If 25.00 mL of 0.30 M NaOH is titrated with 0.30 M HCl, what is the pH of the solution after the addition of 25.00 mL of HCl? a.13.00

b.9.00

c.7.00

d.5.00

e.1.00

(Need Work and Explanation)

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

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