Answer:
E) 2.00
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
Answer:
E) 2.00
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
[tex]K_{\text{a}} = \dfrac{\text{[H}_{3}\text{O}^{+}] \text{F}^{-}]} {\text{[HF]}} = 6.8 \times 10^{-4}\\\\\dfrac{x^{2}}{0.15 - x} = 6.8 \times 10^{4}\\\\\text{Check for negligibility of }x\\\\\dfrac{ 0.15 }{6.8 \times 10^{-4}} = 220 < 400\\\\\therefore x \text{ is not negligible. We must solve a quadratic.}[/tex]
[tex]x^{2} = 6.8 \times 10^{-4} (0.15 - x)\\x^{2} = 1.02 \times 10^{-4} - 6.8 \times 10^{-4}x\\x^{2} + 6.8 \times 10^{-4}x - 1.02 \times 10^{-4} = 0\\\text{Solve the quadratic and get}\\x = 9.77 \times 10^{-3}\\\rm [H_{3}O^{+}]= x \, mol \cdot L^{-1} = 9.77 \times 10^{-3} \, mol\cdot L^{-1}[/tex]
Step 2. Calculate the pH
[tex]\text{pH} = -\log{\rm[H_{3}O^{+}]} = -\log{9.77 \times 10^{-4}} = \boxed{\mathbf{2.01}}[/tex]
