# Chemistry DrBob

What is the pH of the solution created by combining 2.30 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)?

mL NaOH pH wHCl pH wHC2H3O2
2.30 ________ _________

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

mL NaOH pH wHCl pH wHC2H3O2
2.30 _______ ______

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Also for the previous question that i asked, for the second part, ..

How would the volume of base change from problem three if the 8.00 mL of 0.10 M Acid was first diluted with 100 mL of water ? the volume will increase? or stay the same?

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1. Is this two problems? or one?
one with NaOH + HCl
the other one NaOH + acetic acid.

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2. its two sorry the whole thing that i wrote didn't show up again.

What is the pH of the solution created by combining 2.30 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)?

Here is the chart that i am supposed fill out:

mL NaOH pH w/ HCl pH w/ HC2H3O2
2.30 _________ ____________

Here's the second part of the problem.
I need to complete the chart too

mL NaOH pH w/ HCl pH w/ HC2H3O2
2.30 _________ _____________

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For my previous question, for the second part: How would the volume of base change from problem three if the 8.00 mL of 0.10 M Acid was first diluted with 100 mL of water ?

Answer: The volume of base does not change right?

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3. The NaOH + HCl part you know how to do from the previous problem I worked. And you are right about the volume of base not changing if the solution is diluted AFTER putting it in the titration vessel.

For the NaOH + HC2H3O2 problem, you need to write the equation. It is
NaOH + HC2H3O2 ==> NaC2H3O2 + HOH.

Determine mols NaOH from mols = M x L.
Determine mols HC2H3O2 from M x L = ??
This problem will not be the same kind as the NaOH + HCl BECAUSE in NaOH/HCl you are titrating a strong acid with a strong base and both are ionized 100%. With NaOH + acetic acid, you must realize that this is the addition of a strong base to a weak acid. I haven't gone through the numbers but I'm sure acetic acid is in excess which means you are left with some sodium acetate and some unreacted acetic acid. That is a buffer. Solve that with the Henderson-Hasselbalch equation. I'll let you worry about the chart.

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4. me thinks u right

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