Question

2. A 100.00 mL volumetric flask is loaded with 1.60 grams of NaF. To this was added 30.0 mL of a 4.00 M solution of HF. When this flask is filled to the mark and shaken to ensure complete mixing, what is the pH of the final solution

Answer 1

Answer:

pH of the final solution = 3.8

Explanation:

Concentration of NaF = $\frac{1.6 X 1000}{42 X 100}$ molar

                                    = 0.3 molar

NaF → Na⁺ +    F⁻

HF   ⇆ H⁺    +    F⁻

• NaF is strong electrolyte so completely ionized but HF weak acid not completely ionized.

• Since F⁻ is common ion here

         according to common ion effect dissociation of weak acid decreases.

        Ka = $\frac{[H]^{+}[F]^{-} }{[HF]}$

⇒    [H⁺]  = $\frac{K_{a} [HF]}{[F]^{-} }$ ...............(1)

{Ka of HF = 3.5 x 10⁻⁴} & Concentration of HF = 30 x 4 x 10⁻³ = 0.12 molar

from equation 1

⇒ [H⁺]  = $\frac{3.5 X 10^{-4}X 0.12 }{0.3}$     [Concentration of F⁻ ≡ Concentration of NaF]

⇒ [H⁺] = 0.00014

⇒pH = - log 0.00014 = 3.85