To calculate the pH at each point in the titration, we need to consider the reaction that is occurring. In this case, HCl is a strong acid and NaOH is a strong base, so the reaction that takes place is:
HCl + NaOH -> NaCl + H2O
Since we are adding HCl to NaOH, the HCl will react with the NaOH until one of the reactants is completely used up.
(i) At 0.0 mL of titrant added:
- moles of NaOH = 0.050 L * 0.100 mol/L = 0.005 mol
- moles of HCl = 0
Since no HCl has been added yet, the solution only contains NaOH. The pH of the initial NaOH solution can be calculated by using the formula for a strong base:
pH = 14 - pOH = 14 - log[OH-]
pOH = -log(0.100) = 1
pH = 14 - 1 = 13
(ii) At 10.0 mL of titrant added:
- moles of HCl added = 0.010 L * 0.200 mol/L = 0.002 mol
- moles of NaOH left = 0.005 - 0.002 = 0.003 mol
After adding the HCl, the limiting reactant is NaOH, so we need to calculate the concentration of OH- ions and then convert it to pH using the formula above.
(iii) At 25.0 mL of titrant added:
- moles of HCl added = 0.025 L * 0.200 mol/L = 0.005 mol
- moles of NaOH left = 0
Since all the NaOH has reacted with the HCl, the solution only contains salt (NaCl) and water. The pH of a salt solution can be calculated using the formula:
pH = 7
(iv) At 30.0 mL of titrant added:
- moles of HCl added = 0.030 L * 0.200 mol/L = 0.006 mol
- moles of NaOH left = 0
Similar to the previous step, the solution only contains NaCl and water. Therefore, the pH remains at 7.
(ii) pH at 0.0, 10.0, 25.0, and 30.0 mL of titrant in the titration
of 50.0 mL of 0.100 M NaOH with 0.200 M HCl.
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