CHEM 102 Instructional Objectives

Henderson-Hasselbalch Equation

According to the Brønsted-Lowry theory of acids and bases, an acid (HA) is capable of donating a proton (H⁺) and a base (B) is capable of accepting a proton. After the acid (HA) has lost its proton, it is said to exist as the conjugate base (A^-). Similarly, a protonated base is said to exist as the conjugate acid (BH⁺).

The dissociation of an acid can be described by an equilibrium expression:

[H₃⁺O][ A^¯]

Acid: HA + H2O H3O⁺ + A^¯ ; Ka = ---------------

[HA]

Consider the case of acetic acid (CH₃COOH) and acetate anion (CH₃COO^-):

CH₃COOH(aq) + H₂O(l) H₃O⁺(aq) + CH₃COO^¯(aq), for which the equiilibrium constant is

[H₃⁺O ][ CH₃COO^¯] CH₃COOH(aq); K_a = ----------------------- [HC₂H₃O₂]

Acetate is the conjugate base of acetic acid. Acetic acid and acetate are a conjugate acid/base pair. We can describe this relationship with an equilibrium constant:

In this example, we will use K_a for the acid dissociation constant. Taking the negative log of both sides of the equation gives:

This can be rearranged:

By definition, pK_A = -logK_A and pH = -log[H⁺], so

This equation can then be rearranged to give the Henderson-Hasselbalch equation:

The Henderson-Hasselbalch equation can be used to prepare buffer solutions and to estimate charges on ionizable species in solution, such as amino acid side chains in proteins. Caution must be exercised in using this equation because pH is sensitive to changes in temperature and salt concentration in the solution being prepared.

Buffering Capacity

Buffer capacity is defined as the amount of H⁺ or OH^- a buffered solution can absorb without significantly changing the pH. Capacity determined by magnitude of [HA] and [A^-], lager the concentrations of them higher the buffer capacity.

`How is Henderson-Hesselbalch equation is used to calculate pH and pOH of buffer solutions.

If the concentration of the acid and the conjugate base is known in a buffer solution, the pH of the solution can be calculated using a general equation called Henderson-Hesselbalch equation.

[ACID]

pH = pK_a - log -----------

[BASE]

E.g. Buffer solution made from a weak acid/salt: HC₂H₃O₂/NaC₂H₃O₂

[HC₂H₃O₂]

pH = pK_a - log ---------------

[NaC₂H₃O₂]

E.g. Buffer solution made from a weak base/salt: NH₃/NH₄Cl

[NH₄Cl]

pH = pK_a- log ------------

[NH₃]

For a weak base (NH₃) or acid (HC₂H₃O₂) with K_b or K_a ,respectively, the value of K_a or K_b for the corresponding conjugate acid (NH₄⁺) or base (C₂H₃O₂^-) can be calculated from the following expression:

K_b x K_a = K_w

Calculate the pH of a solution prepared by adding 50 g of NH₄Cl and 32.5g of NH₃ and diluted to a liter. (K_b(NH₃); 1.8 x 10^-5

This is a problem where weak base/salt is making a buffer solution. We can apply Henderson-Hesselbalch equation for this condition.

[ACID]

pH = pK_a- log ----------- ;

[BASE]

[ACID] = [NH₄Cl] conjugate acid, [NH₃] base

[NH₄Cl]

pH = pK_a- log -----------

[NH₃]

[NH₄Cl]	=	50.0 ~~g NH₄Cl~~	x l mol NH₄Cl	x 1	=	0.9346 M( mol/L)
			53.3g ~~NH₄Cl~~	1 L

[NH₃]	=	32.5 ~~g NH₃~~	x l mol NH₃	x 1	=	1.9072 M( mol/L)
			17.04 ~~g NH₃~~	1 L

Can be calculated from the following For a weak base (H₃) with K_b the value of K_a for the corresponding conjugate acid (NH₄⁺or NH₄Cl) expression:

K_b x K_a = K_w

K_a

K_w

_-------

K_b

Acid-Base Titrations

A titration is a procedure used in analytical chemistry to determine the amount or concentration of a substance. In a titration one reagent, the titrant, is added to another slowly. As it is added a chemical stoichiometric reaction occurs until one of the reagents is exhausted, and some process or device signals that this has occurred. The purpose of a titration is generally to determine the quantity or concentration of one of the reagents, that of the other being known beforehand. In any titration there must be a rapid quantitative reaction taking place as the titrant is added, and in acid-base titrations this is a stoichiometric neutralization. The type of titration is simply the type of chemical reaction taking place, and so in this section we consider acid-base titrations.

Acid-Base Indicators

Many substances, including litmus, the one dye almost everyone associates with acids and bases, change color in response to acid or base. The pigment in red cabbage is another natural substance very commonly used to show color change. Phenolphthalein is one of the most common indicators used for beginning chemistry, because its color change is very obvious which makes it easy to use. There are many other indicators that change colors at different pH's, and so are useful for different purposes. pH paper commonly contains a mixture of different indicators that change colors at different pH's. The mixture is applied to paper, and then compared to a color chart to see what the pH of a solution is, approximately.

Acid-Base indicators are dyes that are themselves weak acids and bases. However, the conjugate acid-base forms of the dye have different colors. The actual chemical structures of the dyes is often quite complex; however, we can use the generic symbol for the indicator as HIn. The Brönsted-Lowry equation for the indicator is:

HIn + H₂O H₃O⁺ + In^-

where the color of the letters is used to show the differently colored forms of the dyes.

Suppose that we increase the concentration of H₃O⁺. Then if we apply LeChatelier's principle, the result should be more HIn.
There will be more H₃O⁺ (because we added more), but the system will respond to:

use up some of the added H₃O⁺, at the same time
removing some of the In^-, and
making more HIn (and H₂O, though that is pretty irrelevant since the reaction takes place in lots of water anyways).

Thus we will see the color of the HIn form.

Now suppose that we decrease the concentration of H₃O⁺(which we could do by adding more OH^-). Now applying LeChatelier's principle, the result should be more In^-.

There will be less H₃O⁺ (because we removed it), but the system will respond to try and:

make some more H₃O⁺, at the same time
removing some of the HIn, (and some of the water, but there is so much present that this is pretty irrelevant) and
making more In^-.

Thus we will see the color of the In^- form.

In summary: at a low pH, an indicator is almost entirely in the HIn form. As the pH increases, the intensity of the color of In^- increases as the equilibrium shifts to the right.
Different dyes will change color at different pH's (the value can be calculated from the equilibrium constant for the indicator).. Here is a small sample of some common acid-base indicators, and the range at which their pH changes color. One of the difficulties with giving a range of colors is that different person's eyes are not all equally sensitive (also different monitors will display colors differently), so these colors are only approximations.

When conducting a titration, one must select the proper indicator so that its pH range will match the equivalence point of the titration. Also, you must use an indicator that changes color obviously, so that it can be detected easily. This is why phenolphthalein is so often used for strong acid-strong base titrations.

The equivalence point is the point at which the amount of H₃O⁺ and OH^- are equal. It is important to select an indicator which changes color -- whose endpoint -- is near the equivalence point.

Using pH titration curves in section 17.2 page 786 explain the selection process of an indicator for a titration involving the following:

a) strong acid/strong base

b) weak acid/strong base

c) strong acid/weak base

d) weak acid/weak base

Titrations are experiments to calculate the concentration of an unknown solution. This method is one that is used in volumetric analysis (analysis using volume of solutions). There are few terms or definition that you should know to understand the titration procedure.

Titration: A titration experiment uses a burette and an Erlenmeyer flask.

Equivalence Point: The end point when certain volumes of the acid and the base is mixed together to react completely forming water and a salt. The equivalence point is often marked by a color change of an indicator. Indicators change color at a certain pH range. Different indicators are used based on the pH of the salt formed at the end or equivalence point.

a) Strong acid/strong base

E.g. 0.1 M HCl and 50 mL of 0.1 NaOH solution.

At the equivalence point pH =7 since the salt NaCl form is neutral in solution. Last drop of HCl added will change the pH of the solution in wide range 3-11. Therefore, any indicator that changes color in this (pH 3-11 range) could be used to get the equivalence point accurately. The pH rage for the color change for a list of indicators are given in the textbook. Wide range of indicators (pH, 3-11) could be used in this titration.

b) weak acid/strong base

E.g. 1.0 M HC₂H₃O₂ and 50 mL of 1.0 NaOH solution.

At the equivalence point the concentration of the salt solution of NaC₂H₃O can be calculated as shown below:

Concentration of both HC₂H₃O₂and NaOH solutions are equal to 0.10 M. The volumes of HC₂H₃O₂and NaOH that will react at the equivalence point will also be equal since both HC₂H₃O₂and NaOH have one H and OH groups to produce water. Therefore, the final volume of the mixture would be 100 mL (50 mL 0.10 M HC₂H₃O₂ + 50 mL 0.10 M NaOH = 100 mL salt solution) The amount of NaC₂H₃O₂ formed is equal to either HC₂H₃O₂ or NaOH added. Basically volume of the final solution has been doubled. Therefore, the concentration NaC₂H₃O₂ is half the concentration of either HC₂H₃O₂ or NaOH solution. i.e. 0.10 M ) 2 = 0.05 M NaC₂H₃O₂.

To calculate the pH of 0.05 M NaNC₂H₃O₂ (Similar to a previous problem in Chapter 16:What is the pH of a 0.05 M NaC₂H₃O₂ salt solution?)

At the equivalence point pH =8.72 since the salt NaC₂H₃O₂ forms a basic solution. Last drop of HC₂H₃O₂added will change the pH of the solution in the range 8-10. Therefore, any indicator that changes color in this (pH 8-10) range could be used to get the equivalence point accurately. The pH rage for the color change for a list of indicators are given in the textbook

The indicators, thymol blue (8.0-9.6), phenolphthalein (8.2-10.0) and thymolphthalein (9.4-10.6) could be used in this titration.

c) Strong acid/weak base

E.g. 1.0 M HCl and 50 mL of 1.0 NH₄OH (NH₃) solution.

At the equivalence point the concentration of the salt solution of NH₄Cl can be calculated as shown below:

Concentration of both HCland NH₄OH solutions are equal to 1.0 M. The volumes of HCland NH₄OH that will react at the equivalence point will also be equal since both HCland NH₄OH have one H and OH groups to produce water. Therefore, the final volume of the mixture would be 100 mL (50 mL 1.0 M HCl + 50 mL 1.0 M NH₄OH = 100 mL salt solution). The amount of NH₄Cl formed is equal to either HCl or NH₄OH added. Basically volume of the final solution has been doubled. Therefore the concentration NH₄Cl is half the concentration of either HCl or NH₄OH solution.

i.e. (1.0 M )/ 2 = 0.5 M NH₄Cl.

To calculate the pH of 0.5 M NH₄Cl (see problem 5a: What is the pH of a 0.5 M NH₄Cl salt solution)

At the equivalence point pH = 4.78 since the salt NH₄Cl forms an acidic solution. Last drop of HCladded will change the pH of the solution in the range 4-6. Therefore, any indicator that changes color in this (pH 4-6) range could be used to get the equivalence point accurately. The pH rage for the color change for a list of indicators are given in the textbook.

The indicator, methyl red (4.8-6.0) could be used in this titration.

d) weak acid/weak base

E.g. 1.0 M HC₂H₃O₂and 50 mL of 1.0 NH₄OH (NH₃) solution.

The solution is neutral at the equivalence point because the salt of a weak acid and a weak base forms a neutral, a basic or an acidic solution depending on the relative strengths of the acid and the base. Since both HC₂H₃O₂and and NH₄OH (NH₃) have identical Ka and Kb values their strength are equal making the solution neutral. The last drop of HC₂H₃O₂added will change the pH of the solution in the range 6-8. Therefore, any indicator that changes color in this (pH 4-6) range could be used to get the equivalence point accurately. The pH rage for the color change for a list of indicators are given in page 675, section 15.9 of Ebbing.

The indicators, litmus (5.0-8.0) and bromothymole blue (6.0-7.6) could be used in this titration.

`Titration Curves of Polyprotic Acids

Similar to monoprotic acids, except that two equivalence points are seen. The equivalence points of a polyprotic acid/ base reaction can be determined using an indicator. In alternative experiment way monitoring the changes in pH that occurs during the titration of a weak polyprotic acid with a strong base could also be used to find equivalence points. At the equivalence point one should expect to see a dramatic change n pH as the solution goes from acidic to strongly basic.

Depicted on the left is an idealized pH titration curve for a weak diprotic acid such as H₂SO₃. The first thing that you should notice is that there are two regions where we see a significant pH change. These, if you wish, correspond to two separate titrations. Titration 1 is the reaction of the first proton with the base (in this case sodium hydroxide)

Solubility of Salts

Solubility of a solute in a solvent is the number of grams of solute necessary to saturate 100 grams of solvent at a particular temperature.

Solubility product:

Solubility product is defined as the product of ionic concentration when dissolved ions and un-dissolved ions are in equilibrium.

In other words, When a saturated solution of sparingly or slightly soluble salt is in contact with un- dissolved salt, an equilibrium is established between the dissolved ions and the ions in the solid phase of the un-dissolved salt. Ionic product at this stage is called solubility product.

Symbol: It is denoted by K_sp

Determination of solubility product:

M_aX_b(aq) a M^c+(aq) + b X^d-(aq)

K_sp=

[M^c+]^a[X^d-]^b

Consider a slightly soluble salt such as silver chloride (AgCl).

AgCl(s) Ag⁺(aq) + Cl^-(aq)

Applying equilibrium law:

K_c=	[Ag⁺][Cl^-]
	[AgCl]

K_c [AgCl] = K_sp =

[Ag⁺][Cl^-]

Since AgCl is a solid there is no change in the concentration of salt (AgCl) at equilibrium.

Therefore

[AgCl] = C

K_c x C = [Ag⁺][Cl^-]

Let K_c x C = K_sp

Therefore

K_sp = [Ag⁺][Cl^-]

Ionic Product or Reaction Quotient (Q_sp):

As we discussed already under equilibrium reactions the product of ionic concentration other than equilibrium is called reaction quotient or ionic product.

Applications of solubility product:

Knowledge of solubility product is very useful, to determine whether precipitates will be obtained or not by the addition of more amount of solute to the solution. There are three conditions:

When K_sp> ionic product (Q_sp) (unsaturated):

If solubility product is greater than the ionic product then reaction will shift to right and more will be dissolved and no precipitate will form.

When K_sp< ionic product (Q_sp) (supersaturated):

If solubility product is less than the ionic product then reaction will shift to left and precipitate will form.

When K_sp= ionic product (Q_sp):

In this condition solution is in equilibrium with solid salt and ions and further addition will cause precipitates.

The Common Ion Effect and Precipitation Reactions

The solubility of insoluble substances can be decreased or made to precipitate by the presence of a common ion. AgCl will be our example.

Present in silver chloride are silver ions (Ag⁺) and chloride ions (Cl¯). Silver nitrate (which is soluble) has silver ion in common with silver chloride. Sodium chloride (also soluble) has chloride ion in common with silver chloride.

In fact, mixing sufficiently concentrated solutions of AgNO₃ and NaCl will produce a precipitate of AgCl. In order to be sufficiently concentrated, the product of the [Ag⁺] and the [Cl¯] must exceed the K_sp of 1.77 x 10¯¹⁰.

Here is the example problem: AgCl will be dissolved into a solution with is ALREADY 0.0100 M in chloride ion. What is the solubility of AgCl?

In solubility problems, the source of the chloride is from soluble chloride salt, NaCl because very little is coming from the insoluble salt, AgCl. Let us assume the chloride came from some dissolved sodium chloride, sufficient to make the solution 0.0100 M. So, on to the solution .

The dissociation equation for AgCl is:

AgCl (s) Ag⁺ (aq) + Cl¯ (aq)

The K_sp expression is:

K_sp = [Ag⁺] [Cl¯]

This is the equation we must solve. First we put in the K_sp value:

1.77 x 10¯¹⁰ = [Ag⁺] [Cl¯]

Now, we have to reason out the values of the two guys on the right. The problem specifies that [Cl¯] is already 0.0100. I get another 'x' amount from the dissolving AgCl. Of course, [Ag⁺] is 'x.'

Substituting, we get:

1.77 x 10¯¹⁰ = (x) (0.0100 + x)

This will wind up to be a quadratic equation which is solvable via the quadratic formula. However, there is a chemical way to solve this problem. We reason that 'x' is a small number, such that '0.0100 + x' is almost exactly equal to 0.0100. If we were to use 0.0100 rather than '0.0100 + x,' we would get essentially the same answer and do so much faster. So the problem becomes:

1.77 x 10¯¹⁰ = (x) (0.0100)

and

x = 1.77 x 10¯⁸ M

There is another reason why neglecting the 'x' in '0.0100 + x' is OK. It turns out that measuring K_sp values are fairly difficult to do and, hence, have a fair amount of error already built into the value. So the very slight difference between 'x' and '0.0100 + x' really has no bearing on the accuracy of the final answer. Why not? Because the K_sp already has significant error in it to begin with. Our "adding" a bit more error is insignificant compared to the error already there.

Solubility Product (K_sp) of Slightly Soluble Ionic Compounds

A very soluble ionic salt (e.g., NaCl) dissolves and completely dissociates into Na⁺ (aq) and Cl^‑ (aq). Some salts only slightly soluble, (as given by solubility rules) and an equilibrium exists between dissolved and undissolved compound. Consider the addition of PbSO₄ (s) to water.

PbSO₄ (s) ⇌ Pb²⁺ (aq) + SO₄^2‑ (aq)

If the reaction hasn’t reached equilibrium, the reaction quotient Q_sp is

Q_sp = [Pb²⁺] [SO₄^2‑]

Q_sp = “ion-product expression”

At equilibrium, Q_sp = K_sp

K_sp = [Pb²⁺][SO₄^2‑]

K_sp = “Solubility Product Constant”

We normally only consider systems at equilibrium

\ use K_sp (not Q_sp).

Examples:

Cu(OH)₂ (s) ⇌ Cu²⁺ (aq) + 2 OH^‑ (aq) K_sp = [Cu²⁺][OH^‑]²

CaCO₃ (s) ⇌ Ca²⁺ (aq) + CO₃^2‑ (aq) K_sp = [Ca²⁺][CO₃^2‑]

Ca₃(PO₄)₂ (s) ⇌ 3 Ca²⁺ (aq) + 2 PO₄^3‑ (aq) K_sp = [Ca²⁺]³[PO₄^3‑]²

Note: The S^2‑ (aq) is very unstable in water and converts to HS^‑ and gives OH^‑ (aq).

\ MnS (s) + H₂O (l) ⇌ Mn²⁺ (aq) + HS^‑ (aq) + OH^‑ (aq)

[i.e., S^2‑ (aq) + H₂O (l) ® HS^‑ (aq) + OH^‑ (aq)]

\ K_sp = [Mn²⁺][HS^‑][OH^‑]

The greater is K_sp, the more soluble the substance is.

e.g., PbSO₄ K_sp = 1.6 x 10^‑8 insoluble

CoCO₃ K_sp = 1.0 x 10^‑10 more insoluble

Fe(OH)₂ K_sp = 4.1 x 10^‑15 most insoluble

Calculations Involving Solubility Products

Two types: Use K_sp to find conc of dissolved ions

Use concs to find K_sp.

Make sure the equations are balanced!!

Example: The solubility of Ag₂CO₃ is 0.032 M at 20 °C. What is K_sp of Ag₂CO₃?

This is a common type of question—Note that we are told the molar solubility of Ag₂CO₃, but of course it will dissociate into ions.

Therefore,

Ag₂CO₃ (s) ® 2 Ag⁺ (aq) + CO₃^2‑ (aq)

init (solid) 0 0

change -0.032 M +0.064 M +0.032 M

[equil] (solid) 0.064 M 0.032 M

K_sp = [Ag⁺]²[CO₃] = (0.064)²(0.032)

\ K_sp = 1.3 x 10^‑4

Example: The solubility of Zn (oxalate) is 7.9 x 10^‑3 M at 18 °C. What is its K_sp?

Zn (ox) ⇌ Zn²⁺ (aq) + ox^2‑ (aq)

init (solid) - -

change -7.9 x 10^‑3 M +7.9 x 10^‑3 M +7.9 x 10^‑3 M

[equil] (solid) 7.9 x 10^‑3 M 7.9 x 10^‑3 M

K_sp = [Zn²⁺][ox] = (7.9 x 10^‑3)²

K_sp = 6.2 x 10^‑5

Example: What is the molar solubility of SrCO₃? (K_sp = 5.4 x 10^‑10)

SrCO₃ (s) ⇌ Sr²⁺ (aq) + CO₃^2‑ (aq)

init (solid) 0 0

change -x +x +x

[equil] (solid) x x

K_sp = x² \ x =

\ x = 2.3 x 10^‑5 M

\ Solubility of SrCO₃ is 2.3 x 10^‑5 M

Example: What is the molar solubility of Ca(OH)₂ in water? (K_sp = 6.5 x 10^‑6).

Ca(OH)₂ (s) ⇌ Ca²⁺ (aq) + 2 OH^‑ (aq)

init (solid) 0 0

change -x +x +2x

[equil] (solid) x 2x

K_sp = 6.5 x 10^‑6 = x(2x)² = 4x³ (careful!)

\ Solubility of Ca(OH)₂ = 1.2 x 10^‑2 M

To obtain the of a number, learn to use the button on your calculator or take the log, divide by x, then antilog.

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pH for buffer solutions	concentration of ions in buffer solutions	pH for solutions with common ions
preparing a buffer of a desired pH	pH indicators	determine the color of an indicator at certain pH
titration curve	selecting an indicator for an titration	acid rain
Solubility product	K_sp expression	Factors affecting solubility of ionic compounds
Apply Le Chatelier's solubility problems	K_sp to determine precipitation of a salt	Complex ion formation