The pH of solution formed by mixing 40 mL of 0.10 M HCl and 10 mL of 0.45 M of NaOH is

To find the pH of the solution formed by mixing HCl and NaOH, we first need to determine the moles of each substance.

1. Calculate the moles of HCl and NaOH:

   Moles of HCl = Molarity * Volume(L) = 0.10 M * 0.040 L = 0.004 moles Moles of NaOH = Molarity * Volume(L) = 0.45 M * 0.010 L = 0.0045 moles

2. Determine the limiting reactant:

   Since NaOH and HCl react in a 1:1 ratio, the substance with the fewer moles will be the limiting reactant. In this case, HCl is the limiting reactant.

3. Calculate the moles of NaOH remaining after the reaction:

   Moles of NaOH remaining = Moles of NaOH - Moles of HCl = 0.0045 moles - 0.004 moles = 0.0005 moles

4. Calculate the total volume of the solution:

   Total volume = Volume of HCl + Volume of NaOH = 0.040 L + 0.010 L = 0.050 L

5. Calculate the concentration of NaOH:

   Concentration of NaOH = Moles of NaOH / Volume(L) = 0.0005 moles / 0.050 L = 0.01 M

6. Calculate the pOH of the solution:

   pOH = -log[OH-] = -log(0.01) = 2

7. Calculate the pH of the solution:

   pH = 14 - pOH = 14 - 2 = 12

So, the pH of the solution formed by mixing 40 mL of 0.10 M HCl and 10 mL of 0.45 M of NaOH is 12.