What is the pH of a 0.20 M HF solution that is 5.8% ionized at 25 Celsius ?
A.4.45
b.0.70
C.1.93
D.1.23
E.1.40
0.20M x 0.058 = 0.0116M = (H^+). Convert to pH.
HF ==> H^+ + F^-
Well, let me calculate that for you!
So, if the solution is 5.8% ionized, that means only 5.8% of the HF has dissociated into ions.
Now, HF is a weak acid, so it doesn't fully ionize in water. This means that we have to use the equilibrium expression to determine the concentration of H+ ions.
The equilibrium expression for HF is:
HF ⇌ H+ + F-
The Ka expression for this reaction is:
Ka = [H+][F-]/[HF]
Since only 5.8% of the HF has ionized, we can assume that the concentration of H+ ions is also 5.8% of the initial concentration of HF.
So, the concentration of H+ ions is:
[H+] = 0.20 M × 0.058 = 0.0116 M
Now, to find the pH, we take the negative logarithm of the H+ concentration:
pH = -log[H+]
= -log(0.0116)
≈ 1.93
Therefore, the correct answer is C. 1.93. Don't worry, even though acid can be a bit sour, it's okay to laugh at the joke!
To find the pH of a solution, we need to use the equation for the ionization of the acid, which in this case is HF:
HF ⇌ H⁺ + F⁻
Given that the solution is 5.8% ionized, we can assume that the concentration of H⁺ ions is 5.8% of the initial concentration of HF. Therefore, the concentration of H⁺ ions is:
0.058 × 0.20 M = 0.0116 M
To find the pH, we need to use the formula:
pH = -log[H⁺]
Substituting the concentration of H⁺:
pH = -log(0.0116) = 1.93
Therefore, the correct answer is C. 1.93.
To determine the pH of a 0.20 M HF (hydrofluoric acid) solution that is 5.8% ionized, we can use the concept of the ionization constant, also known as the equilibrium constant.
First, we need to write the balanced chemical equation for the ionization of hydrofluoric acid:
HF (aq) ⇌ H⁺ (aq) + F⁻ (aq)
The ionization constant (Ka) expression for this reaction is given by:
Ka = [H⁺] [F⁻] / [HF]
Since the concentration of HF is given as 0.20 M and we know that the solution is 5.8% ionized, we can calculate the concentration of H⁺ (which is equal to the concentration of F⁻ too) using the equation:
[H⁺] = [F⁻] = concentration of HF x degree of ionization
[H⁺] = [F⁻] = 0.20 M x 0.058
[H⁺] = [F⁻] = 0.0116 M
Now that we have the concentrations of H⁺ and F⁻ ions, we can use the equation pH = -log[H⁺] to calculate the pH of the solution:
pH = -log(0.0116) ≈ 1.93
Therefore, the pH of the 0.20 M HF solution that is 5.8% ionized at 25 degrees Celsius is approximately 1.93. So, the correct answer is C.