Complete the table below: Note: Make simplifying assumptions, do not use the quadratic formula.

What is the pH of the solution created by combining 1.90 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: 1.90
pH w/HCl: 1.21
pH w/HC2H3O2:


Complete the table below: Note: Make simplifying assumptions, do not use the quadratic formula.
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?
mL NaOH: 1.90
pH w/HCl:
pH w/HC2H3O2:

Okay. I found the pH for the first HCl, but I'm not sure on how to do the rest of them. help?

Okay I got the second pH w/HCl and it is 2.23. but still can't find the pH of both of the ones w/HC2H3O2. Please help?

1.9 mL x 0.1M NaOH = 0.19 millimols NaOH

8.00 mL x 0.1M HAc = 0.8 mmols HAc.

.........HAc + NaOH ==> NaAc + H2O
I........0.....0.19.........0......0
added...0.8........................
C......-0.19...-0.19....0.19....0.19
E......0.61......0.......0.19....0.19

(HAc) = 0.61mmols/9.9 mL = ?M (9.9 is total volume from 8.00+1.9 = 9.9 total)
(NaAc) = (Ac^-) = 0.19 mmols/9.9 mL = ?
Then pH = pKa + log(base)/(acid)

For the diluted part,
(HAc) = 0.61 mmols/101.9 mL = ?
(Ac^-) = 0.19 mmol/101.9 mL = ?

Sure! Let's start by going through the steps to find the pH values for the solutions.

To find the pH of a solution, we need to first calculate the concentration of hydrogen ions (H+) in the solution. Then, we can use the formula pH = -log[H+].

Let's start with the solution formed by combining 1.90 mL of 0.10 M NaOH with 8.00 mL of 0.10 M HCl.

Step 1: Calculate the moles of NaOH and HCl used.
Moles of NaOH = volume (in L) x concentration (in mol/L)
= 1.90 mL x (1 L/1000 mL) x 0.10 mol/L
= 0.0019 mol
Moles of HCl = volume (in L) x concentration (in mol/L)
= 8.00 mL x (1 L/1000 mL) x 0.10 mol/L
= 0.008 mol

Step 2: Determine the limiting reactant.
Since NaOH and HCl react in a 1:1 ratio based on their balanced chemical equation (NaOH + HCl -> NaCl + H2O), the limiting reactant is the one with fewer moles. In this case, NaOH is the limiting reactant because it has fewer moles compared to HCl.

Step 3: Calculate the moles of HCl reacted with NaOH.
Moles of HCl reacted = Moles of NaOH (limiting reactant) = 0.0019 mol

Step 4: Calculate the concentration of remaining HCl.
Initial moles of HCl - Moles reacted = Moles of HCl remaining
Moles of HCl remaining = 0.008 mol - 0.0019 mol = 0.0061 mol

Step 5: Calculate the concentration of H+ ions in the solution.
Volume of solution = volume of NaOH + volume of HCl = 1.90 mL + 8.00 mL = 9.90 mL = 9.90 x (1 L/1000 mL) = 0.00990 L
Concentration of H+ ions = Moles of HCl remaining / Volume of solution
= 0.0061 mol / 0.00990 L = 0.616 mol/L

Step 6: Calculate the pH using the formula.
pH = -log[H+]
pH = -log(0.616) = 0.210

So, the pH of the solution formed by combining 1.90 mL of 0.10 M NaOH with 8.00 mL of 0.10 M HCl is approximately 0.210.

Now, let's move on to finding the pH value when 8.00 mL of 0.10 M HC2H3O2 is combined with 1.90 mL of 0.10 M NaOH.

The steps are similar to the previous calculation:

Step 1: Calculate the moles of HC2H3O2 and NaOH used.
Moles of HC2H3O2 = volume (in L) x concentration (in mol/L)
= 8.00 mL x (1 L/1000 mL) x 0.10 mol/L
= 0.008 mol
Moles of NaOH = 1.90 mL x (1 L/1000 mL) x 0.10 mol/L
= 0.0019 mol

Step 2: Determine the limiting reactant.
Again, NaOH is the limiting reactant because it has fewer moles compared to HC2H3O2.

Step 3: Calculate the moles of HC2H3O2 reacted with NaOH.
Moles of HC2H3O2 reacted = Moles of NaOH (limiting reactant) = 0.0019 mol

Step 4: Calculate the concentration of remaining HC2H3O2.
Initial moles of HC2H3O2 - Moles reacted = Moles of HC2H3O2 remaining
Moles of HC2H3O2 remaining = 0.008 mol - 0.0019 mol = 0.0061 mol

Step 5: Calculate the concentration of H+ ions in the solution.
Volume of solution = volume of HC2H3O2 + volume of NaOH = 8.00 mL + 1.90 mL = 9.90 mL = 0.00990 L
Concentration of H+ ions = Moles of HC2H3O2 remaining / Volume of solution
= 0.0061 mol / 0.00990 L = 0.616 mol/L

Step 6: Calculate the pH using the formula.
pH = -log[H+]
pH = -log(0.616)

Unfortunately, we need more information to complete the calculation. Please provide the pKa value for acetic acid (HC2H3O2) in order to calculate the pH value for the solution with HC2H3O2.