# chem lab (webwork)

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What is the pH of the solution created by combining 1.80 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
1.80
Complete the table below:

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)?

mL NaOH pH wHCl pH wHC2H3O2
1.80

• chem lab (webwork) -

Your question is tough to decipher.
The basics.
moles = M x L.
Determine moles acid. Determine moles base. Determine which is in excess. Determine the salt produced.
For solutions in which the salt is not hydrolyzed, use the reactant in excess and calculate H^+ or OH^- accordingly. For salts that hydrolyze, you may have a buffer and you should use the Henderson-Hasselbalch equation.
pH = pKa + log [(base)/(acid)]

• chem lab (webwork) -

can you please elaborate i'm still unsure because

.1 M NaOH x .0018 L = 1.8E-4 mol NaOH

and .1 M HC2H3)2 x .008 L = 8E-4 mol HC2H3)2

so that means that HC2H3)2 is in excess.

after this step i am lost so can you please tell me how to find the salt produced and the other steps.