When the reaction equation is:
HF ↔ H+ + F-
and when the Ka expression
= concentration of products/concentration of reactions
so, Ka = [H+][F-]/[HF]
when we assume:
[H+] = [F-] = X
and [HF] = 0.35 - X
So, by substitution:
6.8 x 10^-4 = X^2 / (0.35 - X) by solving for X
∴ X = 0.015 M
∴[H+] = X = 0.015
when PH = -㏒[H+]
∴PH = -㏒0.015
= 1.8
Answer: The final concentration of
at equilibrium is 0.36 M
Explanation:
Moles of
= 0.64 mole
Volume of solution = 1.0 L
Initial concentration of
= 
The given balanced equilibrium reaction is,

Initial conc. 0.64 M 0M 0M
At eqm. conc. (0.64-x) M (x) M (x) M
The expression for equilibrium constant for this reaction will be,
Now put all the given values in this expression, we get

By solving the term 'x', we get :
x = 0.36
Thus, the final concentration of
at equilibrium is x = 0.36 M
boron because the farther down the table you go the stronger the elements get
Answer: D
Explanation:
Reduction involves the gaining of electrons.
- This eliminates choices A and C.
However, charge must be balanced, and this is not the case for choice B, but it is the case for choice D