Answer:
0,040 M
Explanation:
The global reaction of the problem is:
Al(OH) (s) + OH⁻ ⇄ Al(OH)₂⁻(aq) K= 40
The equation of equilibrium is:
K =
The concentration of OH⁻ is:
pOH = 14 - pH = <em>3</em>
pOH = -log [OH⁻]
[OH⁻] = 1x10⁻³
Thus:
40 =
<em>0,04M = </em>
This means that 0,04 M are the number of moles that the solvent can dissolve in 1L, in other words, solubility.
I hope it helps!
<u>Answer:</u> The equilibrium concentration of chlorine gas, fluorine gas and ClF gas is 0.159 M, 0.317 M and 1.002 M respectively.
<u>Explanation:</u>
We are given:
Initial concentration of chlorine gas = 0.307 M
Initial concentration of fluorine gas = 0.465 M
Initial concentration of ClF gas = 0.706 M
The given chemical equation follows:
<u>Initial:</u> 0.307 0.465 0.706
<u>At eqllm:</u> 0.307-x 0.465-x 0.706+2x
The expression of for above equation follows:
We are given:
Putting values in above equation, we get:
Neglecting the value of x = 0.993 because the equilibrium concentrations of chlorine and fluorine gases will become negative, which is not possible
So, equilibrium concentration of chlorine gas = (0.307 - x) = [0.307 - 0.148] = 0.159 M
Equilibrium concentration of fluorine gas = (0.465 - x) = [0.465 - 0.148] = 0.317 M
Equilibrium concentration of ClF gas = (0.706 + 2x) = [0.706 + 2(0.148)] = 1.002 M
Hence, the equilibrium concentration of chlorine gas, fluorine gas and ClF gas is 0.159 M, 0.317 M and 1.002 M respectively.