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

pH of the final solution = 3.8

Step-by-step explanation:

Concentration of NaF =
`(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 =
`([H]^(+)[F]^(-) )/([HF])`

⇒ [H⁺] =
`(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⁺] =
`(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