A reaction caf2(s)ca2+(aq)+2f(aq)caf2(s)ca2+(aq)+2f(aq) is at equilibrium. If the concentration of ca2+ca2+ is increased fourtimes,fourtimes, what will be the change in ff concentration as compared to the initial concentration of ff ?
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HEY MATE HERE'S YOUR ANSWER ;
calcium fluoride will dissociate into calcium ions and fluoride ions
As an equation, we can write that as:
CaF2(s) Ca2+ (aq) + 2F-(aq)
Now that we have the equation, let's put it into a solubility expression:
Ksp = [Ca2+] * [F-]2
When you plug in your known values, you have:
1.9 x 10-10 = [2.4 x 10-4] * [F-]2
Solving for the concentration of F- should give you 8.9 x 10-4 M.
calcium fluoride will dissociate into calcium ions and fluoride ions
As an equation, we can write that as:
CaF2(s) Ca2+ (aq) + 2F-(aq)
Now that we have the equation, let's put it into a solubility expression:
Ksp = [Ca2+] * [F-]2
When you plug in your known values, you have:
1.9 x 10-10 = [2.4 x 10-4] * [F-]2
Solving for the concentration of F- should give you 8.9 x 10-4 M.
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