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.
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