Question

For the reaction: HF(aq)H'(aq)+F(aq) determine the change in the pH (ApH) for the addition of 1.75 M NaF salt to a 5.45 M solution of HF, Ka-7.2 x 104 Note: ApH pHHFtion pHacid no ion View Available Hint(s)

Answer 1

Answer:

Dph=1.46

Explanation:

HF(aq) ↔ H^+(aq) + F-(aq)

Ka = [H+][F-]/[HF]

(7.2×10^-4) = X×X/(5.45-X)

X = 0.0623

[H+] = X = 0.0623 M

pH = -log[H+]

= -log0.0623

= 1.2

after addition of NaF

Ka = [H+][F-]/[HF]

(7.2×10^-4) = X×(1.75+X)/(5.45-X)

x = 0.0022

[H+] = 0.0022

pH = - log0.0022 = 2.66

DpH = 2.66 - 1.2 = 1.46

therefore, DpH ≅ 1.5

Answer 2

Answer:

1.4

Explanation:

Let's consider the acid-dissociation of HF.

HF(aq) ⇄ H⁺(aq) + F⁻(aq)

For a weak acid, we can calculate the concentration of H⁺ ([H⁺]) using the following expression.

H⁺ = √(Ka × Ca) = √(7.2 × 10⁻⁴ × 5.45) = 0.063 M

where,

Ka: acid-dissociation constant

Ca: concentration of the acid

The pH is

pH = -log [H⁺] = -log 0.063 = 1.2

When we add a NaF solution, we have a buffer system, formed by a weak acid (HF) and its conjugate base (F⁻ coming from NaF). We can calculate the pH of a buffer using the Henderson-Hasselbalch's equation.

pH = pKa + log ([F⁻]/[HF])

pH = 3.1 + log (1.75M/5.45M) = 2.6

The change in the pH is

ΔpH = 2.6 - 1.2 = 1.4