Chemical Calculations for Solutions (Ch 12) Dr. Harris Lecture 12 Suggested HW: Ch 12: 1, 10, 15, 21, 53, 67, 81

Solutions As we learned in a previous chapter, solutions are homogenous mixtures, meaning that the components comprising the solution are uniformly dispersed The most common type of solution is a solid dissolved in a liquid. The dissolved solid is the solute, the liquid is the solvent. Solutes and solvents do not react, merely co-exist, as is the case with an aqueous solution like NaCl(aq) NaCl (s) -----> NaCl(aq) H 2 O (L)

Solubility When NaCl is dissolved, the ions are surrounded by water molecules, and dipole interactions disperse causes the ions to separate and disperse As more and more NaCl is added, a point is reached where further dissociation ceases, the salt simply drops to the bottom of the beaker The solution is now saturated The quantity of NaCl dissolved at this point is its solubility. Figure Above: Dissolution of NaCl and uniform distribution of solute and solvent.

Electrolytes Reminder: Strong electrolytes fully dissociate in water. All strong electrolytes are ionic compounds. These include: All salts with group 1 cations All salts with ammonium cations All salts with nitrate, perchlorate, and acetate anions Hydroxides of Ca, Sr, and Ba (plus group 1 cations and ammonium) All sulfates except Ca and Ba

Strong vs. Weak Electrolytes CaCl 2 (s) > Ca 2+ (aq) + 2Cl - (aq) HgCl 2 (s) > > HgCl 2 (aq) > HgCl + (aq) + Cl - (aq) > Hg 2+ (aq) + 2Cl - (aq) H 2 O(l) 100% 99.8% 0.18% 0.02% Full dissociation of strong electrolyte Minimal dissociation of weak electrolyte

Strong Acids and Strong Bases Are Strong Electrolytes Strong ACIDS HCl HBr HI HNO 3 HClO 4 H 2 SO 4 Strong BASES Hydroxides of group 1 metals Hydroxides of Ca, Ba, and Sr

pH Scale At this point, we will not go into full detail of pH. However, it is important to know how acids and bases are distinguished. The pH scale allows us to do this. Bases Acids WATER

Concentration (Molarity)

Examples 30 g of NaCl are dissolved in 450 mL of H 2 O. What is the concentration of NaCl? How many moles of NaCl are there in 500 mL of this solution?

Example 15 g of Aluminum nitrate, Al(NO 3 ) 3, is dissolved in 200 mL of H 2 O. What is the concentration of nitrate in the solution? Aluminum nitrate will dissociate into aluminum and nitrate ions, as according to the chemical formula: Al(NO 3 ) > Al 3+ (aq) + 3NO 3 - (aq) Therefore, every mole of aluminum nitrate yields 3 moles of nitrate H 2 O(L)

Dilution In many instances (especially in lab), you may need to prepare a solution of some desired concentration from a pre-existing stock solution. An example of this would be a water enhancer, like Mio. You wouldn’t drink the Mio directly because it is extremely sweet. Instead, you add a small amount (aliquot) to your water, until you’ve attained the desired level of taste and sweetness. This is dilution. The flavored water is our diluted aqueous solution, and the bottle of Mio is the stock solution

Dilution Keep in mind that dilution does not change the total moles of solute, only the molarity. We know that the moles (n) of solute in V liters of a solution with molarity M is: n = MV Therefore, we know the concentration of a solution before and after dilution: V1V1 V2V2

How to perform a Series Dilution High concentration stock solution of concentration M 1 Aliquot of stock solution with volume V 1 and concentration M 1. Dilute with solvent to desired volume, V 2 After complete mixing, we have a dilute solution with volume V 2 and concentration M 2 Take an aliquot of the stock solution, add it to a new container

Example A concentrated stock solution of NaOH is 19.1 M. How would you prepare 500 mL of a 3.0 M solution? We are given: initial concentration of the NaOH stock (M 1 = 19.1 M), the desired final concentration of NaOH (M 2 = 3.0 M), and the final volume of the solution (V 2 = 0.5 L).  We need to find the volume of the aliquot (V 1 ) A 78.5 mL aliquot of the stock solution is added to mL of water to make a 3.0 M solution

Example a.) Explain how would you make a 500 mL stock solution that is 1.0 M NaBr (aq) ? (molar mass NaBr: g/mol) b.) From this stock solution, you decide to make 100 mL of a 0.10M solution. Explain how you would do this? Dilution

Applying Molarity to Stoichiometric Calculations For reactions of solutions, we can use molarity to calculate product yields Example: MnO 2 (s) + 4HBr(aq) -----> MnBr 2 (aq) + Br 2 (L) + 2H 2 O(L) What volume of an 8.84 M HBr solution is needed to completely react with 3.62 g of MnO 2 ? Convert MnO 2 to moles Determine the required moles of HBr Calculate volume

The Reaction of Strong Acids and Strong Bases is A Double- Replacement Reaction Known as a Neutralization Reaction When acids and bases react, they neutralize each other, and the product is salt and water HCl (aq) + NaOH(aq) H 2 O(L) + NaCl(aq) This is a double replacement reaction. The net ionic equation is: H + (aq) + Cl - (aq) + Na + (aq) + OH - (aq) H 2 O(L) + Na + (aq) + Cl - (aq) H + (aq) + OH - (aq) H 2 O(L)

Titrations Knowing that acids and bases neutralize each other, lets imagine that we have an acid/base of unknown concentration. How can we find the concentration? Perform an acid/base titration

Titrations In a titration, an indicator is added to the basic solution. In the example to the right, as long as the pH is above 7 (basic) the indicator will make the solution pink. An exact volume of an acid solution with a known concentration is added to a buret. The acid solution is added drop-by-drop until the solution just turns clear (neutralized, pH =7 ).

Say we have 100 mL of a basic NaOH solution of an unknown concentration. We titrate with 5 mL of 1.0 M HCl, and the solution just turns clear. Titrations Before titration After titration NaOH(aq) + HCl(aq)  H 2 O(L) + NaCl(aq) We know that the acid and base are completely neutralized, and none is left in solution. Moles of acid added = Stoichiometric equivalent of base Concentration of base solution =