Chem Chem 103103Lecture 3b

Acids and Bases5

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Today:Today:1. Acid base titration2. Polyprotic acids

1. Calculating pH in 2 more scenarios:a) Pure weak baseb) Buffer solutions

Ch 103 Seating arrangementIn preparation for group work and midterm exams:

Seating arrangement enforced for extra credit work.(if in wrong row, no extra credit)

If your last name starts with please sit in row

A, B or C A (front)

D, E, F, G, H, I, J B

K, L, M C

N, O, P, Q, R, Sa D

Si, T, U, V, W, X, Y, Z E

Neutralization reactionsNeutralization reactions

When an acid encounters a base, neutralization occurs:

Generic case: Strong acid + strong base ---> salt + water

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

Ionic eq.: H+ + Cl- + Na+ + OH- --> Na+ + Cl- + H2O

Important is net ionic eq: H+ + OH- --> H2O (spectators: Na+,Cl-)

Example: HA(aq) + NaOH(aq) --> NaA(aq) + H2O

Net ionic eq: HA(aq) + OH-(aq) --> A-(aq) + H2O(l)

Strong acid + strong base ---> 100% completion

Strong + weak ---> 100% completion

Weak + weak ---> not 100% completion

Ways a buffer can resultWays a buffer can result

When you add HCl to NH3 what is the net ionic equation?

Chemical rxn: HCl (aq) + NH3 (aq) ---> NH4Cl

Ionic eq.: H+ + Cl- + NH3 ---> NH4+ + Cl-

Net ionic equation: H+ + NH3 --> NH4+

If you add 0.50 moles of H+ to 1.00 moles of NH3 what remains?

Answer: 0.50 moles of NH4+ and 0.50 moles of NH3

I.e. A buffer containing base and its conjugate acid.

ABCABC’’s of titrations of titrationA=acid, B=base, C=calculationsA=acid, B=base, C=calculations

Titrant

Analyte of known volume+ pH indicator

Ultimate purpose: to determine molarity of analyte

Acid base titrationAcid base titrationIn titration, determine the equivalence point (ep).

Simple case: HCl + NaOH --> H2O + NaCl ( a 1:1 titration)

Titration of 20.0 mLs of NaOH requires 15.0 mLs of 0.120 Mhydrochloric acid, HCl, to reach the equivalence point (ep).What is the concentration of the NaOH analyte?HCl + NaOH --> H2O + NaCl Solution:

[NaOH] = =

But #mol HCl = MHClVHCl so:

finally: [NaOH] = = 0.0900M

!

# mol NaOH

L

!

#mol HCl x 1 mol NaOH

1 mol HCl

L

!

MHCl

VHCl

x 1 mol NaOH

1 mol HCl

L

!

(0.120M)(0.0150 L)(1/1)

0.0200L

Acid base titration, Acid base titration, a quicker versiona quicker versionLook at the same 1:1 titration problem

Titration of 20.0 mLs of NaOH requires 15.0 mLs of 0.120 Mhydrochloric acid, HCl, to reach the equivalence point (ep).What is the concentration of the NaOH analyte?Solution: chem. rxn:HCl + NaOH --> H2O + NaClAt equivalence pt (ep): # equiv HCl = # equiv NaOHBut # equiv HCl = # mol HCl x 1 H+/mole = # mol HClSame for NaOH so: # mol HCl = #mol NaOHSo MHClVHCl = MNaOHVNaOH => “ M1V1 = M2V2 “M2 = M1V1 / V2 = (0.120M)(15.0mL)/(20.0mL)M2 = 0.0900M

Acid base titrationAcid base titration

In titration, determine the equivalence point (ep).

Example: 2 H3PO4 + 3 Mg(OH)2 --> 6 H2O + Mg3(PO4)2

Titration of 20.0 mLs of Mg(OH)2 requires 15.0 mLs of 0.120M phosphoric acid, H3PO4, to reach the equiv. pt (ep). Whatis the concentration of the Mg(OH)2 analyte?

2 H3PO4 + 3 Mg(OH)2 --> 6 H2O + Mg3(PO4)2

Solution:

[Mg(OH)2] = =

!

mol Mg(OH)2

L

!

mol H3PO4

Lx

3 mol Mg(OH)2

2 mol H3PO4

ContinuationContinuation……

In previous titration example: VMg(OH)2 = 20.0 mL, MH3PO4 = 0.120M, Ve=15.0 mL

2 H3PO4 + 3 Mg(OH)2 --> 6 H2O + Mg3(PO4)2

Solution:

[Mg(OH)2] = =

= =

=

!

mol Mg(OH)2

L

!

mol H3PO4

Lx

3 mol Mg(OH)2

2 mol H3PO4

!

MH3PO4VH3PO4

Lx

3 mol Mg(OH)2

2 mol H3PO4

!

(0.120M)(15.0mL)

20.0mLx

3

2

!

0.135M

Using the other approach

In previous titration example: VMg(OH)2 = 20.0 mL, MH3PO4 = 0.120M, Ve=15.0 mL

2 H3PO4 + 3 Mg(OH)2 --> 6 H2O + Mg3(PO4)2

Solution: Always true at equivalenc point:

# equivalents H3PO4 = # equivalents Mg(OH)2

but # eq = # mol x # H+ transferred:

So # mol H3PO4 x 3 = # mol Mg(OH)2 x 2

3 M1V1 = 2 M2V2 where subscript “1” = H3PO4 ; “2” = Mg(OH)2

So: M2 = = = 0.135 M

!

3M1V1

2V2

!

3(0.120)(15.0)

2 (20.0)M

Strong Acid + strong base. pH = ?

Problem:20.0 mLs of 1.00 M HCl + 21.0 mLs of 1.00 M NaOH. pH =?

Solution: HCl + NaOH --> NaCl + H2O

NOTE: THIS IS NOT AT EQUIVALENCE! (Another type of problem!)

(easiest is: “follow the moles”):

mmol HCl=MHClVHCl =(1.00M)(20.0mL)=20.0 mmol HCl

mmol NaOH=MNaOHVNaOH = (1.00M)(21.0mL)=21.0mmol NaOH

Limiting reagent is HCl ; mmol NaOH excess=21.0-20.0=1.0mmol

So [OH-] = mole OH-/total Vol(L) = mmol OH-/total mLs

[OH-]=1.0mmol/(20.0+21.0)mL= 1.0/41.0 M = 0.0244 M

pOH = -log(0.0244)= 1.613; pH=14.00-1.653= 12.38