Posted by michelle on Saturday, November 6, 2010 at 5:04pm.
The pH at the beginning of the titration is just the pH of a 0.2 M HF solution.
b).
moles HF = M x L = ??
moles NaOH = M x L = ??
Which is in excess. Subtract. If HF in in excess, you will have a buffer solution of HF and NaF. If NaOH is in excess the pH will be determined by the concn OH.
Post your work if you get stuck.
for a) i got the pH of HF to be 2.07 and for b) i'm a little stuck.
the mole of HF is 0.2M x 0.05L =0.1mol
the mole of NaOH is 0.2M x 0.02 L= 0.02mols. this is where i'm stuck
Hint write a balnace equation between HF and NaOH.
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