solve Ka for HX => H^+ + X^- given 0.1M LiX at pH=8.9
LiX(aq) ==> Li^+(aq) + X^-(aq)
X^- is a base and hydrolyzes in water to
X^- + HOH ==> HX + OH^-
Make an ICE chart and substitute into the following:
Kb = (Kw/Ka) = (HX)(OH^-)/(X^-)
You know pH, convert to pOH, then to OH^-. HX = OH^-. You know Kw. The only unknown is Ka. Solve for that. Post your work if you get stuck.
Is the X^- in the ICE going to be 0.1 from the initial o.1M LiX?
See above.
i cant understand please elaborate
To solve for the equilibrium constant (Ka) for the dissociation of HX into H^+ and X^-, we need the concentration of H^+ and X^- ions at equilibrium.
Given 0.1 M LiX at pH=8.9, we need to determine the concentration of H^+ and X^- ions.
To find the concentration of H^+ ions, we need to know the pH value. The pH is a measure of the concentration of H^+ ions in a solution. The formula to calculate the concentration of H^+ ions from pH is as follows:
[H^+] = 10^(-pH)
In this case, the pH is 8.9. Plugging in the value:
[H^+] = 10^(-8.9)
Calculating this value will give us the concentration of H^+ ions.
To find the concentration of X^- ions, we need to subtract the concentration of H^+ ions from the initial concentration of LiX. Since the initial concentration of LiX is given as 0.1 M, we can calculate the concentration of X^- ions as follows:
[X^-] = Initial concentration of LiX - [H^+]
Calculating this value will give us the concentration of X^- ions.
Once we have both the concentrations of H^+ and X^- ions, we can use their concentrations to calculate the equilibrium constant (Ka). The expression for Ka is:
Ka = [H^+] * [X^-] / [HX]
Substituting the concentrations we calculated above, we can determine the value of Ka.