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 = ?

Answer 1

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