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.
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.