posted by al on .
During an acid-base titration, 25 mL of NaOH 0.2 M were required to neutralize 20 mL of HCl.
Calculate the pH of the solution for each:
a) Before titration
b) After adding 24.9 mL of NaOH
c) At equivalence pt
d) After adding 25. 1 mL of NaOH
e) The pH of the equivalence pt during a certain titration is 3.5
f) Which indicator would be adequate for this titration?
g) Which indicator would you suggest to be avoided? Explain
First, determine the M of the HCl. That will be 25*0.2/20 = 0.25M
Where is the equivalence point. That is at 25.0 mL, NaOH, of course.
The questions divide the titration curve up nicely into
a. at the beginning.
b. before the eq point.
c. at eq pt
d. just after eq pt
It's a matter of keeping account of the mols.
M HCl = 0.2M so pH = -log(HCl)
millimols HCl initially = 5
millimols NaOH added = 4.98
millimols HCl remaining = 5-4.98 = 0.02
M HCl remaining = mmols/mL = 0.02/44.9
pH = -log(HCl)
c. At the eq you have NaCl and H2O. Neither Na^+ nor Cl^- are hydrolyzed; therefore, the pH is 7.0
d. mmols HCl initially = 5.0
mmols NaOH at 25.1 ix 25.1 x 0.2 = 5.02
mmols excess NaOH = 0.02
pH = mmols NaOH/mL solution = 0.02/45.1
I don't understand e.
Look in your text/notes for pH range of indicators.