The pH of the solution after adding 40.3 mL of 0.15 M NaOH to 20.00 mL of 0.3000 M HBr is approximately 13.18 (rounded to two decimal places).
To determine the pH of the solution after 40.3 mL of 0.15 M NaOH have been added to 20.00 mL of 0.3000 M HBr, follow these steps:
1. Calculate the moles of HBr initially present:
Moles of HBr = Molarity × Volume (L)
Moles of HBr = 0.3000 M × (20.00 mL / 1000 mL/L) = 0.006 moles
2. Calculate the moles of NaOH added:
Moles of NaOH = Molarity × Volume (L)
Moles of NaOH = 0.15 M × (40.3 mL / 1000 mL/L) = 0.006045 moles
3. Determine the limiting reactant:
Since HBr and NaOH react in a 1:1 ratio, the limiting reactant is the one that is completely consumed first. In this case, both HBr and NaOH react completely, so neither is limiting.
4. Calculate the total moles of HBr and NaOH after the reaction:
Total moles = Moles of HBr initially + Moles of NaOH added
Total moles = 0.006 moles + 0.006045 moles = 0.012045 moles
5. Calculate the total volume of the solution after mixing:
The total volume is the sum of the volumes of HBr and NaOH solutions.
Total volume = (20.00 mL + 40.3 mL) / 1000 mL/L = 0.0603 L
6. Calculate the concentration of the resulting solution:
Concentration (M) = Total moles / Total volume
Concentration = 0.012045 moles / 0.0603 L ≈ 0.1997 M
7. Calculate the pOH of the solution:
pOH = -log10[OH-]
Since NaOH is a strong base, it completely dissociates, and [OH-] is equal to the concentration of NaOH added:
pOH = -log10(0.15 M) ≈ 0.8238
8. Calculate the pH of the solution:
pH = 14 - pOH
pH = 14 - 0.8238 ≈ 13.1762
So, the pH of the solution after adding 40.3 mL of 0.15 M NaOH to 20.00 mL of 0.3000 M HBr is approximately 13.18 (rounded to two decimal places).