how do you solve this?

What is the pH of the solution created by combining 12.30 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
12.30

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

To solve this problem, you need to consider the reaction between the acid (HCl or HC2H3O2) and the base (NaOH). The reaction between a strong acid (HCl) and a strong base (NaOH) results in a neutral solution. On the other hand, the reaction between a weak acid (HC2H3O2) and a strong base (NaOH) produces a basic solution.

To determine the pH of the solution, you can use the equation:

pH = -log[H+]

First, let's calculate the moles of each substance:

Moles of NaOH = volume (in L) × molarity
Moles of HCl = volume (in L) × molarity
Moles of HC2H3O2 = volume (in L) × molarity

Given that the volume of NaOH is 12.30 mL and the molarity is 0.10 M, we can calculate the moles of NaOH:

Moles of NaOH = (12.30 mL / 1000 mL/L) × 0.10 M = 0.00123 mol

Since the reaction between NaOH and HCl is 1:1, the moles of HCl will also be 0.00123 mol.

Now, let's find the moles of HC2H3O2. Given the volume of HC2H3O2 is 8.00 mL and the molarity is 0.10 M:

Moles of HC2H3O2 = (8.00 mL / 1000 mL/L) × 0.10 M = 0.0008 mol

Since the reaction between NaOH and HC2H3O2 is not 1:1, we need to consider the dilution of the acid with 100 mL of water before the reaction. The final volume of the HC2H3O2 solution will be 8.00 mL + 100 mL = 108.00 mL.

Now we can calculate the molarity of the diluted HC2H3O2 solution:

Molarity of diluted HC2H3O2 = Moles of HC2H3O2 / Volume (in L)
= (0.0008 mol) / (108.00 mL / 1000 mL/L)
= 0.00741 M

We can now proceed to calculate the pH values for each scenario:

1. For the combination of NaOH with HCl (resulting in a neutral solution):
pH = -log[H+] = -log[0] = undefined (Neutral solution)

2. For the combination of NaOH with HC2H3O2 (resulting in a basic solution):
pH = -log[H+] = -log[0.00741] = 2.13

So, the pH values when taking into account the dilution of the acid with 100 mL of water are:
- For the NaOH + HCl combination: Undefined (Neutral solution)
- For the NaOH + HC2H3O2 combination: 2.13 (Basic solution)