Answer : The ratio of the protonated to the deprotonated form of the acid is, 100
Explanation : Given,

pH = 6.0
To calculate the ratio of the protonated to the deprotonated form of the acid we are using Henderson Hesselbach equation :
![pH=pK_a+\log \frac{[Salt]}{[Acid]}](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%20%5Cfrac%7B%5BSalt%5D%7D%7B%5BAcid%5D%7D)
![pH=pK_a+\log \frac{[Deprotonated]}{[Protonated]}](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%20%5Cfrac%7B%5BDeprotonated%5D%7D%7B%5BProtonated%5D%7D)
Now put all the given values in this expression, we get:
![6.0=8.0+\log \frac{[Deprotonated]}{[Protonated]}](https://tex.z-dn.net/?f=6.0%3D8.0%2B%5Clog%20%5Cfrac%7B%5BDeprotonated%5D%7D%7B%5BProtonated%5D%7D)
As per question, the ratio of the protonated to the deprotonated form of the acid will be:
Therefore, the ratio of the protonated to the deprotonated form of the acid is, 100
The chemical equation for Hydrogen is just H
Answer:
The concentration is 0.036 mg/mL
Explanation:
Concentration = 0.2 mM = 0.2/1000 = 2×10^-4 M = 2×10^-4 mol/L × 180,000 mg/1 mol × 1 L/1000 mL = 0.036 mg/mL
4d7 , 3s2 will be the valence shell electronic configuration of a neutral atom of cobalt!