Find the degree ionization of .75 M HF
HF ==> H^+ + a = (H^+)(F^-)
Set up an ICE chart, substitute into the Ka expression, solve for H^+, then degree of ionization = (H^+)/(0.75)
To find the degree of ionization of a compound, we need to know its initial molar concentration and the equilibrium molar concentration of the ions formed when it dissociates.
The given compound is hydrofluoric acid (HF), and we are given its initial molar concentration as 0.75 M.
HF is a weak acid, meaning it does dissociate to some extent in water. The balanced equation for the dissociation of HF is:
HF(aq) ⇌ H+(aq) + F-(aq)
Let's assume that at equilibrium, x moles of HF have dissociated. Therefore, the molar concentration of H+ (hydrogen ion) and F- (fluoride ion) at equilibrium will both be x M.
Since HF dissociates in a 1:1 ratio, the concentration of H+ and F- ions at equilibrium will also be x M.
To determine the degree of ionization (α), we need to compare the equilibrium concentration of H+ ions (x M) to the initial concentration of HF (0.75 M).
α = (x M) / (0.75 M)
However, we need additional information to determine the value of x and the degree of ionization. This information can be provided in the question, such as the value of the equilibrium constant or any other data needed to solve for x. Without that information, we cannot determine the degree of ionization for the given concentration of 0.75 M HF.
To determine the degree of ionization of a weak electrolyte, such as HF in this case, we need to use the initial concentration and the equilibrium concentration of ions.
HF is a weak acid, and it partially ionizes in water as follows:
HF ⇌ H⁺ + F⁻
Let's assume that the degree of ionization of HF is represented by the symbol "α". This means that "α" represents the fraction of HF that ionizes.
Initially, when HF is dissolved in water, the concentration of HF is 0.75 M. After ionization, the concentration of H⁺ and F⁻ ions will be "α" M each.
The equilibrium concentration of HF will be (0.75 - α) M, and the concentrations of H⁺ and F⁻ ions will each be α M.
Since HF dissolves to produce one H⁺ ion and one F⁻ ion, we can equate the initial concentration of HF (0.75 M) to the sum of the equilibrium concentrations of H⁺ and F⁻ ions:
0.75 M = α M + α M
0.75 M = 2α M
Now, we can solve for "α":
α = 0.75 M / 2
α = 0.375 M
Therefore, the degree of ionization of 0.75 M HF is 0.375 or 37.5%.