two solutions are mixed: 100.0 mL HCl(aq) with p 2.20 and 100.0 mL of NaOH}(aq) with pH} 11.20.

What is the pH} of the resulting solution?

Same process as your next post.

To determine the pH of the resulting solution, we need to consider the reaction between HCl and NaOH, which is a neutralization reaction. In this reaction, HCl reacts with NaOH to form water (H2O) and a salt (NaCl). The pH of the resulting solution will depend on the concentrations of the H+ and OH- ions in the mixture.

To calculate the pH of the resulting solution, we need to determine the concentrations of H+ and OH- ions in the mixture. We can do this by first calculating the moles of HCl and NaOH in the mixture.

Step 1: Calculate the moles of HCl:
- The volume of HCl solution is given as 100.0 mL, which is equal to 0.100 L (since 1 L = 1000 mL).
- The molarity (concentration) of HCl is not given, but we can calculate it using the pH value. The pH of a solution is related to the concentration of H+ ions using the equation: pH = -log[H+]. Rearranging the equation, we get [H+] = 10^(-pH).
- For HCl with a pH of 2.20, the concentration of H+ ions is [H+] = 10^(-2.20) mol/L.

Using the volume and concentration, we can calculate the moles of HCl:
Moles of HCl = Volume of HCl solution (L) × Concentration of HCl (mol/L)
Moles of HCl = 0.100 L × 10^(-2.20) mol/L

Step 2: Calculate the moles of NaOH:
- The volume of NaOH solution is also given as 100.0 mL, which is equal to 0.100 L.
- The molarity (concentration) of NaOH is not given, but we can calculate it using the pH value. The pOH of a solution is related to the concentration of OH- ions using the equation: pOH = -log[OH-]. Since pH + pOH = 14, we can calculate the concentration of OH- ions using the equation: [OH-] = 10^(-pOH).
- For NaOH with a pOH of 11.20, the concentration of OH- ions is [OH-] = 10^(-11.20) mol/L.

Using the volume and concentration, we can calculate the moles of NaOH:
Moles of NaOH = Volume of NaOH solution (L) × Concentration of NaOH (mol/L)
Moles of NaOH = 0.100 L × 10^(-11.20) mol/L

Step 3: Determine the limiting reactant:
- The reaction between HCl and NaOH occurs in a 1:1 ratio. Whichever reactant is present in lower moles will be the limiting reactant, meaning it will be completely consumed, and the other reactant will be in excess.
- To determine the limiting reactant, compare the moles of HCl and NaOH calculated in steps 1 and 2. The reactant with the lower number of moles will be the limiting reactant.

Step 4: Calculate the remaining moles of excess reactant:
- Subtract the moles of the limiting reactant (from step 3) from the moles of the excess reactant (from step 1 or 2, whichever is not the limiting reactant). This will give us the moles of the excess reactant that did not react.

Step 5: Calculate the concentration of OH- ions in the resulting solution:
- Since NaOH is the limiting reactant and it reacts completely, the remaining moles of NaOH will be zero.
- The moles of OH- ions present in the resulting solution will be equal to the initial moles of NaOH since all of it reacted.
- The volume of the resulting solution is the sum of the volumes of HCl and NaOH (100.0 mL + 100.0 mL = 200.0 mL = 0.200 L).
- Therefore, the concentration of OH- ions in the resulting solution is [OH-] = Moles of NaOH / Volume of resulting solution (L).

Step 6: Calculate the pOH of the resulting solution:
- Use the concentration of OH- ions calculated in step 5 to calculate the pOH using the equation: pOH = -log[OH-].

Step 7: Calculate the pH of the resulting solution:
- Use the equation: pH + pOH = 14 to calculate the pH of the resulting solution. Rearranging the equation, we get pH = 14 - pOH.

By following these steps, you can calculate the pH of the resulting solution.