posted by Jason .
What is the pH value if you take into account that the 8.00 mL of 0.10 M HCl was first diluted with 100 mL of water
and titrated with 1.60 mL of the 0.10 M NaOH
i know that the concentration of HCl after diluting is .008M but than im not sure how to convert that to mols which volume do i use
Won't both give you the same answer. After all, the number of mols is not changing. SO M x L = mols.
0.1 M x 0.008 L = ??
0.008 M x 0.1 L (that's 100 mL) = ??
The (HCl) at the for this problem, after the reaction, simply is moles/L but the liters is now 100 mL + 1.60 mL + 8.00 (check these numbers since I'm not looking at them--that's just the way I remember the problem).
No, that is not the new dilution. The new volume is 108ml (you added 100ml) with the .1*8/1000 moles of HCL.
Now, the NaOH will remove (.0016*.1) moles in titration.
New moles of HCL= .1*.008-.0016*.1
volume= 108+1.6 ml
New concentration of H+ is molesHCl/volume in liters.