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posted by  Chuperstar on 7/23/2008 6:11:41 PM  |  status: Live  

Acids & Bases - Urgent Help... Will Rate alll those who contribute

Course Textbook Chapter Problem
General Chemistry N/A N/A N/A
Question Details:
Part A Titration
 
Concentration of NaOH      0.1286 M
Volume of NaOH               1.75 mL
# Moles of NaOH               ?
Initial Conc. of Weak Acid   ?
 
Show Calculation steps and Formula.
 
Part B: Buffers
 
30 mL 0.1M Acetic Acid
15 mL 0.1M Sodium Hydroxide
 
a) Calculate the pH when mixed together.
b) Add 5 mL of Sodium Hydroxide to Buffer; Calculate the pH.
 
Show Calculations.
 
 

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posted by Chuperstar on 7/23/2008 6:27:09 PM  |  status: Live
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Response Details:
Concentration = mol/L
 
0.1286 M = mol / (0.00175L)
 
moles = 0.00022505 mol of NaOH
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posted by Chuperstar on 7/23/2008 6:41:51 PM  |  status: Live
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Response Details:
A.
moles of NaOH = 0.00175 L x 0.1286 M = 2.25x10^-4 mol
a weak monoprotic acid neutralized by 2.25x10^-4 mol NaOH would have a concentration of 2.25x10^-4 moles/ volume of acid solution in liters.

B.
HOAc: 0.030 L x 0.10 M = 0.0030 moles HOAc
NaOH: 0.015 L x 0.10 M = 0.0015 moles of NaOH
After reaction you have 0.0015 moles of HOAc and 0.0015 moles of NaOAc
pH = pKa + log(0.0015/0.0015)
pH = pKa + log(1) = pKa + 0
Ka for HOAc = 1.75x10^-5
pKa = -log(Ka) = 4.76
a) pH = 4.76
add 5 mL 0.10 M NaOH to the buffer
0.005 L x 0.10 M = 0.0005 moles NaOH
after addition HOAc = 0.0015 - 0.0005 = 0.0010 mole
after addition NaOAc = 0.0015 + 0.0005 = 0.0020 mole
pH = pKa + log(0.002/0.001) = pKa + 0.30
b) pH = 4.76+0.30 = 5.06
 
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