Chapter 4 Solution Stoichiometry

Chapter 4Solution Stoichiometry

Dr. Sapna Gupta

Concentrations of Solutions

• A solution is solute dissolved in a solvent.

• To quantify and know exactly how much of a solute is present in a certain amount of solvent, one will need to calculate concentrations.

• Concentrations are given in • percent solutions

• mass/mass % - (g of solute/g of solution) x 100% • mass/volume % - (g of solute/mL of solution) x 100%• volume/volume % - (mL of solute/mL of solution) x 100%

• Molarity (mol/L of solution)- used more in Chemistry • Molality (mol/Kg of solution) – this is used more in Biology

Dr. Sapna Gupta/Solution Stoichiometry 2

Molar Concentration, Molarity (M)In this chapter we will study Molarity – which is moles in a L of solution.

• Molarity is represented by M and the formula is given below

Molarity =𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒

𝐿 𝑜𝑓𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛

• Moles are converted to grams in order to make the solution in the lab.

• To prepare a solution, add the measured amount of solute to a volumetric flask, then add water to bring the solution to the mark on the flask.

• A 3M solution of NaCl means there are 3 moles of NaCl in the solution.

• If you have a 200 mL of 2 M HCl – that means that there are 2 mols of HCl in 1 L solution. If you want to know how many grams of HCl you have in 200 mL then you will have to calculate the amount of moles in 200 mL of that solution using the Molarity equation; then you can calculate the grams from those moles.


Example: You place a 1.52−g of potassium dichromate, K2Cr2O7, into a 50.0−mL volumetric flask. You then add water to bring the solution up to the mark on the neck of the flask. What is the molarity of K2Cr2O7 in the solution?

Molar mass of K2Cr2O7 is 294 g/mol

3

1 mol1.52 g

294 g

50.0 10 L

0.103 M

Example: A solution of sodium chloride used for intravenous transfusion (physiological saline solution) has a concentration of 0.154 M NaCl. How many moles of NaCl are contained in 500.−mL of physiological saline? How many grams of NaCl are in the 500.−mL of solution?

NaClmol0.0770

L0.5000.154

Lmol

M

M

NaClg4.50

mol1

g58.4mol0.0770

g58.4NaClmassMolar


M = 𝑚𝑜𝑙

𝐿

Example: Calculate the molarity of a solution prepared by dissolving 45.00 grams of KI into a total volume of 500.0 mL.

M 5422.0L 1

mL 1000

KI g 166.0

KI mol 1

mL 500.0

KI g 00.45

Example: How many milliliters of 3.50 M NaOH can be prepared from 75.00 grams of the solid?

mL 536L 1

mL 1000

NaOH mol 3.50

L 1

NaOH g 40.00

NaOH mol 1NaOH g 00.75


• When a higher concentration solution is used to make a less−concentration solution, the moles of solute are determined by the amount of the higher−concentration solution.

• The number of moles of solute remains constant when more water is added.

MiVi = MfVf

Note:

The units on Vi and Vf must match.

Dilution


Diluting a solution quantitatively requires specific glassware.

The photo at the right shows a volumetric flask used in dilution.


Example: A saturated stock solution of NaCl is 6.00 M. How much of this stock solution is needed to prepare 1.00−L of physiological saline solution (0.154 M)?

i

ffi

ffii

M

VMV

VMVM

mL25.7orL0.0257

6.00

L))(1.00(0.154

i

i

V

M

MV

Example: For the next experiment the class will need 250. mL of 0.10 MCuCl2. There is a bottle of 2.0 M CuCl2. Describe how to prepare this solution. How much of the 2.0 M solution do we need?

Concentrated: 2.0 M use ? mL (Vc)

Diluted: 250. mL of 0.10 M

McVc = MdVd

(2.0 M) (Vc) = (0.10 M) (250.mL)

Vc = 12.5 mL

12.5 mL of the concentrated solution are needed; add enough distilled water to prepare 250. mL of the solution. Dr. Sapna Gupta/Solution Stoichiometry 8

• In solution stoichiometry you have to presume that soluble ionic compounds dissociate completely in solution.

• Then using mole ratios we can calculate the concentration of all species in solution.

• There are three common types of stoichiometric calculations

• Quantitative Analysis: The determination of the amount of a substance or species present in a material.

• Volumetric Analysis: A type of quantitative analysis based on titration.

• Gravimetric Analysis: A type of quantitative analysis in which the amount of a species in a material is determined by converting the species to a product that can be isolated completely and weighed.

Solution Stoichiometry


A procedure for determining the amount of substance A by adding a carefully measured volume with a known concentration of B until the reaction of A and B is just complete. This can be for precipitation, neutralization or redox reactions.

• Standardization is the determination of the exact concentration of a solution.

• Equivalence point represents completion of the reaction.

• Endpoint is where the titration is stopped.

• An indicator is used to signal the endpoint.

Volumetric Analysis - Titrations


The figure below shows the reaction of Ba(NO3)2 with K2CrO4forming the yellow BaCrO4precipitate.

The BaCrO4 precipitate is being filtered. It can then be dried and weighed. Then concentration of Ba2+ ions can be calculated

Gravimetric Analysis

• In gravimetric analysis precipitation reactions are carried out.

• After the reaction the product is precipitated and collected in a crucible or filter paper.

• The precipitate is weighed and then using mole ratios we can calculate the concentration of all species in original solution.


Example: Find the concentration of all species in a 0.25 M solution of MgCl2

MgCl2 Mg2+ + 2Cl

Given: MgCl2 = 0.25 M

[Mg2+ ] = 0.25 M (1:1 ratio)

[Cl ] = 0.50 M (1:2 ratio)

Example: A soluble silver compound was analyzed for the percentage of silver by adding sodium chloride solution to precipitate the silver ion as silver chloride. If 1.583 g of silver compound gave 1.788 g of silver chloride, what is the mass percent of silver in the compound?

1mol AgCl 1mol Ag 107.9 g Ag1.788 g AgCl

143.32 g AgCl 1mol AgCl 1mol Ag = 1.346 g Ag in the compound

100%compoundsilverg1.583

Agg1.346= 85.03% Ag

Strategy: g AgCl -> mol AgCl -> mol Ag -> g Ag -> % Ag

Molar mass of silver chloride (AgCl) = 143.32 g


Example: Zinc sulfide reacts with hydrochloric acid to produce hydrogen sulfide gas:

ZnS(s) + 2HCl(aq) ZnCl2(aq) + H2S(g)

How many milliliters of 0.0512 M HCl are required to react with 0.392 g ZnS?

Strategy: g ZnS -> mol ZnS -> mol HCl (mol ratio from eq) -> vol HCl (using Molarity)

Molar mass of ZnS = 97.45 g

= 0.157 L = 157 mL HCl solution

1mol ZnS 2 mol HCl 1L solution0.392 g ZnS

97.45 g ZnS 1mol ZnS 0.0512 mol HCl


Example: A dilute solution of hydrogen peroxide is sold in drugstores as a mild antiseptic. A typical solution was analyzed for the percentage of hydrogen peroxide by titrating it with potassium permanganate:

5H2O2(aq) + 2KMnO4(aq) + 6H+(aq) 8H2O(l) + 5O2(g) + 2K+(aq) + 2Mn2+(aq)

What is the mass percent of H2O2 in a solution if 57.5 g of solution required 38.9 mL of 0.534 M KMnO4 for its titration?

4 2 2 2 23

4 2 2

0.534 mol KMnO 5 mol H O 34.01g H O38.9 10 L

1L 2 mol KMnO 1mol H O

Strategy: mols KMnO4 -> mols H2O2 -> mass H2O2 -> % H2O2

Molar mass of H2O2 = 34.01 g

= 1.77 g H2O2

100%solutiong57.5

OHg1.77 22= 3.07% H2O2


Example: A student measured exactly 15.0 mL of an unknown monoproticacidic solution and placed in an Erlenmeyer flask. An indicator was added to the flask. At the end of the titration the student had used 35.0 mL of 0.12 M NaOH to neutralize the acid. Calculate the molarity of the acid.

Example: Calculate the molarity of 25.0 mL of a monoprotic acid if it took 45.50 mL of 0.25 M KOH to neutralize the acid.

acid mol 0.01338KOH mol 1

acid mol 1L 0.04550

L

KOH mol 0.25

M 0.455L 0.0250

acid mol 0.01338


Strategy: mols NaOH -> mols acid (from eq) -> Molarity of acid

0.035 𝑀 𝑁𝑎𝑂𝐻 𝑥0.12 𝑚𝑜𝑙 𝑁𝑎𝑂𝐻

1 𝐿𝑥

1 𝑚𝑜𝑙 𝑎𝑐𝑖𝑑

1 𝑚𝑜𝑙 𝑁𝑎𝑂𝐻= 0.0042 𝑚𝑜𝑙 𝑎𝑐𝑖𝑑

0.0042𝑚𝑜𝑙

0.015 𝐿= 0.28 M acid

Key Words/Concepts

• Solutions

o Solvent

o Solute

• Molarity (mol/L)

• Dilutions (MiVi = MfVf)

• Solution stoichiometry

o Volumetric analysis (titration)

o End point

o Equivalence point

o Gravimetric analysis


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Chapter 4 Solution Stoichiometry