What they want you to do is to take the solution after reaction, dilute to 100 mL, and recalculate the pH.
For the first one, you had 0.00061 moles HCl left over and dilute to 100 mL means that the molarity, instead of being 0.00061/0.0099 = 0.0616 M, it now is 0.00061/0.1 = 0.0061 and pH = 2.21 (as opposed to 1.21 earlier).
For the second one,
you had 0.00019 of the salt, and 0.00061 moles of the acetic acid left. Instead of 0.00019/0.0099 and 0.00061/0.0099 the concns will be 0.00019/0.1 = ?? and 0.00061/0.1 = ?? The final pH, if I didn't goof will be 4.23 (as opposed to 4.23 earlier; i.e., no change).
You may wonder what all of this is about so don't miss the point of these questions. When you took the first one (strong acid/strong base) and diluted the excess acid the pH changed BIG TIME from 1.21 to 2.21 (a change of 10 in acidity) BUT doing the same thing to the weak acid/weak base (the buffered solution) THERE IS NO CHANGE IN pH even though you have added 90 mL MORE water to it. Remember that a buffered solution is supposed to resist a change in pH AND IT DID.
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?
This is in regards to my previous question. We are still calculating the pH of w/HCl and w/HC2H3O2.
I don't really understand what to do with the 100ml.
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thank you
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