LECTURE FourteenCHM 151 ©slg

Topics:

1. Titration Calculations 2. Dilution Problems

How many mL of solution A, .250 M Pb(NO3)2, would be required to react with 30.00 mL of solution B, .150 M KI?

Pb(NO3)2 (aq) + 2 KI (aq) ---> PbI2 (s) + 2 KNO3 (aq)

.250 M .150 M ? mL 30.00 mL

Pathway: mL “B” soltn mol “B” mol “A” mL “A” soltn

30.00 mL B soltn = ? mL A soltn

Pb(NO3)2 (aq) + 2 KI (aq) ---> PbI2 (s) + 2 KNO3 (aq)

.250 M .150 M ? mL 30.00 mL

30.00 mL soltn .150 mol B 1 mol A 1000 mL soltn 1000 mL soltn 2 mol B .250 mol A

= 9.00 mL soltn A

Molarity equation Molarity

TITRATION CALCULATIONS

Titration: Procedure in which measured increments of one reactant are added to a known amount of a second reactant until some indicator signals that the reaction is complete.

This point in the reaction is called “the equivalence point.”

Indicators include many acid/base dyes, potentiometers, color change in one reagent...

Titrations are run with buret and Erlenmeyer flasks as seen on the CD ROM or demo just observed...

Two types of problems are typically encountered: •Standardizing an acid or a base solution•(determining the exact molar concentration ofan acid or base solution)

• Determining amount of acidic or basic material in a sample ( or other substances detectable by some indicator)

Ex 5.13 type: Standardizing an acid solution:

Suppose a pure, dry sample of Na2CO3 weighing 0.379 gis dissolved in water and titrated to the equivalence point with 35.65 mL of HCl solution. What is the molarity of the HCl solution?

Na2CO3(aq) + 2HCl(aq) -----> 2 NaCl(aq) + H2O + CO2(g)

105.99 g/mol

35.65 mL soln0.379 g M= ? = #mol HCl / L soltn

2 Na = 2 X 22.99 = 45.981 C = 1 X 12.01 = 12.013 O = 3 X 16.00 = 48.00 105.99 g/mol

Calculations for Standardization

1) Calculate moles of solute from balanced equation, using information about known reagent 2) Calculate M, using volume of solution required for titration of known to equivalence point

Na2CO3(aq) + 2HCl(aq) -----> 2 NaCl(aq) + H2O + CO2(g)

105.99 g/mol

35.65 mL soln0.379 g M=? = Mol HCl / L soltn

0.379 g Na2CO3

105.99 g Na2CO3

1 mol Na2CO3

1 mol Na2CO3

2 mol HCl= .00715 mol HCl

Step One: Calculate Moles of HCl from equation .379 g Na2CO3 = ? moles HCl

M, HCl soln = ? = # mol HCl = L soln

.00715 mol HCl 1000 mL = .201 mol HCl = .201 M HCl 35.65 mL soltn 1 L L soltn

Na2CO3(aq) + 2HCl(aq) -----> 2 NaCl(aq) + H2O + CO2(g)

105.99 g/mol

35.65 mL soln0.379 g .00715 mol M=?

Step Two: Calculate molarity:

Calculated in Step One

Group Work:

Suppose a pure, dry sample of Na2CO3 weighing 0.437 gis dissolved in water and titrated to the equivalence point with 39.85 mL of HCl solution. What is the molarity of the HCl solution?

Na2CO3(aq) + 2HCl(aq) -----> 2 NaCl(aq) + H2O + CO2(g)

105.99 g/mol

39.85 mL soln0.437 g M= ? = #mol HCl / L soltn

#1: Find moles of HCl from g Na2CO3

#2: Find Molarity (moles HCl / volume soltn)

Na2CO3(aq) + 2HCl(aq) -----> 2 NaCl(aq) + H2O + CO2(g)

105.99 g/mol

39.85 mL soln0.437 g M= ? = #mol HCl / L soltn

#1: Find Moles of HCl

.437 g Na2CO3 1 mol Na2CO3 2 mol HCl =.008246 mol HCl 105.99 g Na2CO3 1 mol Na2CO3

#2: Calculate Molarity

.008246 mol HCl 1000 mL = .2069 mol HCl = .207 M HCl39.85 mL soltn 1 L L soltn

Group Work Solution:

Standardization of a Base:

Let’s use the 0.201 M HCl which we standardized in the first problem to determine the exact molar concentration of a sodium hydroxide solution (“standardize the base”!):

Problem:

If titrating 25.00 mL of NaOH to a phenolphthalein endpoint required 31.25 mL of .201 M HCl, what is themolar concentration of the base?

If titrating 25.00 mL of NaOH to a phenolphthalein endpoint required 31.25 mL of .201 M HCl, what is themolar concentration of the base?

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

25.00 mL 31.25 mL M=? .201 M

Steps:1) calculate moles NaOH reacted2) calculate M, moles NaOH/ volume soln

NaOH(aq) + HCl(aq) ---> H2O(l) + NaCl(aq) 25.00 mL 31.25 mL M=? .201 M

31.25 mL HCl soln

1000 mL soln

.201 mol HCl

1 mol HCl

1 mol NaOH= .00628 mole NaOH

Step 1) 31.25 mL HCl soln = ? mol NaOH

Step 2) Calculate molarity:

.00628 mole NaOH

25.00 mL soln

1000 mL

1 L

= .251 mol NaOH = .251 M NaOH L

A 30.0 mL sample of vinegar requires 39.35 mL of .843 MNaOH solution for titration to the equivalence point. What is the molar concentration of the acetic acid in vinegar?

CH3CO2H(aq) + NaOH(aq) ----> H2O + NaCH3CO2(aq)

30.0 mL 39.35mL .843 M

M=?mol CH3CO2H /L

Step 1) solve for moles, CH3CO2H

Step 2) solve for M CH3CO2H

Group Work

CH3CO2H(aq) + NaOH(aq) ----> H2O + NaCH3CO2(aq)

30.0 mL 39.35mL M=? .843 M

#1:

#2:

39.35 mL soltn .843 mol NaOH 1 mol CH3CO2H = .03317 mol CH3CO2H 1000 mL soltn 1 mol NaOH

.03317 mol CH3CO2H 1000 mL = 1.1056 mol CH3CO2H = 1.11 M30.00 mL soltn 1 L L soltn

Let’s do another base solution, standardizing it with Potassium acid phthalate, KHC8H4O4, a popular solid for this purpose. (Structure, next slide!) It reacts with strong bases according to the following net ionic equation:

KHC8H4O4 (aq) + NaOH (aq) ------> H2O(l) + KNaC8H4O4 (aq)

If a .896 g sample of this compound is dissolved in waterand titrated to the equivalence point with 16.95 mL of NaOH solution, what is the molarity of the NaOH soln?

CC

CC

C

CC

C

O

O

O

O:

K

H

H

H

H

H

+ NaOH

CC

CC

C

CC

C

O

O:

O

O:

H

H

H

H

H2O + Na

K

CC

CC

C

CC

C

O

O

O

O:

H

H

H

H

H

K

H which actslike an acid

Salt cation

"Hydrogen Phthalate" Anion HC8H4O4:-

KHC8H4O4 (aq) + NaOH (aq) ---> H2O (l) + K NaC8H4O4 (aq)

204.22 g/mol.896 g 16.95 mL

M=? mol NaOH /L

1K = 1 X 39.10 = 39.10 8C = 8 X 12.01 = 96.085H = 5 X 1.008 = 5.044O = 4 X 16.00 = 64.00 204.22 g/mol

If a .896 g sample of potassium acid phthalate is dissolved in water and titrated to the equivalence point with 16.95 mL of NaOH solution, what is the molarity of the NaOH soln?

KHC8H4O4 (aq) + NaOH (aq) ---> H2O (l) + K NaC8H4O4 (aq)

204.22 g/mol 16.95 mL.896 g M=? mol NaOH / L

M, NaOH soln= .00439 mol NaOH

16.95 mL soln

1000 mL

1 L= .259 mol/L NaOH

#1: Find moles of NaOH

#2: Calculate Molarity:

.896 g KHC8H4O4 1 mol KHC8H4O4 1 mol NaOH = .00439 mol NaOH

204.22 g KHC8H4O4 1 mol KHC8H4O4

“Redox” titration

You wish to determine the weight percent of copper in a copper containing alloy. After dissolving a sample of an alloy in acid, an excess of KI is added, and the Cu2+ and I- ions undergo the reaction:

2Cu2+ (aq) + 5 I-

(aq) -----> 2 CuI(s) + I3 - (aq)

The I3 - which is produced in this reaction is titrated with sodium thiosulfate according to the equation:

I3 - (aq) + 2 S2O3 2- (aq)

-----> S4O6 2- (aq) + 3 I- (aq)

If 26.32 mL of 0.101 M Na2S2O3 is required for titration to the equivalence point, what is the wt % of Cu in .251 g alloy?

2Cu2+ (aq) + 5 I-

(aq) -----> 2 CuI(s) + I3 - (aq)

63.55 g/mol produced0.251 g alloy by reactiong Cu =? % Cu, alloy=?

I3 - (aq) + 2 S2O3 2- (aq)

-----> S4O6 2- (aq) + 3 I- (aq)

? mol 26.32 mL .101 M Na2S2O3

I3 - (aq) + 2 S2O3 2- (aq)

-----> S4O6 2- (aq) + 3 I- (aq)

? mol 26.32 mL .101 M Na2S2O3

26.32 mL soln

1000 mL

1 L

1 L soln

.101 mol Na2S2O3

1 mol Na2S2O3

1 mol S2O3 2-

2 mol S2O3 2-

1 mol I3 -

= .00133 mol I3 -

2Cu2+ (aq) + 5 I-

(aq) -----> 2 CuI(s) + I3 - (aq)

63.55 g/mol g=? .00133 mol% Cu, alloy=?

0.251 g alloy

.00133 mol I3 -

1 mol I3 -

2 mol Cu2+

1 mol Cu2+

1 mol Cu

1 mol Cu

63.55 g Cu= .169 g Cu

% Cu, alloy=? = 1.69 g Cu X 100 = 67.3% 0.251 g alloy

DILUTION PROBLEMS

Suppose you would like to make up a more “dilute” solution (less moles/L) from a more “concentrated” one (more moles /L). This is how you might proceed:

a) figure out how many total moles of solute you want in the desired volume of the dilute solution

b) figure out what volume of the more concentrated solution will deliver this number of moles

c) measure out the concentrated solution and add water to make up the desired volume of the dilute solution. (DEMO!)

Describe how you might makeup 750. mL of 1.00 MH2SO4 from a concentrated sulfuric acid solution which is 12.0 M H2SO4 .

a) Calculate desired number of moles in dilute soln:

b) Calculate volume of concentrated solution which willcontain this number of moles:

750 mL soln 1.00 mol H2SO4= .750 mol H2SO4

1000 mL soln

.750 mol H2SO4

12.0 mol H2SO4

1000 mL soln= 62.5 mL soln

c) Measure out 62.5 mL of the concentrated soln andcarefully add it to sufficient water to make up 750. mL of solution.

NEVER ADD WATER TO CONCENTRATED ACIDS; ALWAYS ADD CON ACIDS TO WATER, SLOWLY...

Formula for Dilution Problems:

Since we can calculate moles as follows:

mL X # moles = moles of solute 1000 mL

and since: moles, con soln = moles, dil soln

We can say:mL, con X M, con = mL, dil X M, dil

and do the last problem “the easy way”:

Describe how you might makeup 750. mL of 1.00 MH2SO4 from a concentrated sulfuric acid solution which is 12.0 M H2SO4 .

mL, con X M, con = mL, dil X M, dil

? mL X 12.0 M H2SO4 = 750. mL X 1.00 M H2SO4

mL con = 750 mL dil X 1.00 M H2SO4

12.0 M H2SO4

mL con = 62.5 mL

Punch line the same: measure out 62.5 mL con acid and dilute to 750. mL.

Group Work:

How many mL of 2.35 M AgNO3 solution are required tomakeup 2.00 L of .100 M AgNO3 solution?

Describe how you would makeup this new solution.

mL con X M con = mL dil X M dil

? 2.35 M 2.00L .100 M

mL con X M con = mL dil X M dil

? 2.35 M 2.00L .100 M 2000 mL

mL con = mL, dil X M, dil M, con

ml, con = 2000 mL X .100 M 2.35 M

mL, con = 85.1 mL

Measure out 85.1 mL of con soltn, dilute to 2.00 L

Energy Units

The “ calorie”: “The quantity of energy required to raise 1.00 g of water 1oC”. Very small amount of energy, soKcal are generally used: 1000 calories(cal) = 1 kilocalorie, kcal

SI unit of energy is the “joule”, defined in terms ofkinetic energy rather than heat energy. One joule is theamount of kinetic energy involved when a 2.0 kg object is moving with a velocity of 1.0 m/s. Again, a very smallamount of energy, so kilojoules are generally used:

1000 joules (J) = 1 kilojoule kJ (kJ)

Relationship:

1 calorie = 4.184 joules

Dietary or nutritional Calorie:

1 nutritional Calorie = 1 Cal = 1 kcal = 1000 cal

Try the problems assigned in Chapter six!

End, Lecture 14