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Chapter Titrations

CHEM 102 Chapter Notes - Chapter Titrations : Conjugate Acid, Buffer Solution, Mols Premium

4 Pages
46 Views
Spring 2018

Department
Chemistry
Course Code
CHEM 102
Professor
Naleway
Chapter
Titrations

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A beaker with 105 mL of an acetic acid buffer with a pH of 5.000 is sitting on a benchtop.
The total molarity of acid and conjugate base in this buffer is 0.100 M
mL of a 0.470 M
HCl solution to the beaker. How much will the pH change? The pK
a of
acetic acid is 4.740.
ONE: First calculate the number of moles of conjugate acid and conjugate base in the initial buffer. Next,
determine the effect of the addition of the acid, HCl, on the total number of moles of conjugate acid and
conjugate base. Finally, use the Henderson-Hasselbalch equation to determine the new pH,
pH=pK
a+log[conjugate base]/[conjugate acid]
and compare this value to the initial pH of 5.00 to determine the pH change.
ONE STEP ONE: Initial mols of weak acid
Find mols of acid + base : 105 mL = 0.105 L , 0.100M , > 0.0105 mols acid and base in
buffer.
Find the value of nA- / nHA = 10^ pH - pka = 1.82
Find value of nHA : nA- / nHA
acid + base = 0.0105
base/acid = 1.82
Because nA- / nHA , nA- = 1.82nHA
0.0105 = 1.82(nHA) + nA which is like 2x + x = 4
So do 0.0105 / 2.82 = 0.0037 is
moles of
CH
3
COOH (n
HA
)
ONE STEP TWO: initial mols of base
Find mols of base from previsou information.
acid + base = 0.0105 mols
acid mols = 0.0037 mols
Base mols would be = 0.006776
ONE STEP THREE: mols of HCL added to acetic acid buffer
5.50 mL HCL is 0.0055 L * 0.470 M = 0.002585 mols
ONE STEP FOUR: New mols of conjugate base
Initial base - mols of HCl = 0.006776 - 0.002585 = 0.004191 mol!!
ONE STEP FIVE: New mols of acid
Initial acid + mols of HC = 0.0037 + 0.002585 = 0.006285 mol!!!
ONE STEP SIX: New pH
pH = pKa + log conjugate base / conjugate acid
pH = 4.740 + log (0.004191 / 0.006285) = 4.56401
ONE STEP SEVEN: Change in pH due to HCl addition
New pH - initial = 4.56401 - 5 = -0.4398

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Description
A beaker with 105 mL of an acetic acid buffer with a pH of 5.000 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.100 M. A student adds 5.50 mL of a 0.470 M HCl solution to the beaker. How much will the pH change? The pKaof acetic acid is 4.740. ONE: First calculate the number of moles of conjugate acid and conjugate base in the initial buffer. Next, determine the effect of the addition of the aciHCl, on the total number of moles of conjugate acid and conjugate base. Finally, use the HendersonHasselbalch equation to determine the new pH, pH=pK a+log [conjugate base][conjugate acid] and compare this value to the initial pH of 5.00 to determine the pH change. ONE STEP ONE: Initial mols of weak acid Find mols of acid + base : 105 mL = 0.105 L , 0.100M , > 0.0105 mols acid and base in buffer. Find the value of nA nHA = 10^ pH pka = 1.82 Find value of nHA : nA nHA acid + base = 0.0105 baseacid = 1.82 Because nA nHA , nA = 1.82nHA 0.0105 = 1.82(nHA) + nA which is like 2x + x = 4 So do 0.0105 2.82 = 0.0037 is moles of CH 3COOH (n HA) ONE STEP TWO: initial mols of base Find mols of base from previsou information. acid + base = 0.0105 mols acid mols = 0.0037 mols Base mols would be = 0.006776 ONE STEP THREE: mols of HCL added to acetic acid buffer 5.50 mL HCL is 0.0055 L * 0.470 M = 0.002585 mols ONE STEP FOUR: New mols of conjugate base Initial base mols of HCl = 0.006776 0.002585 = 0.004191 mol!! ONE STEP FIVE: New mols of acid Initial acid + mols of HC = 0.0037 + 0.002585 = 0.006285 mol!!! ONE STEP SIX: New pH pH = pKa + log conjugate base conjugate acid pH = 4.740 + log (0.004191 0.006285) = 4.56401 ONE STEP SEVEN: Change in pH due to HCl addition New pH initial = 4.56401 5 = 0.4398
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