Answer:
The ratio [A-]/[HA] increase when the pH increase and the ratio decrease when the pH decrease.
Explanation:
Every weak acid or base is at equilibrium with its conjugate base or acid respectively when it is dissolved in water.
⇄ 
This equilibrium depends on the molecule and it acidic constant (Ka). The Henderson-Hasselbalch equation,
![pH = pKa + Log \frac{[A^{-}]}{[HA]}](https://tex.z-dn.net/?f=pH%20%3D%20pKa%20%2B%20Log%20%5Cfrac%7B%5BA%5E%7B-%7D%5D%7D%7B%5BHA%5D%7D)
shows the dependency between the pH of the solution, the pKa and the concentration of the species. If the pH decreases the concentration of protons will increase and the ratio between A- and AH will decrease. Instead, if the pH increases the concentration of protons will decreases and the ratio between A- and AH will increase.
I would say G sorry if it’s not right
Answer:
HA + KOH → KA + H₂O
Explanation:
The unknown solid acid in water can release its proton as this:
HA + H₂O → H₃O⁺ + A⁻
As we have the anion A⁻, when it bonded to the cation K⁺, salt can be generated, so the reaction of HA and KOH must be a neutralization one, where you form water and a salt
HA + KOH → KA + H₂O
It is a neutralization reaction because H⁺ from the acid and OH⁻ from the base can be neutralized as water
Because that solid is frozen liquid, so when it gets heated up, it will melt. Like Ice.
Hope I Helped! :)