Answer:
Explanation:
Hello,
In this case, ammonia dissociation is:
So the equilibrium expression:
![Kb=\frac{[NH_4^+][OH^-]}{[NH_3]}](https://tex.z-dn.net/?f=Kb%3D%5Cfrac%7B%5BNH_4%5E%2B%5D%5BOH%5E-%5D%7D%7B%5BNH_3%5D%7D)
That in terms of the reaction extent and the initial concentration of ammonia is written as:
Thus, solving by using solver or quadratic equation we find:
Which actually equals the concentration of hydroxyl ion, therefore the pOH is computed:
![pOH=-log([OH^-])=-log(0.00133)=2.88](https://tex.z-dn.net/?f=pOH%3D-log%28%5BOH%5E-%5D%29%3D-log%280.00133%29%3D2.88)
And the pH from the pOH is:
Best regards.
Answer:
halogen
Explanation:
group 17 contains halogens
Answer:
the quantity required can go from 117 ml (for maximum concentration) up to 2900 ml ( if the concentrated solution has molarity =0.420 M)
Explanation:
the amount of water required to dilute a solution V₁ liters of Molarity M₁ to V₂ liters of M₂
moles of hydrochloric acid = M₁ * V₁= M₂ * V₂
V₁ = V₂ * M₂/M₁
where
M₂ = 0.420 M
V₂ =2.90 L
Since the hydrochloric acid can be concentrated up to 38% p/V ( higher concentrations are possible but the evaporation rate is so high that handling and storage require extra precautions, like cooling and pressurisation)
maximum M₁ =38% p/V = 38 gr/ 0.1 L / 36.5 gr/mol = 10.41 M
then
min V₁ = V₂ * M₂/ max M₁ = 2.90 L* 0.420 M/ 10.41 M= 0.117 L = 117 ml
then the quantity required can go from 117 ml up to 2900 ml ( if M₁ = M₂)
A sodium cation is smaller than a sodium atom.
Hopefully this is right...
