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

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)]

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.

To find the pH of the solution created by combining the given volumes and concentrations of NaOH, HCl, and HC2H3O2, we need to consider the reaction that occurs between the acids and bases and then calculate the resulting pH.

Let's start with the reaction between NaOH and HCl. The balanced chemical equation is:

NaOH(aq) + HCl(aq) -> NaCl(aq) + H2O(l)

Since NaOH is a strong base and HCl is a strong acid, they will react completely to form NaCl (a neutral salt) and water.

To calculate the resulting pH, we need to find the concentration of H+ ions (hydrogen ions) in the solution. Since NaCl is a neutral salt, it does not affect the pH of the solution. Therefore, we only need to consider the H+ ions from HCl.

To do this, we can use the equation:

[H+] = (moles of acid) / (total volume of solution in liters)

Given that we have 8.00 mL of 0.10 M HCl, we can calculate the moles of acid as:

moles of acid = (volume of acid) x (concentration of acid)
= (8.00 mL) x (0.10 mol/L)
= 0.80 mmol

Since we diluted the 8.00 mL of 0.10 M HCl with 100 mL of water, the total volume of the solution is 108.00 mL (8.00 mL + 100 mL).

Converting the total volume to liters:

total volume = 108.00 mL / 1000 mL/L
= 0.108 L

Now, we can calculate the concentration of H+ ions:

[H+] = (0.80 mmol) / (0.108 L)
≈ 7.41 M

To find the pH, we can use the equation:

pH = -log[H+]

pH = -log(7.41)
≈ -0.869

Therefore, the pH of the solution created by combining 1.80 mL of 0.10 M NaOH with 8.00 mL of 0.10 M HCl (diluted with 100 mL of water) is approximately -0.869.

You can follow a similar process to calculate the pH when 8.00 mL of 0.10 M HC2H3O2 (also diluted with 100 mL of water) is combined with 1.80 mL of 0.10 M NaOH by considering the reaction between the acid and base and using the appropriate equation.