How did you do part 1. You had moles NaOH and moles HCl (HCl in excess), yo subtracted moles HCl-moles NaOH to arrive at moles HCl in excess. I think that is 0.69 millimoles and that divided by the volume (8 mL + 1.1 mL = 9.1 mL) = 0.0758 M and pH = 1.12. So for part 2, you have the same of everything except the 0.69 millimoles is now in 100 + 1.1 = 101.1 mL. Convert that to pH.
The 3.94 you obtained for the initial part was obtained by using the Henderson-Hasselbalch equation. That is
pH = pKa + log (base)/(acid)
Notice, however, in this one that the pH is determined by a ratio (not like HCl/NaOH). When you dilute the base by what ever factor (8 mL up to 100) that it dilutes BOTH the base AND the acid the same amount and since mole/L is concn, the volume cancels and its just moles/moles and the volume doesn't matter. The end result is that dilution has no effect on the buffer solution.
I got part 1:
What is the pH of the solution created by combining 1.10 mL of the 0.10 M NaOH(aq) with 8.00 mL of the 0.10 M HCl(aq)? with 8.00 mL of the 0.10 M HC2H3O2(aq)?
pH w/HCl =1.12 w/HC2H3O2(aq)= 3.94
But how do you do part two?
What are the pH values if you take into account that the 8.00 mL of 0.10 M Acid was first diluted with 100 mL of water (like it will be in the experiment you perform in lab)?
thanks
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