Consider the titration of 100.0 mL of 0.0835 M HA(a weak acid) with NaOH. Stoich
ID: 878242 • Letter: C
Question
Consider the titration of 100.0 mL of 0.0835 M HA(a weak acid) with NaOH.
Stoichiometric Reaction: HA+NaOH ->NaA+H20
Equilibrium Reaction: HA+H20 <-->A^-+H30+
The Ka of HA=6.9E-07 and concentration of NaOE=0.250 M.
Determine:
a)Initial pH of HA solution before any base has been added (I believe this answer is 3.620)
b) pH after 29 mL base has been added (I believe this answer is 7.079)
c) Volume of NaOH base needed to reach equivalence point (I believe this is 33.4 mL)
d) pH at equivalence point
e) pH after 75.0 mL of naOH has been added
Thank you!! Please explain
Explanation / Answer
d-At the equivalence point the pH is greater then 7 because all of the acid (HA) has been converted to its conjugate base (A-) by the addition of NaOH and now the equilibrium moves backwards towards HA and produces hydroxide, that is: A+H2OAH+OH
pH= pKa= -log Ka = 7+0.83885=7.83885
a-HA+ H2O -----> H3O+ + A-
0.0835M 0 0 I
0.0835-x x x E
k= x^2/0.0835-x
6.9x10^-7(0.0835-x) -x^2=0
0.5886x10^-7-6.9x10^-7x -x^2=0
x=0.24227e-3
pH= -log[H+]= - log [.24227x10^-3]= 3-(-6157) = 3.6157
b-volume of NaOH= 29ml
The number of millimoles of HA to be neutralized is 100x0.0835mmol/1ml= 8.35mmol of HA
The number of millimoles of OH_ that will be added within 29 mL is 29x 0.250mmol/1ml=7.25 mmol
HA+ OH- ----> H2O + A-
8.35 7.25 0 0 I
1.1 7.25
concentration of HA= 1.1/129= 8.53x10^-3M
concentration of A- = 7.25 /129= 0.0562M
pH = pKa + log[-A]/[HA]
pH= - log 6.9x10^-7 + log[0.0562/ 8.53x10^-3]=7-0.84+ 0.82= 7.02
c-The equivalence point occurs when equal moles of acid react with equal moles of base.
mmol of HA= 100x0.0835=8.35mmol
mmol of NaOH= 8.35
8.35mmol = 0.250 M x V of NaOH
volume of NaOH = 33.4ml
e- NaOH= 75 ml
The millimoles of OH- added in the 75ml = 75 x 0.25 mmol/1ml=18.75 mmol
mmoles of hydroxide in excess: 18.75 mmol - 8.35 mmol= 10.4 mmol OH-
concentration of OH = 10.4 mmol /175= =0.05942 M
pOH = - log [OH-]= - log [0.05942]= 1.226
pH = 14- 1.226= 12.774