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 = ![\frac{[Al(OH)_{2} ^-]}{[Al(OH)][OH^-]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BAl%28OH%29_%7B2%7D%20%5E-%5D%7D%7B%5BAl%28OH%29%5D%5BOH%5E-%5D%7D)
The concentration of OH⁻ is:
pOH = 14 - pH = <em>3</em>
pOH = -log [OH⁻]
[OH⁻] = 1x10⁻³
Thus:
40 = ![\frac{[Al(OH)_{2} ^-]}{[Al(OH)][1x10^{-3}]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BAl%28OH%29_%7B2%7D%20%5E-%5D%7D%7B%5BAl%28OH%29%5D%5B1x10%5E%7B-3%7D%5D%7D)
<em>0,04M = ![\frac{[Al(OH)_{2} ^-]}{[Al(OH)]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BAl%28OH%29_%7B2%7D%20%5E-%5D%7D%7B%5BAl%28OH%29%5D%7D)
</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!
Answer:
H2O molecules are thus able to form an average of 4 H-bonds. H2O has a higher boiling point than NH3 because (i) the H-bonds are stronger and (ii) it contains twice as many H-bonds. H2O has a higher boiling point than HF because it contains twice as many H- bonds, despite these being individually weaker.
Explanation:
hope this is helpful
An ionic compound is always an electrolyte
Answer:
Glucose also called(Energy)
