Chem Chem 103 103 Lecture 3b Acids and Bases5 Last time: Last time: Today: Today: 1. Acid base titration 2. Polyprotic acids 1. Calculating pH in 2 more scenarios: a) Pure weak base b) Buffer solutions Ch 103 Seating arrangement In 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 reactions Neutralization reactions When an acid encounters a base, neutralization occurs: Generic case: Strong acid + strong base ---> salt + water Example: HCl(aq) + NaOH(aq) --> NaCl(aq) + H 2 O(l) Ionic eq.: H + + Cl - + Na + + OH - --> Na + + Cl - + H 2 O Important is net ionic eq: H + + OH - --> H 2 O (spectators: Na + ,Cl - ) Example: HA(aq) + NaOH(aq) --> NaA(aq) + H 2 O Net ionic eq: HA(aq) + OH - (aq) --> A - (aq) + H 2 O(l) Strong acid + strong base ---> 100% completion Strong + weak ---> 100% completion Weak + weak ---> not 100% completion Ways a buffer can result Ways a buffer can result When you add HCl to NH 3 what is the net ionic equation? Chemical rxn: HCl (aq) + NH 3 (aq) ---> NH 4 Cl Ionic eq.: H + + Cl - + NH 3 ---> NH 4 + + Cl - Net ionic equation: H + + NH 3 --> NH 4 + If you add 0.50 moles of H + to 1.00 moles of NH 3 what remains? Answer: 0.50 moles of NH 4 + and 0.50 moles of NH 3 I.e. A buffer containing base and its conjugate acid. ABC ABC’ s of titration s of titration A=acid, B=base, C=calculations A=acid, B=base, C=calculations Titrant Analyte of known volume + pH indicator Ultimate purpose: to determine molarity of analyte
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Ch 103 Seating arrangement Neutralization reactions...Simple case: HCl + NaOH --> H 2 O + NaCl ( a 1:1 titration) Titration of 20.0 mLs of NaOH requires h15.0 mLs of 0.120 M hydrochloric
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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
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).
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