Chemical Analysis in Solutions II

Outline

• Determining and Describing Solution Concentrations
  • Determining Concentrations
  • Solution Stoichiometry
  • Solutions and Preparations
  • Titrations
• Discussion Questions
• Optional Website Reading

Determining and Describing Solution Concentrations

Solutions are liquid mixtures of chemicals. The solute is the chemical which is dissolved, and is sometimes solid outside of the solution. The solvent is the dissolving agent, which causes the solute to dissociate, and is water in all the solutions discussion in this chapter (hence aqueous, from aqua). The solution is the combination of both the solvent and the solute. Get these terms straight, because determining concentrations of solutes (and we could have more than one) and solvents depends on your understanding not only which is which, but how they relate to the whole solution. We can use these relationships to create solutions of specific concentrations for different experiments or industrial purposes.

Determining Concentrations

Note that molarity is the quantity of solute in the whole solution, not the ratio of solute to the solvent. The symbol for molarity is M, usually in italics, and there are several conventions for indicating concentrations this way. We can say that a solution with a molarity of 12 is an 12-Molar solution, or that M = 12. The shorthand for "concentration of" is brackets around the molecule, so [HCl] = 12 M means "the concentration of HCl is 12 moles of HCl per liter of solution".

If we know any two of the three quantities, we can determine the other two. For example, if we have 12 M hydrochloric acid and we need .10 moles of HCl for our reaction, we have molarity and moles of solute, and we need to figure out the liters of solution to use. Rearranging the equation first gives

The amount of solution necessary is 8.3 mL of 12 M HCl.

Or we can determine the mass of a compound or solute. Suppose we have 125 mL of a 6 M sodium hydroxide solution. How much NaOH (molecular weight = 22.99 + 16.01 + 1.008 = 40.008g/mole NaOH) is in the solution? We have to remember to convert mL to liters along the way, because molarity is in moles/liter:

So if we wanted to make 125 mL of a 6 M solution of NaOH, we would need to combine 30g NaOH with enough water to make 125 mL altogether. Notice that we do not combine the 30g NaOH with 125 mL water; that would give us more than 125 mL of total solution, and the concentration would be less than 6 M.

We could also start from a compound which dissociates in solution, and determine the molar concentrations of the dissociated ions. Given .100 M Al₂(SO₄)₃, how how much Al³⁺ and SO₄^2- are produced? Here we have to remember to write the dissociation reaction formula so that we can get the correct number of moles:

There is .100 moles/liter of the compound in the solution, but 2 moles of silver ion for every one of the compound. So the solution has .200 moles/liter of silver ion, or [Al³⁺] = .200 M. Likewise, 3 ions of sulfate are produced for each molecule of silver sulfate put into the solution, so [SO₄^2-] = .300 M.

In most chemistry labs, the stockroom contains very concentrated forms of solutions, from which the researcher can make solutions of any lesser concentration by diluting the stock solution with distilled (pure) water. Now, by definition,

We solve this for moles of solute:

The moles of solute is unchanged when we perform a dilution —only the amount of solvent changes. So we can set up an equivalence, where C is the concentrated solution and D is the dilute solution:

or in math notation, using V for volume in liters, c_c * V_c = c_d * V_d.

You may also see this written as initial molarity * initial volume = final molarity * final volume, M_i*V_i = M_f*V_f. Use whichever notation makes the most sense to you; just be consistent!

Suppose we have stock (essentially, an unlimited amount) of 18 M HCl and we need 100ml of 12M HCl for an experiment. We would would have the equivalence

We would need take .067L or 67mL of the 18M HCl and add enough water to make 100mL of solution, which would have the required 12M concentration.

Molality

The concept molality is sometimes used when determining the total solution may be difficult. Molality is the ratio of the moles of solute to volume of solvent. In most cases, molality will be slight less than molarity. If the addition of the solute to solvent does not increase the volume of the resulting solution much over the original solvent volume, molality may be used as a close approximation to molarity.

Solution Stoichiometry

We can get from concentration and volume information (molarity and liters) to mass amounts by first determining the moles of compounds involved and then converting moles to gram weights from the molecular or atomic weights of the compounds and atoms. Suppose we combine two aqueous solutions and a precipitate forms -- I'll expand on example 4.8 from the book, in which NaOH and Fe(NO₃3 form a red gel. We want to know how much precipitate, in grams. We know the concentrations of each solution, and the amount of each that we used: 50 mL of 0.200 M NaOH and 30.00 mL of 0.125 M Fe(NO₃3. What are the steps for determining the mass of the precipitate?

1. Write the net ionic equation for the reaction. That is, determine which ions from the solutions combined to form the precipitate, and balance the equation. We must do this in order to determine the moles of each ion which are available to form the precipitate. Nitrates are always soluble (rules of solution, p.76). So the precipitate forms from the iron and hydroxide, since the later is always insoluble with iron. Balancing the 3 positive charges on the iron ion against the hydroxide ion's single negative charge gives:
   Fe³⁺ + 3 OH^- --> Fe(OH)₃
2. Now we can determine the number of moles of iron ion present. We start with the amount of solution (given in mL), and multiply it by L/mL to get liters so that we can use molarity (always in listers) to determine the moles involved. Then we multiply that by the moles of iron ions per mole of compound to get the moles of iron ions available to form the precipitate (assuming that the iron nitrate has completely dissolved in solution):
   30 mL Fe(NO₃3 * (1L/1000mL) * (0.125 moles Fe(NO₃3/1 L Fe(NO₃3) * (1 mole Fe³⁺/mole Fe(NO₃3) = 3.75 * 10^-3 moles Fe³⁺
3. We do the same thing for the hydroxide ion, using the concentrations of NaOH:
   50 mL NaOH * (1L/1000mL) * (0.200 moles NaOH/1 L NaOH) * (1 mole OH^-/mole NaOH) = 1.00 * 10^-2 moles OH^-
4. Be sure to double check: do all the units cancel so we come out with moles of the ion?
5. Now we have to determine which is the limiting ion. If all the iron were used, how much iron hydroxide would form?
   3.75 * 10^-3 moles Fe³⁺ * mole Fe(OH)₃/mole Fe³⁺ = 3.75 * 10^-3 * mole Fe(OH)
6. If all the hydroxide were used, how much iron hydroxide would form?
   >1.00 * 10^-2 moles OH * mole Fe(OH)₃/3 moles OH^- = 3.33 * 10^-3 moles Fe(OH)₃
7. Fewer moles of precipitate form when we consider the amount of OH^- available, so it is the limiting reactant. We can get at most 3.33 * 10^-3 moles Fe(OH)₃; the unused iron ions will remain in the solution.
8. Now we can determine the grams of precipitate formed, since we can calculate the molar mass and we know the number of moles of the precipitate.
   55.85 + 3 * 16.00 + 3 * 1.008 = 106.87 g/mole Fe(OH)₃ 3.33 * 10^-3 moles Fe(OH)₃ * 106.87 g/mole Fe(OH)₃ = .356 g Fe(OH)₃

Solutions and Preparations

We often want to to determine the amount of a solution required to react with some specific amount of compound or another solution. For example, suppose we want to make copper hydroxide. The net ionic equation is

Cu²⁺ + 2 OH^- ---> Cu(OH)₂

We have on hand 50.0 mL of 0.100 M NaOH, and a supply of 0.200 M CuSO₄. How much of the copper sulfate must we use?

1. First, determine the moles of OH^- available. This is our limiting reactant, because we have "enough" copper sulfate available in the lab.
   50.0 mL * 1L/1000mL * 0.100 mole NaOH/Lister * mole OH^-/mole NaOH = 5 * 10^-3 moles OH_-
2. We need half as many copper ions as hydroxide ions: so 2.5 * 10^-3 moles copper.
3. We need to find the volume of 0.200 M copper sulfate solution which contains 2.5 * 10^-3 moles copper. So we multiply to moles of copper we need times the moles copper sulfate per moles copper ion, and then multiply that by the inverse of the molarity of solution we have, so all the units come out: 2.5 * 10_-3 moles Cu²⁺ * (mole CuSO₄/mole Cu²⁺) * liter/.200 mole CuSO₄ = .0125 L CuSO₄ solution, or 12.5 mL of solution.

Titrations

In a titration experiment, you add a solution of known concentration drop by drop to a solution of unknown concentration until a reaction occurs. The reaction must be dependent on the relative concentrations of the two solutions, so that we can make some assumptions about how the concentrations are related in order to determine the unknown one. The point at which the reaction occurs is the end of your titration activity (endpoint) and the beginning of your calculation activity.

Consider the titration of a known volume of basic solution, NaOH, which has an unknown concentration. To this we add, a drop at a time, some 0.150 M (fairly weak) HCl. The basic solution also contains a litmus solution as well; this is currently blue but will turn red as soon as the concentration of H⁺ ions from the HCl overbalances the OH^- ion excess due to the NaOH. We want to add just enough HCl to turn the solution as close to clear (faint pink) as possible; then the solution is neutralized and we have added as many H⁺ as there were excess OH^- to start with.

1. First we write the equation for the reaction:

   NaOH(aq) + HCl(aq) --> NaCl(aq) + H₂O(l)

2. Then we perform the titration. 27.25 ml of the HCl solution appears to neutralize 30 mL of NaOH.
3. We determine the moles of HCl involved:
   .150 m/L * (27.25 * 10^-3L) = 4.09 * 10^-3 moles
4. Since there is a 1:1 relationship in the net ionic equation between the OH and H involved, we can simply write 4.09 * 10^-3 moles for the NaOH present in the 30 mL of solution.
5. We know how many moles of NaOH, and we know how much solution of NaOH were used, so we can now determine the concentration:
   [NaOH] = 4.09*10^-3 moles/30mL * ( 1000mL/L ) = .136 moles/Liter or .136M

Practice with the Concepts

Discussion Questions

• Why doesn't the empirical formula equal the molecular formula in all cases?
• What is the equivalence point of a titration? Why do we care?
• How can we use indicators?

Optional Readings

• Practice with more chemical analysis at the University of Wisconsin Stoichiometry site.
• A very brief introduction to the basic methods of chemical analysis from the University of Vermont Chemistry department gives you a better idea of the complexity of this part of chemistry.

© 2005 - 2024 This course is offered through Scholars Online, a non-profit organization supporting classical Christian education through online courses. Permission to copy course content (lessons and labs) for personal study is granted to students currently or formerly enrolled in the course through Scholars Online. Reproduction for any other purpose, without the express written consent of the author, is prohibited.