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

This is what I know, what should I do next:

.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.

  • chem lab (webwork) -

    Subtract HAc moles - NaOH moles.
    The difference = moles HAc (acetic acid) remaining. Amount NaOH added = moles Ac^-(acetate ion) produced.
    pH = pKa + log ([base)/(acid)]
    where base is the acetate and acid is the acetic acid.

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