Answer:
c. 7.00
Explanation:
The neutralization reaction between NaOH and HCl is the following:
NaOH + HCl → NaCl + H₂O
NaOH dissociates into OH⁻ ions and HCl dissociates into H⁺ ions. At the equivalence point, the total number of moles of OH⁻ is neutralized with the moles of H⁺ added:
OH ⁻ + H⁺ ⇄ H₂O
Moles of NaOH = M x V = 0.025 L x 0.30 mol/L = 7.5 x 10⁻³ mol OH⁻
Since NaOH is a strong base, the moles of NaOH is equal to the moles of OH⁻.
Moles of HCl = M x V = 0.025 L x 0.30 mol/L = 7.5 x 10⁻³ mol H⁺
As we can see, the moles of HCl added are equal to the moles of H⁺ because HCl is a strong acid.
At the equivalence point: 7.5 x 10⁻³ mol OH⁻ = 7.5 x 10⁻³ mol H⁺ (the total amount of OH⁻ reacted with H⁺)
Thus, in the medium we have only the H⁺ from the water equilibrium:
Kw = [H⁺] [OH⁻] = 1 x 10⁻¹⁴
[H⁺] = 1 x 10⁻⁷
⇒ pH = -log [H⁺] = -log (1 x 10⁻⁷) = 7
Therefore, the correct answer is c. 7.00
