I posted this exact question a few days ago and some doofus said it wasn\'t clea
ID: 951519 • Letter: I
Question
I posted this exact question a few days ago and some doofus said it wasn't clear. I don't know how to make it any more clear. This is all the information the question gives. There are 5 parts, A-E. Please answer them all. Thank you.
Consider the titration of a 20.0 mL sample of 0.110 M HC2H3O2 with 0.125 M NaOH. Determine each quantity:
Part A)
The initial pH = ?
Part B)
The volume of added base required to reach the equivalence point.
V = ? mL
Part C)
The pH at 4.00 mL of added base.
pH = ?
Part D)
the pH at one-half of the equivalence point.
pH = ?
Part E)
The pH at the equivalence point.
pH = ?
Explanation / Answer
Ka for acetic acid =1.8*10-5
pKa= -log(1.8/105) =4.74
CH3COOHà CH3COO- +H+
a)Let x= drop in concentration of CH3COOH
At Equilibrium [CH3COOH] = 0.11-x and [CH3COO-]=[H+] =x
Ka= x2/(0.11-x)= 1.8*10-5 , when solved using solver of x
[H+] = 0.000141 , pH= -log(0.000141) =3.85. This is the initial PH
Part B :
Moles of acetic acid = 0.11*20/1000 =0.0022
Moles of NaOH= 0.0022 at equivalence point
Volume of NaOH= 0.0022/0.125 =0.0176 L =17.6 ml
Part C)
Moles of acetic acid =0.0022
Moles of NaOH in 4ml of 0.125M= 0.125*4/1000 =0.0005
The reaction is CH3COOH+ NaOH--à CH3COONa+H2O
1 mole of NaOH requires 1mole of CH3COOH and this will reduce the moles of acetic acid and increases the moles of acetate ion
Moles of acetate formed= 0.0005 moles of excess acetic acid =0.0022-0.0005=0.0017
PH=PKa+log (0.0005/0.0017) (since total volume gets cancelled during taking the ratio)
pH= 4.74-0.53=4.21
Part D)
the pH at one-half of the equivalence point.
pH = ?
at one half equivalence point moles acid = moles acetate
pH = 4.74
moles NaOH required to reach the equivalence point = 0.0022
volume NaOH = 0.0022/ 0.125 M=0.176 L
total volume = 0.0176+ 0.0200 =0.0376 L
moles acetate formed = 0.0022
concentration acetate = 0.0022/ 0.0376 =0.0585 M
CH3COO- + H2O ------ CH3COOH + OH-
Kb= Kw//Ka= 10-14/ 1.8*10-5 = 5.56*10-10 =x2/(0.0596-x)
Where x= [OH-] =5.76*10-6
pOH= 5.24
pH= 14-5.24= 8.76
Part E)
The pH at the equivalence point.
Since moles of acid= moles of acetate
pH= 4.74