Calculate the pH of a solution obtained by mixing 100ml of 0.2M HCl with 100ml of 0.1M NaOH

Step 1: Calculate the moles of HCl and NaOH used in the solution.

Moles of HCl = concentration (M) × volume (L) = 0.2 mol/L × 0.1 L = 0.02 mol
Moles of NaOH = concentration (M) × volume (L) = 0.1 mol/L × 0.1 L = 0.01 mol

Step 2: Determine which reactant is in excess.

Since the moles of HCl (0.02 mol) are greater than the moles of NaOH (0.01 mol), HCl is in excess.

Step 3: Calculate the moles of HCl remaining after the reaction.

Moles of HCl remaining = moles of HCl initially - moles of NaOH used
Moles of HCl remaining = 0.02 mol - 0.01 mol = 0.01 mol

Step 4: Calculate the concentration of HCl in the solution after the reaction.

The volume of the solution is 100 mL (0.1 L) + 100 mL (0.1 L) = 0.2 L

Concentration of HCl = moles of HCl remaining / volume of solution
Concentration of HCl = 0.01 mol / 0.2 L = 0.05 M

Step 5: Calculate the pOH of the solution.

pOH = -log[OH-]
Since the moles of NaOH added were 0.01 mol and the volume of the solution is 0.2 L, the concentration of OH- is:
Concentration of OH- = moles of OH- / volume of solution = 0.01 mol / 0.2 L = 0.05 M

pOH = -log(0.05) ≈ 1.30

Step 6: Calculate the pH of the solution.

pH + pOH = 14
pH = 14 - pOH = 14 - 1.30 ≈ 12.70

Answer: The pH of the solution obtained by mixing 100 mL of 0.2 M HCl with 100 mL of 0.1 M NaOH is approximately 12.70.