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elisabeth-morris
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Titrations Titration Curve – always calculate
equivalent point first Strong Acid/Strong Base
Regions that require “different” calculations B/F any base is added Half-way point region At the equivalence point After the equivalence point
Strong Acid/Strong Base 50 mL of 0.02000 M KOH Titrated with 0.1000 M HBr
First -find Volume at equivalence M1V1 = M2V2
(0.050 L)(0.02000M) = 0.1000 V V = 10.0 mL
Strong Acid/Strong Base 50.00 mL of 0.02000 M KOH Titrated with 0.1000 M HBr
Second – find initial pH pH = - logAH ~ -log [H+]
pOH = -logAOH ~ -log [OH-]
pH = 12.30
Strong Acid/Strong Base 50 mL of 0.02000 M KOH Titrated with 0.1000 M HBr
Third– find pH at mid-way volume KOH (aq) + HBr (aq) -> H2O (l) +
KBr(aq) BeforeAfter
0.001000 mol 0.0006000 mol0.000400 mol 0 mol
Limiting Reactant
0.0006000 mol0.0006000 mol
pH = 11.8
(~6 ml)
Strong Acid/Strong Base 50 mL of 0.02000 M KOH Titrated with 0.1000 M HBr
Fourth – find pH at equivalence point KOH (aq) + HBr (aq) -> H2O (l) + KBr(aq)
BeforeAfter
0.001000 mol 0.0010000 mol0 mol 0 mol 0.0010000 mol0.0010000 mol
pH = 7.0
Strong Acid/Strong Base 50 mL of 0.02000 M KOH Titrated with 0.1000 M HBr
Finally – find pH after equivalence point KOH (aq) + HBr (aq) -> H2O (l) + KBr(aq)
BeforeAfter
0.001000 mol 0.001200 mol0 mol 0.0002000 mol 0.0010000 mol
pH = 2.5
12 ml
Limiting Reactant
Titration of a weak acid solution with a strong base.
25.0 mL of 0.1000M acetic acid Ka = 1.8 x 10-5 Titrant = 0.100 M NaOH
First, calculate the volume at the equivalence-point M1V1 = M2V2
(0.0250 L) 0.1000 M = 0.1000 M (V2) V2 = 0.0250 L or 25.0 mL
Titration of a weak acid solution with a strong base.
25.0 mL of 0.1000M acetic acid Ka = 1.8 x 10-5 Titrant = 0.100 M NaOH
Second, Calculate the initial pH of the acetic acid solution
Titration of a weak acid solution with a strong base.
25.0 mL of 0.1000M acetic acid Ka = 1.8 x 10-5 Titrant = 0.100 M NaOH
Third, Calculate the pH at some intermediate volume
Titration of a weak acid solution with a strong base.
25.0 mL of 0.1000M acetic acid Ka = 1.8 x 10-5 Titrant = 0.100 M NaOH
Fourth, Calculate the pH at equivalence
Titration of a weak acid solution with a strong base.
25.0 mL of 0.1000M acetic acid Ka = 1.8 x 10-5 Titrant = 0.100 M NaOH
Finally calculate the pH after the addition 26.0 mL of NaOH
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0 10 20 30 40 50
Volume of NaOH, mL
pH
Initial pH
aa CKH ][
Buffer Region
][
][log
acid
basepKpH a
Equivalence pointM1V1=M2V2
pH @ equivalence?
pH after equivalence?
0
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10
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0 10 20 30 40 50
Volume of NaOH, mL
pH
Initial pH
aa CKH ][
Buffer Region
][
][log
acid
basepKpH a
Equivalence pointM1V1=M2V2
pH @ equivalence
pH after equivalence?
bb CKOH ][
0
2
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6
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10
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14
0 10 20 30 40 50
Volume of NaOH, mL
pH
Initial pH
aa CKH ][
Buffer Region
][
][log
acid
basepKpH a
Equivalence pointM1V1=M2V2
pH @ equivalence
pH after equivalenceDominated by remaining[OH-]
bb CKOH ][
Weak Base titrated with strong acid Consider a 100 ml of a 0.0100 M
base with 0.0500 M HCl Kb = 1 x 10-5
Initial pH
bb CKOH ][
Buffer Region
][
][log
acid
basepKpH a
pH @ equivalence
aa CKH ][
pH after equivalenceDominated by remaining[H+]