Chemistry 30

Unit 6: Acids and Bases

General Info about Acids and BasesAcids:

React with bases

Taste _________________________ (don’t taste in chemistry)

React with certain active metals to produce _________________________

Turn litmus paper _________________________

Bases:

Reacts with acids

_________________________ taste

Feel _________________________

Turn litmus paper _________________________

Arrhenius TheoryAcids vs Bases

All acids produce _________________________ ions when they are dissolved in water. These ions are responsible for the acidic properties of these solutions.

All bases produce _________________________ ions when they are dissolved in water. These ions are responsible for the basic properties of these solutions.

Strong acids and bases have _________________________ K values (because the products are favoured) and weak acids and bases have _________________________ K values because they do not produce a lot of ions (reactants are favoured).

Since strong acids and bases have lots of ions in solution (favour products) they conduct electricity _________________________.

General Idea

The acidic properties of aqueous solutions of HCl (hydrochloric acid) and HNO3(nitric acid) must be due to the hydrogen ions since they are the only ions _________________________ to both solutions.

The Arrhenius Model of Neutralization

When equal molar quantities of an acid and a base are mixed in water, a solution is obtained that does not act as an acid or a base; it is _________________________ in terms of its acid-base properties. This kind of chemical reaction is a _________________________ reaction.

ACID + BASE SALT + WATER

Ex. Write the net ionic equation for the reaction below:

HCl(aq) + NaOH(aq)

Problems with Arrhenius’ Theory

The Solvent Problem: The nature of the solvent is very important in determining the acidic behaviour of a substance.

HCl is a strong acid and when it dissolves in water it _________________________ electricity well because there are lots of H+ and Cl- ions, however when HCl dissolves in toluene it _________________________ conduct electricity which means there are few ions present. If there are few H+ ions present it would no longer be a strong acid.

The Salt Problem: Salts solutions should not change the colour of litmus paper if they do not contain H+ ions or OH- ions because they would not be acids or bases (should be neutral).

Salt Effect on Litmus ConclusionNaClNa3PO4

Na2CO3

NaNO3

NH4ClPb(NO3)2

*How can Pb(NO3)2 be an acid if it needs to produce H+ ions and how can Na2CO3 be a base if it needs to produce OH- ions?

The Brønsted-Lowry TheoryAcids vs Bases

An acid is a molecule or ion that can _________________________a hydrogen ion. Any ion containing H atoms is a potential acid.

A base is any molecule or ion that can react with (_________________________) a hydrogen ion. Any ion with a pair of valence electrons available for bonding is a potential base.

General Idea

An acid is a hydrogen-ion _________________________, and a base is a hydrogen-ion _________________________.

An acid can only act as an acid if a base is present and is willing/able to accept a hydrogen ion.

This theory was able to explain the role of the solvent as well as the existence of acidic and basic salt solutions.

Dissociation of Brønsted-Lowry Acids

Acids donate a proton (H+) which combines with water to form _________________________ (H3O+). Often H+ is used for convenience but the meaning is H3O+.

Ex. HCl(g) + H2O(l)

Water functions as a hydrogen-ion acceptor, or _________________________, in this reaction.

H3O+ is an acid itself and can release an H+ ion if the right _________________________ is present.

Ex. H3O+(aq) + H2O(l) H⇌ 2O(l) + H3O+

(aq)

Dissociation of Brønsted-Lowry Bases

Metal hydroxides dissolve in water and produces _________________________directly which can accept an H+ ion.

Ex. NaOH(s) Na+(aq) + OH−

(aq)

Ba(OH)2(s) Ba2+(aq) + 2 OH−

(aq)

There are bases that do not contain OH- ions. With these bases H2O can act as an acid by _________________________ a hydrogen ion and leaving an OH- ready to react.

Ex. NH3(g) + H2O(l) NH4+

(aq) + OH−(aq)

Example: Liquid hydrogen perchlorate, HClO4, dissolves in water to form a solution of perchloric acid. Identify the hydrogen ion donor and acceptor, and write an equation for the solution process.

Example: A solution of gaseous methylamine, CH3NH2, turns red litmus blue. Write the balanced equation.

Weak and Strong Acids and Bases The strongest acid that can exist in water is the _________________________ ion. The strongest base

that can exist in water is the _________________________ ion.

To differentiate acids that ionize completely from those that do not, we use the terms strong and weak. A _________________________ is completely ionized in solution. A _________________________is only partially ionized in solution. Hydrochloric acid is a strong acid even when it is dilute. Acetic acid is a weak acid even when it is concentrated.

Bases can also be weak or strong. Sodium hydroxide is an example of a _________________________. It completely dissociates in water. Ammonia, however, is a _________________________.

Using a table of acid-dissociation constants is the surest way to quantify relative strengths of weak acids, but you can often classify acids and bases qualitatively as strong or weak just from their formulas:

Strong acids. Two types of strong acids, that you should memorize are:

1. The _________________________ acids HCl, HBr, and HI

2. Oxoacids in which the number of O atoms exceeds the number of ionizable protons by ______ or more, such as HNO3, H2SO4, HClO4; for example, in H2SO4, 4 O’s – 2 H’s = __

Weak acids. There are many more weak acids than strong ones. Four types are:

1. The _________________________ acid HF

2. Acids in which H is _________________________ to O or to a halogen, such as HCN and H2S

3. Oxoacids in which the number of O atoms equals or exceeds by _______________________ the number of ionizable protons, such as HClO, HNO2, and H3PO4

4. _________________________ acids (general formula RCOOH, with the ionizable proton shown in italics), such as CH3COOH and C6H5COOH

Strong bases. Water-soluble compounds containing _________________________ions are strong bases. The cations are usually those of the most active metals:

1. M2O or MOH, where M = _________________________ (1) metal (Li, Na, K, Rb, Cs)

2. MO or M(OH)2, where M = _________________________ (2) metal (Ca, Sr, Ba)

Weak bases. Many compounds with an electron-rich _________________________ atom are weak bases. The common structural feature is an N atom with a lone pair (shown here in italics in the formulas):

1. Ammonia ( H3)

2. Amines (general formula R H2, R2 H, or R3 ), such as CH3CH2 H2, (CH3)2 H,

and (C3H7)3

Examples H2SeO4

(CH3)2CHCOOH

KOH

(CH3)2CHNH2

Salts of Weak Acids and Bases

Ions of Neutral Salts

Cations

Na+ K+ Rb+ Cs+

Mg2+ Ca2+ Sr2+ Ba2+

Anions

Cl- Br- I-

ClO4- BrO4

- ClO3- NO3

-

Acidic Ions

NH4+ Al3+ Pb2+ Sn2+

HSO4- H2PO4

- HSO3-

Basic Ions

F- C2H3O2- NO2

- HCO3-

CN- CO32- S2- SO4

2-

HPO42- PO4

3-

Example: Using an equation, show why a sodium nitrite solution is basic.

The sodium ions are _________________________, but the basic nitrite ions will react with water

The formation of some _________________________________ will render the solution ___________________.

Example: Account for the fact that sodium hydrogen carbonate produces a basic solution when it is dissolved in water.

The sodium ions are _________________________, but the _________________________ hydrogen carbonate ions will react with _________________________:

The formation of some __________________________________will render the solution _________________________.

Example: Account for the fact that sodium hydrogen sulfite produces an acidic solution when it is dissolved in water.

Example: Account for the fact that salts containing ammonium produce acidic solutions when they are dissolved in water.

Ammonium reacts with water (which acts as a base, accepting a hydrogen ion).

Amphoteric Substances We`ve learned that water can act as an acid or a base:

o NH3 + H2O NH⇌ 4+ + OH- here water acts as a(n) _________________________.

o HCl + H2O → H3O+ +Cl- here water acts as a(n) _________________________. Substances that can act as an acid in one reaction and a base in another are called

_________________________ or _________________________ substances. Another amphoteric substance is HSO4

- (bisulphate ion).

Polyprotic Acids Acids that can give up more than one hydrogen ion per molecule are called

_________________________ acids. Diprotic acids, which can release _________________________ hydrogen ions, are much more common than triprotic acids, which can release _________________________ hydrogen ions.

Ex. Ionization of sulfuric acid:

H2SO4(aq) + H2O(l) H3O+(aq) + HSO4

−(aq)

HSO4−

(aq) + H2O(l) H3O+(aq) + SO4

2−(aq)

Ex. Ionization of Phosphoric acid

H3PO4(aq) + H2O(l)

Example: Write equations for the ionization of sulfurous acid in water.

See Acids and Bases Assignment

Conjugate Acid-Base Pairs Acid-base reactions involve the exchange of hydrogen ions. When an acid loses a hydrogen ion

it becomes a _________________________, and when a base accepts a hydrogen ion it becomes an _________________________.

Structures that differ in only one hydrogen ion are called _________________________ acid-base pairs. The stronger the acid, the _________________________ will be its conjugate base, and vice versa. In these reactions, not only the reactants but also the _________________________ are acids and bases.

H2O(l) + NH3(aq) NH4+

(aq) + OH−(aq)

acid1 base2 acid2 base1

HNO3(aq) + H2O(l) H3O+(aq) + NO3

−(aq)

o Acid1 is the conjugate acid of base1, and base2 is the conjugate base of acid2.o Notice, as well, that the formula for the conjugate base always has one more

_________________________ charge than the formula for the corresponding acid.

*Notice that water is acting as an acid in the first reaction, but as a base in the second one.*

Acid Base Conjugate Base Acid ConjugateHCl OH-

H2O HSO4-

H3O+ NO3-

NH4+ H2O

See Conjugate Acids and Bases Assignment

Dissociation of Water Write out the chemical equation for the dissociation of water:

The equilibrium constant expression for the self-ionization of water can be written as follows:

Since Kw is so small, H2O is favored over H+/OH- production (ie. Not a lot of water dissociates)

Determine [H3O+] and [OH-] for pure water

The ion product constant is valid for all aqueous solutions. If an acid is added to pure water, the hydronium ion concentration will increase. Since the value of Kw is constant, the concentration of the hydroxide ion will decrease.

[H3O+] does not have to equal [ OH−], they just need to multiply to give 1.0

× 10-14

Example: What are the hydronium ion and hydroxide ion concentrations in a 0.050 M aqueous solution of hydrogen chloride at 25°C?

Equation:

HCl almost completely dissociates,

Use Kw expression to calculate [OH-]

Example: What are the hydroxide ion and hydronium ion concentrations in an aqueous solution containing 0.010 M barium hydroxide?

Equation:

Since Ba(OH)2 is a strong base, it completely dissociates:

Use Kw expression to determine [H3O+]

Ex. What is [H+] in an aqueous solution in which [OH-] = 1.0 x 10-3?

See Dissociation of Water Assignment

pH Scale and Indicators In 1909 the Danish biochemist, Søren P. Sørenson, introduced the pH scale as a way to express

the acidity and basicity of an aqueous solution (pH is a measurement that expresses how acidic or basic a solution is).

pH scale ranges from __________________ when pH = 7 the solution is neutral

and moles of H3O+ = moles of OH-

if the pH < 7 the solution is acidic and [H3O+]>[OH-]

if the pH > 7 the solution is basic and [H3O+]<[OH-]

The pH of a solution is defined as the negative logarithm, to the base ten, of the hydronium ion concentration.

In a neutral solution at 25°C, the hydronium ion and the hydroxide ion concentrations are both 1.0 × 10-7 mol/L. Thus, the pH of a neutral solution is 7.

pH = −log10(1.0 × 10-7) = 7.00

Example: What will be the pH of an aqueous solution containing 0.040 M sodium hydroxide?

Example: What is the hydronium ion concentration of a solution with a pH of 2.50?

Determining pOH pOH is another way to express how acidic or basic a solution is. It’s basically just the opposite of pH.

pH + pOH = 14.00 pOH = −log10[OH−] pOH = 14.00 – pH

Example: A solution has a pH of 3.00, is it an acid or base? What would be the pOH of the same solution?

Example: An aqueous solution containing 0.040 M sodium hydroxide has the following pOH:

See pH Assignment

Acid-Base Indicators

Indicators are chemicals that appear different _________________________ in acids and bases. Indicators are weak acids or bases Indicators change colours when the pH (or pOH) of a solution reaches a certain

_________________________, so they can be used to determine the pH of solutions.

Acid-Base Titrations Recall: neutralization refers to the double displacement reaction which occurs when an acid is

mixed with a base, the products of neutralization are a _____________________and ___________________.

To achieve neutralization we use a process called titration.

o One reactant is carefully added to another reactant until the two have combined in their exact stoichiometric proportions.

The objective of a titration is usually to find the number of moles or grams, the concentration, or the percentage of the analyte (the substance we are looking for) in a sample. This is usually done by measuring the precise volume and concentration of a titrant (the solution being added) needed to react completely with the analyte. Stoichiometric relationships are then used to determine the quantity of analyte present from the number of moles (volume × molarity) of titrant added.

o During titration, the solution reaches an _________________________ point where the moles of acid equals the moles of the base (this does not necessarily mean pH = 7).

o When doing titrations we use a _________________________ because it is really accurate at measuring volumes.

o When performing a titration you add your titrant to the analyte (with indicator) until there is ONE DROP that has a _________________________ colour change.

When you hear the term ‘neutralizatoin’ you may think that results in a final pH of 7, but this is not the case. The final pH depends on the strength of the acid and base being combined.

o Strong acid + Strong base pH = 7 (neutral) when neutralized.

o Strong acid + weak base pH < 7 (acidic) when neutralized.

o Weak acid + strong base pH > 7 (basic) when neutralized.

Titration Calculations: When neutralized, moles H+ = moles OH-, so we can use the formula:

o C (MA x VA ) = D (MB x VB), where C and D represent the coefficients from the balanced equation.

o Alternatively the formula could be MA x VA = MB x VB where MA represents the molarity of H+ and MB represents the molarity of OH+

Example: In a titration, a few drops of bromothymol blue indicator are added to 16.80 mL of an aqueous sodium hydroxide solution of unknown concentration and is neutralized by 25.00 mL of a 0.190 M solution of sulfuric acid. What is the concentration (initial) of the sodium hydroxide solution? Remember: when neutralized, moles H+ = moles OH-

Example: A 20.0 mL aqueous solution of strontium hydroxide, has a drop of indicator added to it. The solution turns colour after 25.0 mL of a standard 0.0500 M HCl solution is added. What was the original concentration of the strontium hydroxide?

See Titration Assignment

Primary Standards To perform a titration, the concentration of one of the solutions must be precisely and

accurately known. These substances are called primary standards. A primary standard must meet the following criteria:

1. It should be obtainable as a very pure solid at reasonable cost.

2. The substance should be air stable. That is, it should not react with any component in the air, such as oxygen, carbon dioxide, or water vapour.

3. The substance should be stable in solution for a reasonable length of time.

4. A substance with a high molar mass is preferred. This minimizes weighing errors.

Ionization Constants for Acids and Bases Since weak acids and bases _________________________ dissolve in water and reach equilibrium,

equilibrium law expressions can be written.

The larger the acid ionization constant, the _________________________ the acid (really strong acids almost completely dissociate so no equilibrium is reached).

The base ionization constant is the equilibrium constant expression for a weak base. Conjugate Acid-Base Pairs:

Ka x Kb = Kw here`s why:

We can calculate the value of an acid ionization constant if we know the acid concentration and the pH of the solution. Conversely, if we know the concentration and acid ionization constant, we can calculate the expected pH of the solution.

Example: Acetylsalicylic acid (aspirin) is a weak monoprotic acid. A 0.100 M solution of the acid has a pH of 2.24. Calculate the acid ionization constant of the acid (Ka).

Example: Ascorbic acid (vitamin C) is a weak monoprotic acid. It has a K a = 8.0 × 10-5. Calculate the pH of a 0.100 M solution.

[HAsc]

[H3O+] [Asc−]

Initial

Change

Equilibrium

Example: Caffeine is a weak base that is related to ammonia. It has a Kb = 4.1 × 10-4. Calculate the pH of a 0.70 M solution.

Buffers Buffers are mixtures of chemicals that make a solution resist a change in pH. Solutions that

have a resistance to changes in their pH because of the presence of buffers are called buffer solutions.

In general, buffers are made up of (1) a weak acid and one of its soluble salts, or (2) a weak base and one of its soluble salts.

Buffers of the first type buffer a solution in the acid range. Buffers of the second type buffer a solution in the base range. Both kinds of buffers make solutions in which an equilibrium exists that shifts in response to the addition of an acid or base so as to allow only small changes in the hydrogen-ion concentration.

[cafN] [cafNH+] [OH−]

Initial

Change

Equilibrium

Acids & Bases http://www.saskschools.ca/curr_content/chem30_05/5_acids_bases/labs/labs.htm

Acid-Base Titration

OVERVIEW

Often we want to determine the concentration of a solution. One way to do so is to carrying out an analytical procedure known as a titration. During a titration a carefully measured volume of the solution with the unknown concentration (called the analyte) is reacted with a second solution (the titrant) whose concentration is known (a standard solution). By knowing how much of the standard solution is required to react completely – no more, no less – with the solution with the unknown concentration we can calculate that solution’s concentration.

The point at which stoichiometrically equal amounts of the two solutions have been combined is called the equivalence point. When we neutralize an acid with a base, this will occur when [H+] = [OH-]. By using an appropriate indicator we can detect this point by noting when the indicator changes colour. This will be used to signify the end point of the titration. A balanced equation and simple calculations will then allow us to determine the concentration of the solution.

PURPOSE

To determine the concentration of a solution of NaOH by titration with a standard solution of HCl.

To determine the concentration of a sample of white vinegar by titration with a standard solution of NaOH

SAFETY

Acids and bases are corrosive substances. Safety goggles must be worn. Be sure to report any spills to your teacher so they may be cleaned up properly.

EQUIPMENT AND MATERIALS

two 50-mL buretsburet stand and clampsErlenmeyer flask, 125-mLErlenmeyer flask, 250-mLwash bottledistilled water

10-mL graduated cylinder10-mL volumetric pipette (optional)

0.100M HCl standard solutionNaOH solution with unknown concentrationvinegar (acetic acid, HC2H3O2)phenolphthaleindistilled water

PROCEDURE

Part A. Titration of Base of Unknown Concentration

1. Wash two burets with detergent solution. Rinse them thoroughly.

2. With a grease marking pencil or tape identify which buret is to hold each solution, the acid or base.

Rinse each buret with about 10 mL of solution that it is to hold – rinse the acid-containing buret with the HCl solution and the base-containing buret with the NaOH solution. Allow the acid or base to run out of the buret tip to rinse them.

3. Fill each buret with the proper solution and allow the some of each solution to run out of the buret tip. Make sure no drop remains hanging on the buret. Be sure there are no air bubbles in the tips.

It is very important that you accurately read and record the initial and final volumes. It is not necessary that the burets be filled to the very top mark (0.0 mL) at the start of the titration, but it is important that the level never go below the bottom mark (50.0 mL). Be sure to read the bottom of the meniscus at eye level. You may find it helpful to hold a white card with a large black streak or rectangle behind the buret to make it easier to read.

4. Place the 125-mL Erlenmeyer flask beneath the acid buret. Add 10.0 mL of acid to the flask. Use your rinse bottle to make sure all drops make it to the bottom of the flask; rinse any drops that remain on the sides of the flask. Read the buret carefully and record both the initial and final volumes from the buret into your data table.

5. Add 10-mL of distilled water to the flask.

6. Add three drops of phenolphthalein to the flask, and swirl the flask to mix thoroughly.

7. Move the flask so it it beneath the base buret. Place the flask on a sheet of white paper so a colour change will be more readily observed.

8. After recording the initial volume of base in the buret, begin the titration by adding NaOH to the flask. For your initial trial you may want to add the base fairly quickly until you notice a pink colour appearing in the flask. Swirling the flask should make the pink colour disappear. At that point begin adding the NaOH more slowly, swirling the flask after each drop is added. As soon as a faint pink colour becomes permanent, stop the titration – the end point has been reached. Do NOT continue until a darker pink colour has been reached – if that happens you’ve gone past the end point.

If you do go past the end point, add a few drops of acid (be sure to record the new volume used), then add more base.

Record the final volume of base in the buret.

9. Repeat the titration, performing at least four trials. Be sure to rinse the Erlenmeyer flaks well between trials.

For your other trials add the base more slowly as you near the end point in order to get more accurate readings. You do not need to refill burets between trials.

Part B. Titration of Vinegar

1. Using the volumetric pipette (or another clean buret), add exactly 10.0 mL of vinegar to a clean 250-mL Erlenmeyer flask.

2. Add 100 mL of distilled water to the flask and three drops of phenolphthalein.

3. Titrate the vinegar with the NaOH solution used in Part A. If necessary add more NaOH to the buret before beginning the titration. Record the initial volume of base in the buret.

4. As before, the end point will be reached as soon as a permanent, pale pink colour appears in the flask. Record the final volume of base.

5. Repeat the titration at least two more times.

RESULTS

Copy Data tables 1 and 2, as shown on the last page of this lab, into your data notebook.

CALCULATIONS

Part A. Titration of Base of Unknown Concentration

To calculate the concentration of the unknown base we must begin with a balanced equation. The reaction between hydrochloric acid and sodium hydroxide is:

HCl + NaOH → NaCl + H2O

Stoichiometrically we see that one mole of the acid reacts with one mole of the base. Because of this one-to-one relationship we can use the following formula to calculate the unknown concentration:

Macid Vacid = Mbase Vbase

Rearrange the equation to solve for the unknown concentration of the base:

M base=M acid×V acid

V base

For each trial in Part A, determine the molarity of the NaOH solution, [NaOH]. Show your calculations in a table similar to the one shown below. Calculate the average for your trials.

Table 3. Calculating the concentration of the sodium hydroxide solution.

Trial Calculations [NaOH]

M base=M acid×V acid

V base

1

2

3

4

Average -----

Collect the data from the rest of the class. Calculate the class average:

Table 4. Class data for the concentration of the sodium hydroxide solution.

Group [NaOH]

1

2

3

etc.

Average

Part B. Titration of Vinegar

The reaction between sodium hydroxide and vinegar – acetic acid, HC2H3O2 – is represented by:

HC2H3O2 + NaOH NaC2H3O2 + H2O

Again there is a 1:1 relationship between the acid and the base. As before we can determine the concentration of the unknown solution – in this case the acetic acid – if we know the volume and molarity of the base and the volume of the acid used:

M acid=M base×V base

V acid

Using the molarity of the base you calculated in Part A of the lab, determine the molarity of the acetic acid. Show your calculations in Table 5, which you should copy into your notebook.

Collect the data from the rest of the class. Calculate the class average:

Table 5. Calculating the concentration of the vinegar solution.

Trial Calculations [HC2H3O2]

M acid=M base×V base

V acid

1

2

3

4

Average -----

Table 6. Class data for the concentration of the vinegar solution.

Group [NaOH]

1

2

3

etc.

Average

CONCLUSIONS AND QUESTIONS

1. How did the results for each of your trials for the titration of the sodium hydroxide compare? Were the results similar or did they vary a great deal?

2. What are some of the major sources of error with this experiment?

3. The volume of water added during this experiment – to rinse droplets of acid from the buret or as water dded to the acid in the flask – does not affect the calculations and thus does not need to be accounted for. Why not?

Data Tables.

Table 1. Titration of NaOH with Unknown Concentration

Trial 1 Trial 2 Trial 3 Trial 4

HCl NaOH HCl NaOH HCl NaOH HCl NaOH

initial volume

final volume

volume used

Table 2. Titration of Vinegar

Trial 1 Trial 2 Trial 3 Trial 4

Vinegar NaOH Vinegar NaOH Vinegar NaOH Vinegar NaOH

initial volume - - - -

final volume - - - -

volume used 10.0 10.0 10.0 10.0

Name:______________________________________

Acids and Bases1. Identify the hydrogen donor and acceptor in each of the following reactions:

2. Which of the following would you expect to act as Brønsted-Lowry bases:

3. Write the overall neutralization reaction equation and the net ionic equation for the reaction in aqueous solution between the following acids and bases:

a) sulfuric acid and sodium hydroxide

b) hydrochloric acid and calcium hydroxide

4. A light bulb conductivity test was carried out on 1.0 M solutions of various bases. The bulb glowed brightly for potassium hydroxide, barium hydroxide, and lithium hydroxide solutions, but gave only weak flickers for methylamine and caffeine solutions. From this information determine which of these bases are strong and which are weak.

5. For each of the following, which one is the stronger acid:

a) HNO2 or HNO3

b) HIO2 or HIOc) H3AsO4 or H3AsO3

6. Which compound in each group is the stronger acid in aqueous solution:

a) H2SO3 or H2SO4

b) HBrO2 or HBrO3

c) H3PO4 or H3PO3

7. Which compound in each of the following pairs is the stronger acid:

a) H2O or H2Sb) H2O or NH3

8. Write the Brønsted-Lowry equations for the reactions with water of the following acids and bases, assuming that only one hydrogen ion is exchanged.

a) fluorosulfonic acid, FSO3H, strong acid

b) sulfurous acid, H2SO3, weak acid

c) hydrogen bromide, HBr, a strong acid

d) perchloric acid, HClO4, a strong acid

e) hydrogen cyanide, HCN, a weak acid

f) hydrogen sulfide, H2S, a weak acid

g) formic acid, HCO2H, a weak acid

9. Write the reaction equations for the dissociation of each of the following substances in water and explain, with the use of Brønsted-Lowry reaction equations, the acidity or basicity of the resulting solutions:

a) potassium bisulfate, KHSO4, weak acid

b) sodium ethoxide, C2H5ONa, strong base

c) sodium phosphate, Na3PO4, weak base

d) sodium hydrogen carbonate, weak base

10. Write the equations for the ionization of the weak diprotic acid, carbonic acid, H2CO3.

11. For the reaction in aqueous solution between the diprotic weak acid, sulfurous acid, and sodium hydroxide write the overall neutralization equation

12. Write the reaction equations for the ionization in water of the weak triprotic acid, citric acid (C6H5O7H3).

13. Predict whether aqueous solutions of the following salts are neutral, acidic, or basic:

a) potassium iodide b) sodium cyanidec) ammonium dihydrogen phosphate

14. Show, using equations, why sodium acetate produces a basic solution in water.

15. Write the reaction equations for the dissociation of each of the following substances in water and explain, with the use of Brønsted-Lowry reaction equations, the acidity or basicity of the resulting solutions:

a) potassium bisulfate, KHSO4, weak acid

b) ammonium chloride, weak acid

c) sodium ethoxide, C2H5ONa, strong base

d) sodium phosphate, weak base

Name: ______________________________________

Conjugate Acid/Base Pairs1. What is the conjugate base for each of the following acids?

a) NH3 c) HSO3− e) H2PO4

−

b) H3O+ d) NH4+

2. What is the conjugate acid for each of the following bases?

a) PO43− b) NH3 c) HCO3

− d) CH3OH

3. For each of the following reactions identify, where possible, the acid-base conjugate pairs:

a) CH3CO2−

(aq) + H2O(l) ↔ CH3CO2H(aq) + OH−(aq)

b) HClO(aq) + CH3NH2(aq) ↔ CH3NH3+

(aq) + ClO−(aq)

c) ClCH2CO2H(aq) + H2O(l) ↔ ClCH2CO2−

(aq) + H3O+(aq)

4. Complete the following acid-base reactions, assuming that only one hydrogen ion is exchanged. Identify the conjugate acid-base pairs by writing them below the equations.

5. Which of the following do not represent a conjugate acid-base pair:

a) SO32− and SO2

b) CO32− and CO

c) H3O+ and H2

d) NH4+ and NH3

Name: __________________________________

Dissociation of Water1. Calculate the H3O+ and OH− concentrations in the following solutions. Each is either a

strong acid or a strong base.a. 0.05 M sodium hydroxide

b. 0.0025 M sulfuric acid

c. 0.013 M lithium hydroxide

d. 0.150 M nitric acid

e. 0.0200 M calcium hydroxide

f. 0.390 M perchloric acid

2. What will [H+] be if 0.010 mole of solid NaOH is added to 1.0L of water?

3. What will [OH-] be if 0.010 mole HCl is added 1.0L of water?

4. (HARDER)Determine the [H+] and [OH-] if 50.0mL 0.200M HCl and 49.0mL 0.200M NaOH are combined.

5. Find the [H+] and [OH-] of a solution made by dissolving 12.00g of Ba(OH)2 in enough water to make 350mL of solution.

Name: __________________________________

pH1. Calculate the pH of each of the following:

a) an aqueous solution that has a hydronium ion concentration of 3.0 × 10−3 M

b) an aqueous solution that has a hydroxide ion concentration of 6.0 × 10−4 M

c) an aqueous solution containing 0.0020 M barium hydroxide

d) 250.0 mL of an aqueous solution containing 1.26 g of nitric acid

2. Calculate the hydronium concentration of

a) 100.0 mL of an aqueous solution containing 0.60 g of sodium hydroxide

b) a blood sample with a pH of 7.40

c) orange juice with a pH of 3.20

3. Calculate the pH of a solution created by dissolving one pellet (0.10 g) of sodium hydroxide in 1.0 L of pure water.

4. The pH of an aqueous solution of NaOH is 12.9. What is the molarity of the solution?

5. What is the pH of a 0.00125 M HBr solution? If 175 mL of this solution is diluted to a total volume of 3.00 L, what is the pH of the diluted solution?

6. A container is labeled 500.0 mL of 0.00157 M nitric acid solution. A chemist finds that the container was not sealed and that some evaporation has taken place. The volume of solution is now 447.0 mL.

a. What was the original pH of the solution?

b. What is the pH of the solution now?

7. An acetic acid solution has a pH of 4.0. What are the [H3O+] and [OH−] in this solution?

8. What is the pH of a 0.000460 M solution of Ca(OH)2?

9. A solution of HCl has a pH of 1.50. Determine the pH of the solutions made in each of the following ways.

a. 1.00 mL of the solution is diluted to 1000.0 mL with water.

b. 25.00 mL is diluted to 200.0 mL with distilled water.

c. 18.83 mL of the solution is diluted to 4.000 L with distilled water.

d. 1.50 L is diluted to 20.0 kL with distilled water.

10. A hydrochloric acid solution has a pH of 1.70. What is the [H3O+] in this solution? Considering that HCl is a strong acid, what is the HCl concentration of the solution?

11. Calculate the pH of the following aqueous solutions:a. an aqueous solution containing 3.0 × 10−5 M hydronium ions

b. an aqueous solution containing 6.5 × 10−4 M hydroxide ions

c. the solution obtained when 7.2 g hydrogen chloride is dissolved in 5.0 L of water

d. a 500.0 mL aqueous solution containing 1.0 g sodium hydroxide

12. What are the hydronium ion and the hydroxide ion concentrations of the following:

a) a solution with a pH = 5.00

b) a solution with a pH = 9.55

13. Fill in the blanks in the following table. For each solution indicate whether the solution is acidic or basic.

pHpOH [H3O+] [OH−] A or B

4.0

11.6

1.8 × 10−9

3.5 × 10−2

14. Calculate the pH of the solution obtained when 50.00 mL of 0.150 M Hydrochloric acid is added to 75.0 mL of a 0.111 M aqueous sodium hydroxide solution

Name: ______________________________

Titration1. The following aqueous solutions were titrated with a 0.150 M aqueous sodium hydroxide

solution. Write the overall reaction equations and calculate the concentrations of the acid solutions.

a) 25.00 mL of hydrochloric acid requiring 16.50 mL of base solution

b) 25.00 mL of sulfuric acid solution requiring 42.00 mL of the base solution

c) 10.00 mL of vinegar, containing acetic acid, requiring 55.0 mL of the base solution

2. A sample of powdered vitamin C (ascorbic acid) tablet was suspended in water and titrated with a 0.150 M aqueous solution of sodium hydroxide. A total of 21.50 mL of the base solution was required to change the colour of the indicator. How many grams of ascorbic acid, C6H8O6, did the tablet contain assuming the presence of one acidic hydrogen atom per molecule?

3. The following aqueous acid solutions were all titrated with a 0.180 M sodium hydroxide solution. In each case write the overall reaction equation for the neutralization reaction and calculate the concentration of the acid solution.

a) 25.00 mL of formic acid, HCO2H (with one acidic hydrogen per molecule) requiring 37.50 mL of the base solution

b) 10.00 M of a sodium dihydrogen phosphate solution requiring 25.00 mL of the base solution

c) 25.00 mL of a phosphoric acid solution requiring 62.00 mL of the base solution

4. The following aqueous base solutions were all titrated with a 0.126 M hydrochloric acid solution. Write the overall reaction equation for the neutralization reactions and calculate the concentration of the base solutions.

a) 10.00 mL of an ammonia solution requiring 22.50 mL of the acid solution

b) 10.00 mL of an diaminoethane, H2NCH2CH2NH2 requiring 36.00 mL of the acid solution

c) 25.00 mL of an sodium carbonate solution requiring 45.50 mL of the acid solution

5. A 50.00 mL sample of a potassium hydroxide is titrated with a 0.8186 M HCl solution. The titration requires 27.87 mL of the HCl solution to reach the equivalence point. What is the molarity of the KOH solution?

6. A solution of citric acid, a triprotic acid, is titrated with a sodium hydroxide solution. A 20.00 mL sample of the citric acid solution requires 17.03 mL of a 2.025 M solution of NaOH to reach the equivalence point. What is the molarity of the acid solution?

7. Calculate the pH of an aqueous solution of strong acid prepared by adding 50.00 mL of a 1.50 M hydrochloric acid to 100.0 mL of 0.500 M nitric acid.

8. A 15.00 mL sample of acetic acid is titrated with 34.13 mL of 0.9940 M NaOH. Determine the molarity of the acetic acid.

9. A 12.00 mL sample of an ammonia solution is titrated with 1.499 M HNO3 solution. A total of 19.48 mL of acid is required to reach the equivalence point. What is the molarity of the ammonia solution?

10. A H2SO4 solution of unknown molarity is titrated with a 1.209 M NaOH solution. The titration requires 42.27 mL of the NaOH solution to reach the equivalent point with 25.00 mL of the H2SO4 solution. What is the molarity of the acid solution?

11. A flask contains 41.04 mL of a solution of potassium hydroxide. The solution is titrated and reaches an equivalence point when 21.65 mL of a 0.6515 M solution of HNO3 is added. Calculate the molarity of the base solution.

12. A bottle is labeled 2.00 M H2SO4 . You decide to titrate a 20.00 mL sample with a 1.85 M NaOH solution. What volume of NaOH solution would you expect to use if the label is correct? 43.2 mL

13. What volume of a 0.5200 M solution of H2SO4 would be needed to titrate 100.00 mL of a 0.1225 M solution of Sr(OH)2 ?

14. An ammonia solution of unknown concentration is titrated with a solution of hydrochloric acid. The HCl solution is 1.25 M, and 5.19 mL are required to titrate 12.61 mL of the ammonia solution. What is the molarity of the ammonia solution?

Name: ____________________________________

Ionization Constants Acids and Bases1. The reaction of oxalate ions, C2O4

2−, with water is an equilibrium reaction. Write the equation for the reaction and the equilibrium expression for the ionization constant. Given that the Kb of oxalate ions is 1.6 × 10−10, calculate the value of Ka for the hydrogen oxalate ions.

2. An aqueous solution containing 1.00 M boric acid (H3BO4) is found to have a pH of 4.57. From this information, calculate the Ka value of boric acid. Consider only the first ionization step of boric acid.

3. Using the Ka value for nitrous acid, determine Kb for the basic anion, NO2−. Use this value to

calculate the pH of a 0.15 M solution of sodium nitrite, NaNO2.

4. A solution of a weak acid requires 25.40 mL of base for neutralization. After 12.70 mL of this base had been added during the titration, the pH of the solution was 4.82. Determine the Ka for the weak acid

5. Calculate the ammonium ion concentration in 500.0 mL of a 0.050 M aqueous solution of ammonia to which 4.0 g of sodium hydroxide has been added. Kb for ammonia is1.8 × 10−5.

6. A solution of a weak acid requires 25.40 mL of base for neutralization. After 12.70 mL of this base had been added during the titration, the pH of the solution was 4.82. Determine the Ka for the weak acid.