SAT Chemistry Acids, Bases, and Salts - Indicators
SAT Chemistry Acids, Bases, and Salts - IndicatorsSome indicators can be used to determine pH because of their color changes somewhere along this pH scale. Some common indicators and their respective color changes are given below.
Here is an example of how to read this chart: At pH values below 4.5, litmus is red; above 8.3, it is blue. Between these values, it is a mixture of the two colors.
In choosing an indicator for a titration, we need to consider if the solution formed when the end point is reached has a pH of 7. Depending on the type of acid and base used, the resulting hydrolysis of the salt formed may cause it to be slightly acidic, slightly basic, or neutral. If the titration is of a strong acid and a strong base, the end point will be at pH 7 and practically any indicator can be used. This is because the addition of 1 drop of either reagent will change the pH at the end point by about 6 units. For titrations of strong acids and weak bases, we need an indicator, such as methyl orange, that changes color between 3.1 and 4.4 in the acid region. In the titration of a weak acid and a strong base, we should use an indicator that changes in the basic range. Phenolphthalein is a suitable choice for this type of titration because it changes color in the pH 8.3 to 10.0 range.
Knowledge of the concentrations of solutions and the reactions they take part in can be used to determine the concentrations of “unknown” solutions or solids. The use of volume measurement in solving these problems is called titration.
A common example of a titration uses acid-base reactions. If you are given a base of known concentration, that is, a standard solution, let us say 0.10 M NaOH, and you want to determine the concentration of an HC1 solution, you could titrate the solutions in the following manner.
First, introduce a measured quantity, 25.0 milliliters, of the NaOH into a flask by using a pipet or burette in a setup like the one in the accompanying diagram. Next, introduce 2 drops of a suitable indicator. Because NaOH and HC1 are considered a strong base and a strong acid, respectively, an indicator that changes color in the middle pH range would be appropriate. Litmus solution would be one choice. It is blue in a basic solution but changes to red when the solution becomes acidic. Slowly introduce the HC1 until the color change occurs. This point is called the end point. The point at which enough acid is added to neutralize all the standard solution in the flask is called the equivalence point.
Suppose 21.5 milliliters of HCl was needed to produce the color change. The reaction that occurred was
H+(aq) + OH-(aq) → H2O(l)
until all the OH- was neutralized; then the excess H+ caused the litmus paper to change color. To solve the question of the concentration of NaOH, this equation is used:
Macid × Vacid = Mbase × Vbase
Substituting the known amounts in this equation gives
x Macid × 21.5 mL = 0.1 M × 25.0 mL
x = 0.116M
In choosing an indicator for a titration, we need to consider whether the solution formed when the end point is reached has a pH of 7. Depending on the types of acid and base used, the resulting hydrolysis of the salt formed may cause the solution to be slightly acidic, slightly basic, or neutral. If a strong acid and a strong base are titrated, the end point will be at pH 7, and practically any indicator can be used because adding 1 drop of either reagent will change the pH at the end point by about 6 units. For titrations of strong acids and weak bases, we need an indicator, such as methyl orange, that changes color between 3.1 and 4.4 in the acid region. When titrating a weak acid and a strong base, we should use an indicator that changes in the basic range. Phenolphthalein is the suitable choice for this type of titration because it changes color in the pH 8.3 to 10.0 range.
The process of the neutralization reaction can be represented by a titration curve like the one below, which shows the titration of a strong acid with a strong base.
Example 1 _______________________________________________
Find the concentration of acetic acid in vinegar if 21.6 milliliters of 0.20 M NaOH is needed to titrate a 25-milliliter sample of the vinegar.
Using the equation Macid × Vacid = Mbase × Vbase, we have
x Macid × 25 mL = 0.20 × 21.6 mL
x = 0.17 Macid
Another type of titration problem involves a solid and a titrated solution.
Example 2 _______________________________________________
A solid mixture contains NaOH and NaCl. If 10.0 milliliters of 0.100 M HC1 is required to titrate a 0.100-gram sample of this mixture to its end point, what is the percent of NaOH in the sample?
Since 1 mol of HCl neutralizes 1 mol of NaOH, 0.001 mol of NaOH must be present in the mixture. Since 1 mol NaOH = 40.0 g then
- 01 mol × 40.0 g/mol = 0.04 g NaOH
Therefore, 0.04 g of NaOH was in the 0.100-g sample of the solid mixture. The percent is 04 g/0.100 gx 100 = 40%.
In the explanations given to this point, the reactions that took place were between monoprotic acids (single hydrogen ions) and monobasic bases (one hydroxide ion per base). This means that each mole of acid had 1 mole of hydrogen ions available, and each mole of base had 1 mole of hydroxide ions available, to interact in the following reaction until the end point was reached:
H+(aq) + OH-(aq) → 2H2O(l)
This is not always the case, however, and it is important to know how to deal with acids and bases that have more than one hydrogen ion and more than one hydroxide ion per formula. The following is an example of such a problem.
Example 3 __________________________________________________
If 20.0 milliliters of an aqueous solution of calcium hydroxide, Ca(OH)2, is used in a titration, and an appropriate indicator is added to show the neutralization point (end point), the few drops of indicator that are added can be ignored in the volume considerations. Therefore, if 25.0 milliliters of standard 0.050 M HC1 is required to reach the end point, what was the original concentration of the Ca(OH)2 solution?
The balanced equation for the reaction gives the relationship between the number of moles of acid reacting and the number of moles of base:
The mole relationship here is that the number of moles of acid is twice the number of moles of base:
No. of moles of acid = 2 × No. of moles of base
↑ mole factor
Since the molar concentration of the acid times the volume of the acid gives the number of moles of .acid:
Ma × Va = moles of acid
and the molar concentration of the base times the volume of the base gives the number of moles of base:
Mb × Vb = moles of base
then, substituting these products into the mole relationship, we get
MaVa = 2MbVb
Solving for Mb gives
Buffer solutions are equilibrium systems that resist changes in acidity and maintain constant pH when acids or bases are added to them. A typical laboratory buffer can be prepared by mixing equal molar quantities of a weak acid such as HC2H3O2 and its salt, NaC2H3O2. When a small amount of a strong base such as NaOH is added to the buffer, the acetic acid reacts (and consumes) most of the excess OH- ion. The OH- ion reacts with the H+ ion from the acetic acid, thus reducing the H+ ion concentration in this equilibrium:
HC2H3O2(aq) ⇌ H+(aq) + C2H3O2-(aq)
This reduction of H+ causes a shift to the right, forming additional C2H3O2- ions and H+ ions. For practical purposes, each mole of OH- added consumes 1 mole of HC2H3O2 and produces 1 mole of C2H3O2- ions.
When a strong acid such as HCl is added to the buffer, the H+ ions react with the C2H3O2- ions of the salt and form more undissociated HC2H3O2. This does not alter the H+ ion concentration. Proportional increases and decreases in the concentrations of C2H3O2- and HC2H3O2 do not significantly affect the acidity of the solution.
A salt is an ionic compound containing positive ions other than hydrogen ions and negative ions other than hydroxide ions. The usual method of preparing a particular salt is by neutralizing the appropriate acid and base to form the salt and water.
Five methods for preparing salts are as follows:
1. Neutralization reaction. An acid and a base neutralize each other to form the appropriate salt and water. For example:
2HCl(aq) + Ca(OH)2(aq) → CaCl2(aq) + 2H2O(l)
acid + base → salt + water
2. Single replacement reaction. An active metal replaces hydrogen in an acid. For example:
Mg(s) + H2SO4(aq) → MgSO4(aq) + H2(g)
3. Direct combination of elements. An example of this method is the combination of iron and sulfur. In this reaction small pieces of iron are heated with powdered sulfur:
Fe(s) + S(s) → FeS(s)
iron (II) sulfide
4. Double replacement. When solutions of two soluble salts are mixed, they form an insoluble salt compound. For example:
AgNO3(aq) + NaCl(aq) → NaNO3(aq) + AgCl(s)
5. Reaction of a metallic oxide with a nonmetallic oxide. For example:
MgO(s) + SiO2(s) → MgSiO3(s)
The naming of salts is discussed on page 117.
Some substances, such as the HCO3- ion, the HSO4- ion, the H2O molecule, and the NH3 molecule, can act as either proton donors (acids) or proton receivers (bases), depending upon which other substances they come into contact with. These substances are said to be amphoteric. Amphoteric substances donate protons in the presence of strong bases and accept protons in the presence of strong acids.
Examples are the reactions of the bisulfate ion, HS04-:
With a strong acid, HSO4 accepts a proton:
HSO4-(aq) + H+(aq) → H2SO4(aq)
With a strong base, HSO4 donates a proton:
HSO4-(aq) + OH-(aq) → H2O(ℓ ) + SO42- (aq)
ACID RAIN—AN ENVIRONMENTAL CONCERN
Acid rain is currently a subject of great concern in many countries around the world because of the widespread environmental damage it reportedly causes. It forms when the oxides of sulfur and nitrogen combine with atmospheric moisture to yield sulfuric and nitric acids— both known to be highly corrosive, especially to metals. Once formed in the atmosphere, these acids can be carried long distances from their source before being deposited by rain. The pollution may also take the form of snow or fog or be precipitated in dry form. This dry form is just as damaging to the environment as the liquid form.
The problem of acid rain can be traced back to the beginning of the industrial revolution, and it has been growing ever since. The term “acid rain” has been in use for more than a century and is derived from atmospheric studies made in the region of Manchester, England.
In 1988, as part of the Long-Range Transboundary Air Pollution Agreement sponsored by the United Nations and the United States, along with 24 other countries, a protocol freezing the rate of nitrogen oxide emissions at 1987 levels was ratified. The 1990 amendments to the Clean Air Act of 1967 put in place regulations to reduce the release of sulfur dioxide from power plants to 10 million tons per year by 2000. That achieved a 20 percent decrease in sulfur dioxide. The attempts continue through international organizations to further clean the air.
These equations show the most common reactions of sulfur- and nitrogen-containing gases with rainwater. The sulfur dioxide reacts with rainwater to form sulfuric acid solutions:
2SO2(g) + O2(g) → 2SO3(g)
SO3(g) + H2O(ℓ ) → H2SO4(aq)
The oxides of nitrogen react to form nitrous and nitric acid:
2NO2(g) + H2O(ℓ ) → HNO2(aq) + HNO3(aq)