100.0 ml of 0.100 M acetic acid is placed in a flask and is titrated with 0.100
ID: 839191 • Letter: 1
Question
100.0 ml of 0.100 M acetic acid is placed in a flask and is titrated with 0.100 Msodium hydroxide . An appropriate indicator is used. Ka for acetic acid is 1.7 x 10^-5
Calculate the pH in the flask at the following points in the titration.a. when no NaOH has been added.b. after 25.0 ml of NaOH is addedc. after 50.0 ml of NaOH is addedd. after 75.0 ml of NaOH is addede. after 100.0 ml of NaOH is addedf. what would be the appropriate indicator that was used ? consult fig 16.7 inBLB and explain your choice.
g. What would the pH be after 300 ml of 0.100 M NaOH was added?
h. How could the equivalence point be detected without the use of a visualindicator?
Explanation / Answer
a) when no NaOH is added
CH3C00H ----> H+ + CH3C00-
Ka = x2 / ( C -x)
x = sqrt ( Ka C)
[H+] = x = sqrt ( ka C)
[H+] = sqrt ( 1.7 *10-5 *0.1 )
[H+] = 0.0013
pH = -log [H+]
pH = -log 0.0013
pH = 2.884
so the pH is 2.884
b) moles of acetic acid = molarity x volume / 1000
moles of acetic acid = 0.1 x 100/1000
moles of acetic acid = 10 x 10-3
moles of NaOH = 0.1 x 25 /1000
moles of NaOH = 2.5 x 10-3
the reaction is given by
CH3COOH + NaOH ------> CH3COONa + H20
from the reaction
moles of CH3COONa formed = moles of NaOh reacted
moles of CH3COONa formed = 2.5 x 10-3
CH3COOH and CH3COONa form a buffer solution
according to hasselbach hendersen equation
pH = pKa + log[salt / acid ]
pH = -log Ka + log[CH3COONa / CH3COOH ]
pH = -log 1.7 x 10-5 + log { 2.5 x 10-3 / 10 x 10-3 ]
pH = 4.76955 + log 0.25
pH = 4.167
c)
moles of acetic acid = molarity x volume / 1000
moles of acetic acid = 0.1 x 100/1000
moles of acetic acid = 10 x 10-3
moles of NaOH = 0.1 x 50 /1000
moles of NaOH = 5 x 10-3
the reaction is given by
CH3COOH + NaOH ------> CH3COONa + H20
from the reaction
moles of CH3COONa formed = moles of NaOh reacted
moles of CH3COONa formed = 5 x 10-3
CH3COOH and CH3COONa form a buffer solution
according to hasselbach hendersen equation
pH = pKa + log[salt / acid ]
pH = -log Ka + log[CH3COONa / CH3COOH ]
pH = -log 1.7 x 10-5 + log { 5 x 10-3 / 10 x 10-3 ]
pH = 4.76955 + log 0.5
pH = 4.4685
d)
moles of acetic acid = molarity x volume / 1000
moles of acetic acid = 0.1 x 100/1000
moles of acetic acid = 10 x 10-3
moles of NaOH = 0.1 x 75 /1000
moles of NaOH = 7.5 x 10-3
the reaction is given by
CH3COOH + NaOH ------> CH3COONa + H20
from the reaction
moles of CH3COONa formed = moles of NaOh reacted
moles of CH3COONa formed = 7.5 x 10-3
CH3COOH and CH3COONa form a buffer solution
according to hasselbach hendersen equation
pH = pKa + log[salt / acid ]
pH = -log Ka + log[CH3COONa / CH3COOH ]
pH = -log 1.7 x 10-5 + log { 7.5 x 10-3 / 10 x 10-3 ]
pH = 4.76955 + log 0.75
pH = 4.6446
e)
moles of acetic acid = molarity x volume / 1000
moles of acetic acid = 0.1 x 100/1000
moles of acetic acid = 10 x 10-3
moles of NaOH = 0.1 x 100 /1000
moles of NaOH = 10x 10-3
the reaction is given by
CH3COOH + NaOH ------> CH3COONa + H20
from the reaction
moles of CH3COONa formed = moles of NaOh reacted
moles of CH3COONa formed = 10 x 10-3
CH3COOH and CH3COONa form a buffer solution
according to hasselbach hendersen equation
pH = pKa + log[salt / acid ]
pH = -log Ka + log[CH3COONa / CH3COOH ]
pH = -log 1.7 x 10-5 + log { 10 x 10-3 / 10 x 10-3 ]
pH = 4.76955 + log 1
pH = 4.76955
f) Methyl orange
g)
moles of acetic acid = molarity x volume / 1000
moles of acetic acid = 0.1 x 100/1000
moles of acetic acid = 10 x 10-3
moles of NaOH = 0.1 x 300 /1000
moles of NaOH = 30 x 10-3
the reaction is given by
CH3COOH + NaOH ------> CH3COONa + H20
from the reaction
moles of CH3COONa formed = moles of CH3COOH reacted
moles of CH3COONa formed = 10x 10-3
moles of NaOH remaining = 30 x 10-3 - 10 x 10-3
moles of NaOh remaining = 20 x 10-3
total volume = 100 + 300
total volume = 400 ml
conc of NaOH = 20 x 10-3 x 1000 / 400
[NaOH] = 0.05
only NaOH and CH3COONa remain in the solution
As naOH is a very strong base , the pH is determined by NaOH
NaOH ----> Na+ + OH-
so
[OH-] = [ NaOH]
pOH = -log [OH-]
pOH = -log 0.05
pOH = 1.3
pH = 14 -pOH
pH = 14 -1.3
pH = 12.70
so the pH is 12.70
h) you may use
N1V1 = N2V2 to find the equivalnce point
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.