molarities are so small, it is useful to represent them as p-functions such as pH or pOH, as you learned in CHEM 151 or your first semester course. A good review of pH and pOH can be found in OpenStax Section 14.2. The following equations show the relationships between [H+], pH, [OH-], and pOH.

$pH=-\log \left[ {{H}^{+}} \right]$

$pOH=-\log \left[ O{{H}^{-}} \right]$

$[{{H}^{+}}]={{10}^{-p{{H}^{{}}}}}$

$[O{{H}^{-}}]={{10}^{-pOH}}$

$\left[ O{{H}^{-}} \right]\left[ {{H}^{+}} \right]\ =\ 1\times {{10}^{-14}}$

$pH+pOH=14$

Given these equations, if you know any one of the four( [H+], pH, [OH-], and pOH) you should be able to calculate all of them.

Also, note that the lower the pH, the more acidic the solution, and the higher the pH, the more basic the solution. This is illustrated nicely here.

Objective 6 Practice

Answer the following three questions for a pH 3.20 solution:

Calculating the pH of Strong Acids or Bases

Strong acids and bases ionize completely.

$\text{HI }\!\!~\!\!\text{ (}aq\text{) }\to \text{ }\!\!~\!\!\text{ }{{\text{H}}^{\text{+}}}(aq)\text{+ }\!\!~\!\!\text{ }{{\text{I}}^{\text{-}}}(aq)$

$\text{NaOH }~\text{ }~\text{ }\to \text{ }~\text{ N}{{\text{a}}^{\text{+}}}\text{+ }~\text{ O}{{\text{H}}^{-}}$

$\text{Ca(OH}{{\text{)}}_{2}}~\text{ }~\text{ }\to \text{ }~\text{ C}{{\text{a}}^{2\text{+}}}\text{+ 2}~\text{ O}{{\text{H}}^{-}}$

In the acid dissociation above, notice that the mole ratio of HI to H⁺ is 1:1. Therefore, the [H⁺] = the original molarity of the acid. Bases can be treated similarly (but remember some bases, like Ca(OH)₂, are not 1:1).

This simple approach does not work for weak acids or bases since they do not ionize completely. An ICE table is necessary to determine pH for weak acids or bases. We’ll review that in the next objective (after more practice problems).

More Objective 6 Practice

Objective 7: Calculate all equilibrium concentrations and pH or pOH in a solution of a weak acid or base given the K_aor K_b

In the last objective, we saw that in the case of a strong monoprotic acid or a strong base with one hydroxide, the [H⁺] (for an acid) or the [OH^–] (for a base) was equal to the molarity of the strong acid or base. This is not the case for a weak acid or base as they do not ionize completely.

$\text{HN}{{\text{O}}_{2}}(aq)\text{ }\!\!~\!\!\text{ }\rightleftarrows \text{ }\!\!~\!\!\text{ }{{\text{H}}^{\text{+}}}(aq)\text{+ }\!\!~\!\!\text{ N}{{\text{O}}_{2}}^{\text{-}}(aq)$

$\text{N}{{\text{H}}_{3}}(aq\text{)}+{{\text{H}}_{2}}\text{O}(l\text{)}\rightleftarrows \text{N}{{\text{H}}_{4}}^{+}(aq\text{)}+\text{O}{{\text{H}}^{-}}(aq\text{)}$

For the nitrous acid equation above, [H⁺] will be less than the original acid molarity due to the incomplete dissociation. Similarly for the weak base ammonia, the [OH^–] will be less than the original base molarity. How do you determine the concentrations (and therefore the pH)? By using an ICE (initial, change, equilibrium) table.

An example of this technique is shown in Example 14.12 of Section 14.3 of OpenStax (scroll down to Example 14.12 after opening the link) or in the video below:

The similar problems for bases are demonstrated in Example 14.13 in the same OpenStax section.

Here is a link to another video example for a pH of a weak acid calculation and a weak base calculation from the same video..

Objective 7 Practice

Objective 8: Calculate the K_a/K_b for a weak acid/base given the pH of a solution of known concentration.

Objective 8 is essentially Objective 7 in reverse – instead of calculating the [H⁺] and the pH using the given Ka we calculate the [H⁺] from the given pH then calculate the Ka using the ICE table.

Examples of these calculations are found in Section 14.3 of OpenStax (scroll to examples 14.11 and 14.10). The video below will also illustrate a solution for this example:

The pH of a 0.10 M solution of an unknown weak acid is 2.17. Calculate the Ka of this unknown weak acid.

Objective 8 Practice

Objective 9:Calculate percent ionization of an acid or base from initial concentrations and K_a or K_b values or vice-versa.

Consider a 0.125-M solution of nitrous acid (a weak acid), which has a Ka of 4.6 X 10^-4.

The dissociation equation is: $\text{HN}{{\text{O}}_{2}}(aq)\text{ }\!\!~\!\!\text{ }\rightleftarrows \text{ }\!\!~\!\!\text{ }{{\text{H}}^{\text{+}}}(aq)\text{+ }\!\!~\!\!\text{ N}{{\text{O}}_{2}}^{\text{-}}(aq)$

If you were to do an ICE table calculation to determine its pH (see Objective 7), the table would be:

	HNO₂	↔	H⁺	NO₂^–
I	0.125		0	0
C	-x		+x	+x
E	0.125-x		x	x

Using the expression for Ka and solving for x (see Objective 7 above for how to do this) yields x= 7.5 X 10^-3.

Percent dissociation or percent ionization is defined as the percentage of the original acid or base that ionizes. It can be expressed as:

$\%dissociation=\frac{{{x}_{\text{from the ICE table}}}}{\text{initial molarity}}\times 100$

For the above example, $\%dissociation=\frac{7.5\times {{10}^{-3}}}{0.125}\times 100=6.1\%$ .

Given the percent ionization and initial molarity , you should also be able to solve for the x in the ICE table using the equation

$\%dissociation=\frac{{{x}_{\text{from the ICE table}}}}{\text{initial molarity}}\times 100$ .

From there, you could calculate pH. The following practice problems will let you try these calculations.

You can read more about percent dissociation in OpenStax Section 14.3.

Objective 9 Practice

Objective 11: Write the balanced chemical equations for hydrolysis of a salt then predict whether the salt solution will be acidic, basic or neutral.

Hydrolysis is the reaction of an ion with water. When a salt dissolves in water, it breaks up into a positive cation and a negative anion. One or both of these ions may react with the water, undergoing hydrolysis.

Hydrolysis of cations:

A cation may react with water. If it does, it acts as a Bronsted- Lowry acid, donating a proton to water. So one of two things will happen with the cation in water:

It does not have a proton to donate to water and therefore will not react (will not undergo hydrolysis). There are typically metal cations. For example: Na⁺+ H₂O → no reaction. These cations will be neutral in water.
It does have a proton to donate to water and therefore will react (will undergo hydrolysis). There are typically cations that are conjugate acids of weak bases. For example

$\text{N}{{\text{H}}_{4}}^{+}(\text{aq)}+{{\text{H}}_{2}}\text{O}(\text{l)}\rightleftarrows \text{N}{{\text{H}}_{3}}(\text{aq)}+{{\text{H}}_{3}}{{\text{O}}^{+}}(\text{aq)}$ These cations will be acidic in water.

Hydrolysis of anions:

A anion may react with water. If it does, it acts as a Bronsted- Lowry base, accepting a proton from water. So one of two things will happen with the anion in water:

It will not accept proton from water and therefore will not react (will not undergo hydrolysis). These anions are typically anions of strong acids. For example: Cl-1, the anion of the strong acid HCl, does not react with water. $\text{C}{{\text{l}}^{-}}(\text{aq)}+{{\text{H}}_{2}}\text{O}(\text{l)}\rightleftarrows \text{no reaction}$ . These anions are neutral in water.
It does have a proton to donate to water and therefore will react (will undergo hydrolysis). There are typically anions of weak acids. For example: $\text{C}{{\text{N}}^{-}}(\text{aq)}+{{\text{H}}_{2}}\text{O}(\text{l)}\rightleftarrows \text{HCN}(\text{aq)}+\text{O}{{\text{H}}^{-}}(\text{aq)}$ . These anions are basic in water.

For further reading, see OpenStax Section 14.4. A description is also shown in this video.

Objective 11 Example

Is an aqueous solution of potassium fluoride (KF) acidic, basic, or neutral?

The cation of the salt is K⁺. It does not have a hydrogen to donate to water and will be a neutral cation. K⁺ + H2O → no reaction.
The anion of the salt is F^–. This anion is the anion of a weak acid (HF is a weak acid) so hydrolysis will occur (producing hydroxide): ${{\text{F}}^{-}}(\text{aq)}+{{\text{H}}_{2}}\text{O}(\text{l)}\rightleftarrows \text{HF}(\text{aq)}+\text{O}{{\text{H}}^{-}}(\text{aq)}$ . F^– is a basic anion.

Since K⁺ is neutral and F- is basic, the salt solution will be basic.

Objective 11 Practice

Objective 12: Calculate the pH of salt solutions.

We just saw that in a solution of KF that the K⁺cation is neutral and the F^– anion is basic, the salt solution will be basic. It is basic due to the hydrolysis reaction: ${{\text{F}}^{-}}(\text{aq)}+{{\text{H}}_{2}}\text{O}(\text{l)}\rightleftarrows \text{HF}(\text{aq)}+\text{O}{{\text{H}}^{-}}(\text{aq)}$ .

Note this is an ionization reaction of the weak conjugate base F^–, with the following K_b expression:

${{K}_{b}}=\frac{\left[ HF \right]\left[ O{{H}^{-}} \right]}{\left[ {{F}^{-}} \right]}$

We could calculate the Kb value using the tabulated Ka of HF:

${{K}_{b}}=\frac{{{K}_{w}}}{{{K}_{a}}}=\frac{{{10}^{-14}}}{6.4\times {{10}^{-4}}}=1.6\times {{10}^{-11}}$

Let’s say we wished to calculate the pH of a 0.10 M KF solution. Since KF dissociates completely, the initial F^– molarity is 0.10 M. So now you can do a calculation using an ICE table just like those in Objective 7.

	F^–	H₂O	↔	HF	OH^–
I	0.10			0	0
C	-x			+x	+x
E	0.10-x			x	x

${{K}_{b}}=\frac{[HF][O{{H}^{-}}]}{[{{F}^{-}}]}$

$1.6\times {{10}^{-11}}=\frac{{{x}^{2}}}{0.10-x}$

Solving for x (which is [OH^–]), we get x=[OH^–]=1.3 X 10^-6, which results in a pH of 8.10.

Example 14.15 in Section 14.4of OpenStax shows an example with an acidic salt. Additional examples are available in this video.

Objective 12 Practice

Next Steps

This concludes the review of acids and bases. If you haven’t already, make sure to look at the reading guide and additional resources available on this site and complete the online homework assignment before going to the next topic, Common Ion Effect and Buffer Solutions.

If you have a good understanding of this, you are well set to begin the new material! If not, please do not hesitate to contact me for help!

Objective 1: Write the equation for an acid/base reaction (either ionization or neutralization) identifying the Brønsted – Lowry acids or bases present.

Bronsted Lowry acids and bases

Objective 1 Example 1

Objective 1 Example 2

Objective 1 Practice

Identifying Acids and Bases by Chemical Formula

Objective 4: Recognize the common strong acids

Objective 5: Recognize the common strong bases.

Strong acids

Weak acids

Ka (acid dissociation constant)

Strong bases

Weak bases

Objective 4 and 5 Practice

Objective 2: Given an acid or base, write the conjugate base or acid.

Objective 2 Practice

Autoionization of water and Kw

Objective 3: Explain the relationship between the strength of an acid and the strength of its conjugate base.

Objective 10: Interconvert between the Ka of an acid and the Kb of its conjugate base.

Objective 3 and 10 example

Objective 3 and 10 practice

Objective 6: Calculate [H+], pH, [OH-], and/or pOH given the value of any one of the variables or the concentration of a strong acid or base.

Objective 6 Practice

Calculating the pH of Strong Acids or Bases

More Objective 6 Practice

Objective 7: Calculate all equilibrium concentrations and pH or pOH in a solution of a weak acid or base given the Ka or Kb

Objective 7 Practice

Objective 8: Calculate the Ka/Kb for a weak acid/base given the pH of a solution of known concentration.

Objective 8 Practice

Objective 9:Calculate percent ionization of an acid or base from initial concentrations and Ka or Kb values or vice-versa.

Objective 9 Practice

Objective 11: Write the balanced chemical equations for hydrolysis of a salt then predict whether the salt solution will be acidic, basic or neutral.

Hydrolysis of cations:

Hydrolysis of anions:

Objective 11 Example

Objective 11 Practice

Objective 12: Calculate the pH of salt solutions.

Objective 12 Practice

Next Steps

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K_a (acid dissociation constant)

Autoionization of water and K_w

Objective 10: Interconvert between the K_a of an acid and the K_b of its conjugate base.

Objective 7: Calculate all equilibrium concentrations and pH or pOH in a solution of a weak acid or base given the K_aor K_b

Objective 8: Calculate the K_a/K_b for a weak acid/base given the pH of a solution of known concentration.

Objective 9:Calculate percent ionization of an acid or base from initial concentrations and K_a or K_b values or vice-versa.