Ch 103 Seating arrangement Neutralization reactions...Simple case: HCl + NaOH --> H 2 O + NaCl ( a...
Transcript of Ch 103 Seating arrangement Neutralization reactions...Simple case: HCl + NaOH --> H 2 O + NaCl ( a...
Chem Chem 103103Lecture 3b
Acids and Bases5
Last time:Last time:
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