Question

Determine the pH of the equivalence point for a reaction in which 0.100 mol/L NaOH neutralizes 10.0 mL of 0.250 mol/L HF.

Answer 1

Question #58559, Chemistry, Other

Determine the pH of the equivalence point for a reaction in which 0.100 mol/L NaOH neutralizes 10.0 mL of 0.250 mol/L HF.

Answer:

HF is weak acid and NaOH is strong base. Alkaline pH is anticipated as a result of their reaction at the equivalence point.

Moles HF = 0.0100 L × 0.250 mol/L = 0.00250

Moles NaOH required to reach the equivalence point = 0.00250

Volume NaOH = 0.00250 mol/ 0.100 mol/L = 0.0250 L

Total volume = 0.0250 + 0.0100 = 0.0350 L

The reaction of neutralization is

$$\mathrm{HF} + \mathrm{NaOH} = \mathrm{NaF} + \mathrm{H_2O}$$

Moles NaF formed = 0.00250

NaF is a strong salt: NaF = Na⁺ + F⁻

$$[F^-] = 0.00250 \text{ mol} / 0.0350 \text{ L} = 0.0714 \text{ M}$$

The equilibrium is: $F^- + H_2O \leftrightarrow HF + OH^-$

$$K_b = K_w / K_a = 1.0 \times 10^{-14} / 6.8 \times 10^{-4} = 1.5 \times 10^{-11} = x^2 / 0.0714 - x$$ $$x = [\mathrm{OH^-}] = 1.03 \times 10^{-6} \text{ M}$$ $$pOH = 5.99$$ $$pH = 14 - 5.99 = 8.01$$

http://www.AssignmentExpert.com/