Answer:
(a) H₃O⁺(aq) + H₂PO₄⁻(aq) ⟶ H₃PO₄(aq) + H₂O(ℓ)
(b) OH⁻(aq) + H₃O⁺(aq) ⟶ 2H₂O(ℓ)
Explanation:
The equation for your buffer equilibrium is:
H₃PO₄(aq) + H₂O(ℓ) ⇌ H₃O⁺(aq)+ H₂PO₄⁻(aq)
(a) Adding H₃O⁺
The hydronium ions react with the basic dihydrogen phosphate ions.
H₃O⁺(aq) + H₂PO₄⁻(aq) ⟶ H₃PO₄(aq) + H₂O(ℓ)
(b) Adding OH⁻
The OH⁻ ions react with the more acidic hydronium ions.
OH⁻(aq) + H₃O⁺(aq) ⟶ 2H₂O(ℓ)
Hello!
After the addition of a small amount of acid, a reasonable value of buffer pH would be
5,15.
If initially there are equal amounts of a weak acid and its conjugate base, the pH would be equal to the pKa, according to the
Henderson-Hasselbach equation:
![pH=pKa+log( \frac{[A^{-}] }{[HA]} ) \\ \\ if [A^{-}]=[HA] \\ \\ pH=pKa + log (1) \\ \\ pH=pKa + 0=5,25](https://tex.z-dn.net/?f=pH%3DpKa%2Blog%28%20%5Cfrac%7B%5BA%5E%7B-%7D%5D%20%7D%7B%5BHA%5D%7D%20%29%20%5C%5C%20%5C%5C%20if%20%5BA%5E%7B-%7D%5D%3D%5BHA%5D%20%5C%5C%20%5C%5C%20pH%3DpKa%20%2B%20log%20%281%29%20%5C%5C%20%5C%5C%20pH%3DpKa%20%2B%200%3D5%2C25)
So, when adding a little amount of acid the pH should be only a little lower than the pKa. The value from the list that is a little lower than the pKa is
5.15
Have a nice day!
Answer:
Ester Linkages
Explanation:
In a fat molecule, the fatty acids are attached to each of the three carbons of the glycerol molecule with an ester bond through the oxygen atom.
You're welcome
The specific heat capacity of sand is much less than water because of which sands require less energy to cool down and similarly less energy to be heated... because of this it feels like sands absrob heat faster than water
Answer:
You have a trait (plant color) and you have two possible outcomes for it (purple and white). You know that some alleles for white were present in one of the parent plants, so you know they must be present in the offspring too. But they're expression is covered up by the purple alleles from the other plant, so you know that purple is dominant to white.