ACID BASE REACTIONS - MIXING STRONG ACIDS...

Chemistry 12UNIT V - APPLICATION OF ACIDS AND BASES ver 5.3

ACID BASE REACTIONS - MIXING STRONG ACIDS AND BASES (NOT @ EQUIL)

When a strong acid and base are mixed, a neutralization reaction occurs. The resulting solution may be acidic, basic, or neutral depending on the relative amounts of each of the reactants that were used.

(recall limiting reactant calculations from Chemistry 11)

Ex. What is the pH that results when 25.0 mL of 0.250 M HCl is mixed with 35.0 mL of 0.200 M NaOH?

1. write rxn HCl + NaOH NaCl + H2O2. setup ICE table (modified)3. calculate moles of each4. compare values (LR)5. determine [OH-]6. answer

Try: What is the resulting pH of a solution made by mixing 45.0 mL of 0.450 M KOH with 75.0 mL of 0.275 M HClO4?

HCl + NaOH NaCl + H2O

Initial

Change

End

Chemistry 12 – Applications of Acids and Bases 5.3

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CALCULATING MOLES OF ACID NEEDED TO REACH A DESIRED pH

Ex. How many moles of HCl(g) must be added to 40.0 mL of 0.180 M NaOH to produce a solution that has a pH of 12.500? Assume that there is no change in volume when the HCl is added.

F1. ACID BASE TITRATIONS

A titration is a process in which a measured amount of a solution is reacted with a known volume of another solution (and one of the solutions has an unknown concentration) until the ‘equivalence point’ (also known as stoichiometric point) is reached.

EQUIVALENCE POINT

ALL titration problems involve at least FIVE PARAMETERS

Note – pay special attention to significant digits with titration calculations, as the purpose of titration calculations is to get accurate and precise values (often dealing with low concentrations). As such, watch your rounding. Try not to round too much until the entire question is completed.

When the amount of reactants in a reaction is exactly equal (according to the stoichiometric ratio)

Ex.

Concentration of Acid Concentration of Base Chemical ReactionVolume of Acid Volume of Base


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STANDARD SOLUTION

Standard solutions can be prepared in two main ways:

1. Use a substance that can be obtained in a pure and stable form (doesn’t absorb water or carbon dioxide from the air) and has a known molar mass so a solution of known concentration can be made. These are called PRIMARY STANDARDS. (ex. KHC8H4O4, Na2CO3)

2. Titrate a base with an c primary standard (and calculate the concentration of the base accurately) then use the calculated solution as the standard.

TITRATIONS – MULTIPLE TRIALS

CALCULATING CONCENTRATION FROM A TITRATION.

Ex. In the reaction H2SO4 + 2 NaOH Na2SO4 + 2 H2O

An average of 23.10 mL of 0.2055 M NaOH was needed to titrate a 25.00 mL sample of H2SO4 to it’s equivalence point over three trials. What is the [H2SO4]?

Often when titrations are done, several trials (usually 3) are necessary to check the accuracy.

The titration is performed three times and ideally the results should be very similar. If they are, the average is taken as the volume to be used in the calculations.

If one is obviously off, it is omitted (assumed due to experimental error). Then the other two are used for the average of the volume.

In order to carry out a titration, you need to have a standard solution with an accurately known concentration.

A solution of ACCURATELY known concentration that is used to titrate another solution and determine its concentration


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CALCULATING PERCENT PURITY

Titrations can also be used to calculate the purity of an unknown sample. Purity titration calculations follow these steps:

Ex. A 3.4786 g sample of impure NaHSO4 is diluted to 250.0 mL. A 25.00 mL sample of the solution is titrated with 26.77 mL of 0.09975 M NaOH. What is the percent purity of the NaHSO4?

CALCULATING MOLAR MASS

Titrations can also be used to calculate the molar mass of an unknown acid or base, as long as one piece of information is given if the acid or base is monoprotic,diprotic, triprotic in order to determine the balanced reaction . By using the titration data it is possible to calculate

Calculate the [ ] of the acid base (use titration data) – this is the actual

Use the mass if the impure solid to calculate the expected (theoretical) [ ].

Calculate percent purity = [Actual] x 100 [Expected]


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the moles of the acid or the base. The molar mass can then be determined using the mass of sample of moles of sample determined.

Ex. A 3.2357 g sample of unknown monoprotic acid is diluted to 250.0 mL. A 25.00 mL sample of the acid solution is titrated with 16.94 mL of 0.1208 M KOH. What is the molar mass of the acid?

F2. INDICATORS

Indicators tend to be quite complex molecules, so their names are represented by an abbreviation. The acid form of an indicator is symbolized as HIn, and the base form as In-. Since an indicator is a weak acid or base, the equilibrium can be written as:

HIn + H2O ↔ In- + H3O+

Ex. Yellow redIN ACID

When an indicator is put into an acid, the excess [H3O+] shifts the equilibrium of the indicator according to Le Chatalier’s principle.


An indicator is a weak organic acid or base with different colours for its conjugate acid and base forms.

An indicator is in its CONJUGATE ACID FORM when in highly ACIDIC conditions, due to a Le Chatelier shift.


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IN BASE

When an indicator is put into a base, the [H3O+] is so low that it cause a shift in the equilibrium of the indicator.


When a base is added to an acidic solution, then eventually at some point [Hin] = [In-] therefore there is an equal number of each molecule.

Ex. [HIn] = [In-]Yellow Red

So the colour of the solution is ORANGE This point is called the END POINT or TRANSITION POINT for the indicator, which refers to the indicator only (not the overall reaction. (Halfway through the colour change)

At the END POINT:

At the transition point of any indicator, the following are true:

Note – You will be given a table called ‘Acid-Base’ indicators for all exams, and one is

An indicator is in its CONJUGATE BASE FORM when in highly BASIC conditions, due to a Le Chatelier shift.

[HIn] = [In-] Kin = [H3O+] pKin = pH


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present in your text. There are numerous indicators, each of which changes colour at a different pH ([H3O+]). The colour however, doesn’t instantly change at a certain pH, rather it changes colour over a range of about 2 pH units (doesn’t list K in values).

Ex. Bromothymol blue changes from yellow (acid form) to blue (base form) at a pH range of 6.0 to 7.6.

The midpoint is considered to be the average of the two values.

6.0 + 7.6 = 6.8 2

– The END POINT/TRANSITION POINT is the point in the titration where the colour of the indicator changes. The EQUIVALENCE POINT is the point is the titration where the stoichiometry of the reaction is exactly equal (according to the coefficient ratio). An indicator needs to be chosen so that the TRANSITION POINT OF THE INDICATOR happens at or near the pH of the EQUIVALENCE POINT OF THE REACTION.

- If the indicator is chosen poorly, it will change colour at a substantially different point than the equivalence point.

USING THE INDICATOR TABLE TO CALCULATE THE Kin OF AN INDICATOR

Ex. What is the Kin for phenolphthalein?

USING INDICATORS TO DETERMINE THE pH OF A SOLUTION

Ex. What is the approximate pH range of a solution that will change methyl red yellow, and neutral red red?


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PREDICTING THE COLOUR OF INDICATORS AT VARIOUS pH

Ex. What colour will each of the following indicators be in a solution of pH 3.5?

methyl Violet

methyl Orange

phenol red

thymol blue

Try: What colour is alizarin yellow in 1 x 10-5 M NaOH?

F4. SALT HYDROLYSIS

Salts consist of a cation (positive ion) and an anion (negative ion). Many salts contain an ion that can react with water.

HYDROLYSIS

In this section we will deal only with the reactions between ions and water. Reactions between ion may occur, but they are not being considered at the moment.

Before looking at ions that will react we will first look at ions that won’t react with water:

Recall SPECTATOR IONS

Hydrolysis of a salt is a reaction between the water and the cation or anion (or both) of the salt, so as to produce an acidic or basic solution.

Ions that do not participate in the reaction being considered. (Solutions – spectators don’t form precipitatesAcid/Base – spectators do not react to produce acidic or basic sol’n)


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When considering the hydrolysis of ions, the following ions do NOT hydrolyze:

SPECTATORCATIONS

SPECTATOR ANIONS

When an ion hydrolyzes it is simply acting as a Bronsted-Lowry acid or base with water.

ANIONIC HYDROLYSIS

If the anion (- ion) of the salt hydrolyzes:

It acts as a base, accepting a proton from water in an equilibrium reaction.Ex. F-(aq) + H2O(l) HF(aq) + OH-(aq)

CATIONIC HYDROLYSIS

If the cation (+ ion) of the salt hydrolyzes:It acts as an acid, donating a proton to water in an equilibrium reaction.

Ex. HA+(aq) + H2O(l) A(aq) + H3O+(aq)

PROCEDURE TO DETERMINE THE BEHAVIOUR OF A SALT IN WATER

Ions from Group I (alkali metals) and II (alkaline earth metals)

Conjugates of strong acids (except HSO4- which is a weak acid)

1. Determine the ions produced when the salt dissociates2. Identify and disregard any spectator ions3. Any remaining ions will act as acids if they are the acid side of the table (left)

Or bases if they are on the base side of the table (right)

SALT

ANION (-)

CATION (+)

YES

NOConjugate of strong acid? BASE

DOES NOT HYDROLYZE

ACIDFrom Group I or II? NO


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Ex. Predict if each of the following salts will hydrolyze in water. If so, write the hydrolysis equation for the reaction.

a. NaCl

b. NH4Cl

c. KF

d. NH4NO2

e. NaHCO3

SALT HYDROLYSIS CALCULATIONS

Ex. What is the pH of a .150M NaNO2 solution?


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SALTS CONTAINING METAL IONSSome metal ions tend to form hydrated complexes, where water molecules join up and form a polyatomic ion consisting of a central metal ion with water molecules attached. This is because:

Any time you see a metal ion with water molecules attached, it will be located on the Table of Acid Strengths. NEVER split off the water molecules and attempt to show the metal ion separately.

Ex.

Metal ions from Group 1 and 2 (except Be2+) do not hydrolyze. The most common metal ions that you will see in a hydrolysis reaction are:

Fe3+ forms Fe(H2O)63+

Cr3+ forms Cr(H2O)63+

Al3+ forms Al(H2O)63+

TITRATION CURVES

You will need to be able to look at the details of the shapes of a tritration curve and be able to give information about each one. There are 3 main types.

A. TITRATION OF A STRONG ACID WITH A STRONG BASE

Titrations of a strong acid with a strong base follow the general curve:


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VB = Volume of base required to reach the equivalence pointpH = rises almost vertically around VB . 7.0 at the equivalence pointThe salt = Neutral in solution. (Doesn’t hydrolyze)

The indicator = Should have a pKa near 7.0(Often phenolphthalein is used, even though it has a pKa of 9.1 due to its distinct colour change and since the pH rises is so sharp at the equivalence point, the VB at 7.0 is very close to the VB at 9.1.)

B. TITRATION OF A WEAK ACID WITH A STRONG BASEThe following curve is typical when a weak acid is titrated with a strong base.

VB = Volume of base required to reach the equivalence point

pH = Rises almost vertically around VB . > 7.0 at the equivalence point due to the pH @ equivalence point.

Equivalence point = For weak acid/strong base pH > 7.0, due to the salt that is formed. (Salt hydrolyzes to form OH-)

The Indicator = Should have a pKa > 7.0

C. TITRATION OF A WEAK BASE WITH A STRONG ACIDThe following titration curve is typical of when a weak base is titrated with a strong acid.

Looks very similar to a weak acid and a strong base with the curve flipped upside down.


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The graph gives pH values, not pOH, so they need to be converted.

VA = Volume of acid needed to reach equivalence point

pH = Rises almost vertically around VA . < 7.0 at the equivalence point due to the pH @ equivalence point

Equivalence point = For weak base/strong acid, pH < 7.0, due to the salt that is formed hydrolyzing to form an acidic solution.

Indicator = Needs to have a pKa < 7

ACIDS BASE REACTIONS – WEAK ACIDS AND BASES

If solutions containing H2CO3 and SO32- are mixed, the SO3

2- can only act as a base since it has no protons that it can donate.

H2CO3(aq) + SO32-(aq) ↔ HCO3

-(aq) + HSO3-(aq)

-Note that all Bronsted-Lowry equilibrium reactions that are studied in Chemistry 12 will only involve the transfer of a single proton.

(There won’t be two or three proton transfers for weak acids such asH2SO4 + Mg(OH)2 ↔ MgSO4 + 2H2O)

If solutions containing amphiprotic ions such as HCO3- and H2PO4

- are mixed:

The STRONGER of the two acids will donate a proton while the other accepts. Compare Ka values.

HCO3- Ka = 5.6 x 10-11

H2PO4- Ka = 6.2 x 10-8 since this is larger, it will act as the acid

HCO3- + H2PO4

- H2CO3 + and HPO42-

In the following equilibrium:

H2CO3 + SO32- ↔ HCO3

- + HSO3-

AT EQUIVALENCE POINT (at 25oC)STRONG ACID + STRONG BASE = NEUTRAL (pH =7)WEAK ACID + STRONG BASE = BASIC (pH >7)STRONG ACID + WEAK BASE = ACIDIC (pH < 7)

BECAUSE OF THE HYDROLYSIS OF THE SALT FORMED


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There are two conjugate acid base pairs and there is a ‘competition’ set up as to which of the acids (H2CO3 or HSO3

-) will donate their proton.

Which direction is favoured (which way does it primarily go?)Compare Ka values

H2CO3 Ka = 4.3 x 10-7 larger, so it is the stronger acid and will ‘drive’ or ‘push’ the rxn in the forward direction, so the products will be

HSO3- Ka = 1.0 x 10-7 favoured.

If the products are favoured, the Keq will be > 1

F6. DESCRIBE BUFFERS AS EQUILIBRIUM SYSTEMS

BUFFER

It may seem strange that a solution can contain large amounts of an acid and a base (since they ionize very little).

The ionization of acetic acid gives: CH3COOH + H2O CH3COO- + H3O+

If a solution is made by adding acetic acid to water, there will be a small amount of both CH3COO- and H3O+ present. Adding extra CH3COO- to the acetic acid solution will cause the equilibrium to shift to the CH3COOH side according to Le Chatalier’s principle. As a result, the [H3O+] decreases (relatively) a large amount, but only a small amount of CH3COO- will be used up as there wasn’t much hydronium ion available to react with it. Overall, a buffer will

In an equilibrium between Bronsted-Lowry acids and bases, the position of the resulting equilibrium can be predicted based on the strengths of the conjugate acids.

General Rule – Stronger acids and bases will react to produce their weaker conjugates.

A buffer is a solution containing APPRECIABLE amounts of BOTH A WEAK ACID AND ITS CONJUGATE BASE. (Usually close to equal concentrations)

Ex. CH3COOH + H2O CH3COO- + H3O+

Ka =

When equal concentrations of a weak acid and its conjugate are added to water, the pH of the resulting buffer will equal the pKa value of the weak acid.


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have a relatively large concentration of a weak acid and it’s conjugate base, and a small concentration of H3O+.

Buffers usually consist of two solutes: (BOTH MUST be added)

1. A weak acid or base

2. A soluble salt containing its conjugate

Ex. H2CO3 + H2O H3O+ + HCO3-

Note –The equation that describes a buffer is simply a Bronsted-Lowry equilibrium equation. However, a solution of a weak acid or a weak base is not a buffer by itself. Large amount of CH3COOH are present in a solution of a weak acid, but the [acetate ion] is very small. A buffer requires substantial (close to equal) amounts of both the conjugate acid and base.

Example of an ACIDIC BUFFERMix 1.0 mol of acetic acid and 1.0 mol of sodium acetate, dilute to 1.0 L

Example of a BASIC BUFFERMix 1.0 mol of ammonia and 1.0 mol of ammonium nitrate, dilute to 1.0 L

DILUTING A BUFFER

WHAT IS THE PURPOSE OF A BUFFER?

Ex. You start with two 1 L solutions of 1 M acetic acid.

1. To the first, you add .1 mol strong acid (H3O+). The pH will drop by 3.74 units. H3COOH.

2. To the second, you add a salt to create a buffer (1 mol of NaCH3COO). Then you add .1 mol strong acid (H3O+). The pH will drop by .08 units.

DILUTING A BUFFER HAS NO EFFECT ON ITS pH


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GRAPH OF A BUFFER Acetic Acid/Acetate Ion

The capacity of a buffer is:

Its ability to withstand (counteract) the addition of an acid or a base (ability to maintainA relatively constant pH)

The actual number of moles of conjugate acid and base that are used to prepare the buffer is what determines its capacity.

When selecting an acid for a buffer, it is best to choose an acid that:

Has a pKa within 1 unit of the desired pH

BIOLOGICAL BUFFERS

1. Haemoglobin is the oxygen carrier in the blood and is involved in the equilibrium:

HHb + O2 + H2O H3O+ + HbO2-

Buffers prevent the addition of either an acid or base from changing the pH of a solution to any great extent. A Le Chatalier shift will occur to use up most of the acid or base that is added (Will always change slightly) – recall Unit 2

Whenever a weak acid or base is titrated, a buffer solution will occur in the middle portion of the titration curve.


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The optimum pH for blood is 7.35 (slightly basic). If the pH is too low (Acidosis), the equilibrium shifts so that [oxyhaemoglobin] is too low, and as a result O2 won’t bind properly to haemoglobin.If the pH is too high (alkalosis) then the [H3O+] is too low and the equilibrium shifts to the right, increasing the [oxyhaemoglobin], preventing the release of O2.

2. CO2/HCO3- is a buffer system present in the blood.

1. CO2 (aq) + 2H2O(l) H3O+(aq) + HCO3-(aq)

2. CO2 (g) CO2(aq)Breathing out in (2) upsets the [H3O+] in (1). Since there is a buffer present, the loss of CO2 has a minimal effect on blood pH. (Hyperventilating lowers the [CO2] to such an extent that the pH of the blood is raised to the point where a person can black out or have hallucinations.

F8. PREDICT WHAT WILL HAPPEN WHEN OXIDES DISSOLVE

METAL OXIDESWhen a metal oxide is added to water, there is an initial dissociation of ions.

Ex. Na2O(s) 2 Na+(aq) + O2-(aq)

Ex. CaO(s) Ca2+(aq) + O2-(aq)

The metal ions are spectator ions and the oxide ion is a STRONG BASE. The reaction for the hydrolysis of the oxide ion is:

O2-(aq) + H2O(l) 2OH-(aq)

Since both metal and hydroxide ions are present in solution, the hydrolysis can be written as:Na2O(s) + H2O(l) 2NaOH(aq)

CaO(s) + H2O(l) Ca(OH)2(aq)

NON-METAL OXIDESWhen a nonmetal oxide reacts with water, the water tends to bond to the existing oxide molecule to produce AN ACIDIC SOLUTION.

Ex. CO2(g) + H2O(l) H2CO3(aq)

Ex. SO2(g) + H2O(l) H2SO3(aq)

Ex. SO3(g) + H2O(l) H2SO4(aq)


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NORMAL RAINBecause of dissolved carbon dioxide, rain is slightly acidic (pH = 5.0)

CO2 + 2 H2O ↔ H3O+ + HCO3-

ACID RAINAcid rain has a pH of less than 5, due to other elements that react with water (typically Sox or NOx)

Many fuels (coal, oil) contain a small amount of sulphur, which when burned forms sulphur dioxide along with the regular products of combustion. This SO2 can then react with the air to produce sulphur trioxide.

S + O2 SO2

2 SO2 + O2 ↔ 2 SO3 (dust and water catalyze this reaction)

When SO2 and SO2 join with water, acids are formed. The mixture of SO2 and SO3 are called SOx or sulphur oxides.

SO2 + H2O H2SO3 (sulphurous acid) SO3 + H2O H2SO4 (sulphuric acid)

Simlarily, combustion reactions (such as automobiles) can cause small amounts of nitrogen gas in the air to react with oxygen to form HNO3 – nitric acid.

N2 + O2 2NO then 2NO + O2 ↔ 2NO2 then 2NO2 + H2O HNO2 + HNO3

N2 + 2O2 2NO2

ENVIRONMENTAL CONCERNS DUE TO ACID RAIN

1. Fish and plant growth are affected in acidic water and soil. (Tomatoes, apples…)

2. Acid rain leaches minerals out of rocks and soil – some poisonous substances are released from rocks (such as aluminum ions) while nutrients are washed out of the topsoil down to the subsoil making them unavailable for plants.


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3. Metals and stone structures (esp. limestone) are damaged by acid rain.

Most lakes have a moderate buffering system of CO2/HCO3-, which can slowly reverse the

effects of acid rain. In addition, some lakes have limestone which can neutralize the acidityH2SO4(aq) + CaCO3(s) CaSO4(s) + CO2(aq) + H2O(l)

But only while limestone is present, as it can quickly be used up. Some lakes have even had powdered limestone added to reverse the effects of acid rain.

ACID BASE REACTIONS - MIXING STRONG ACIDS...

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