Question

The Ka of hydrofluoric acid (HF) at 25.0 degrees celsius is 6.8*10-4. What is the pH of a 0.15M aqueous solution of HF?The Ka of hydrofluoric acid (HF) at 25.0 C is 6.8 x10-4. What is the pH of a 0.15 M aqueous 8) solution of HF? A) 4.60 B) 0.82 C)3.64 D) 1.17 E) 2.00

Answer 1

E) 2.00

Step-by-step explanation:

Step 1. Calculate the concentration of hydronium ion

We can use an ICE table to organize the calculations.

HF + H₂O ⇌ H₃O⁺ + F⁻

I/mol·L⁻¹: 0.15 0 0

C/mol·L⁻¹: -x +x +x

E/mol·L⁻¹: 0.15 - x x x

`K_{\text{a}} = \frac{\text{[H}_(3)\text{O}^(+)] \text{F}^(-)]} {\text{[HF]}} = 6.8 * 10^(-4)\\\\(x^(2))/(0.15 - x) = 6.8 * 10^(4)\\\\\text{Check for negligibility of }x\\\\( 0.15 )/(6.8 * 10^(-4)) = 220 < 400\\\\\therefore x \text{ is not negligible. We must solve a quadratic.}`

`x^(2) = 6.8 * 10^(-4) (0.15 - x)\\x^(2) = 1.02 * 10^(-4) - 6.8 * 10^(-4)x\\x^(2) + 6.8 * 10^(-4)x - 1.02 * 10^(-4) = 0\\\text{Solve the quadratic and get}\\x = 9.77 * 10^(-3)\\\rm [H_(3)O^(+)]= x \, mol \cdot L^(-1) = 9.77 * 10^(-3) \, mol\cdot L^(-1)`

Step 2. Calculate the pH

`\text{pH} = -\log{\rm[H_(3)O^(+)]} = -\log{9.77 * 10^(-4)} = \boxed{\mathbf{2.01}}`

Answer 2

E) 2.00

Step-by-step explanation:

The hydrofluoric acid is moderate, so, it must be in equilibrium at the solution. Making an equilibrium table for the reaction:

HF ⇄ H⁺ + F⁻

0.15 0 0 Initial

-x +x +x Reacts (stoichiometry is 1:1:1)

0.15-x x x Equilibrium

The equilibrium constant is the multiplication of the products' concentrations elevated by their coefficients, divided by the multiplication of the reactants' concentrations elevated by their coefficients:

Ka = [H⁺]x[F⁻]/[HF]

6.8*10⁻⁴= x*x/(0.15-x)

6.8*10⁻⁴ = x²/(0.15-x)

x² = 1.02*10⁻⁴ - 6.8*10⁻⁴x

x² + 6.8*10⁻⁴x - 1.02*10⁻⁴ = 0

Using Bhaskara's equation:

Δ = (6.8*10⁻⁴)² - 4*1*(- 1.02*10⁻⁴)

Δ = 4.08*10⁻⁴

x = (-6.8*10⁻⁴±√Δ)/2

x must be positive, so let's calculate only the positive:

x = (-6.8*10⁻⁴ + √4.08*10⁻⁴)/2

x = 9.76*10⁻³ M

[H⁺] = 9.76*10⁻³ M

pH = -log[H⁺] = -log(9.76*10⁻³)

pH = 2.00