Answer:
i) pH = 0.6990
ii) pH = 2.389
Explanation:
i) Before adding aqueous NaOH, there are 25.00 mL of 0.2000 M HCl. HCl reacts with the water in the aqueous solution as follows:
HCl + H₂O ⇒ H₃O⁺ + Cl⁻
The HCl and H₃O⁺ are related to each other through a 1:1 molar ratio, so the concentration of H₃O⁺ is equal to the HCl concentration.
The pH is related to the hydronium ion concentration as follows:
pH = -log([H₃O⁺]) = -log(0.2000) = 0.699
ii) Addition of NaOH causes the following reaction:
H₃O⁺ + NaOH ⇒ 2H₂O + Na⁺
The H₃O⁺ and NaOH react in a 1:1 molar ratio. The amount of NaOH added is calculated:
n = CV = (0.2000 mol/L)(24.00 mL) = 4.800 mmol NaOH
Thus, 4.800 mmol of H₃O⁺ were neutralized.
The initial amount of H₃O⁺ present was:
n = CV = (0.2000 mol/L)(25.00 mL) = 5.000 mmol H₃O⁺
The amount of H₃O⁺ that remains after addition of NaOH is:
(5.000 mmol) - (4.800 mmol) = 0.2000 mmol
The concentration of H₃O⁺ is the amount of H₃O⁺ divided by the total volume. The total volume is (25.00 mL) + (24.00 mL ) = 49.00 mL
C = n/V = (0.2000 mmol) / (49.00 mL) = 0.004082 M
The pH is finally calculated:
pH = -log([H₃O⁺]) = -log(0.004082) = 2.389
