Answer:
<u>H2PO4- is a proton donor and HPO42_ is a proton acceptor</u>
Explanation:
Step 1: What are hydrogen ion donor and acceptor
in the following reaction we see that:
⇒ H2PO4- is more likely to give a H+ ion to form HPO42-.
⇒HPO42- is more likely to take a H+ ion, to form H2PO4-
The reaction of an acid in water solvent is described as a dissociation
:
HA ⇔ H+ + A-
⇒where HA is a proton acid
So, H2PO4- = HA and HPO42- = A-
Acids are proton donors. So, <u>H2PO4- is a proton donor and HPO42_ is a proton acceptor</u>
The third shell has 3 subshells: the subshell, which has 1 orbital with 2 electrons, the subshell, which has 3 orbitals with 6 electrons, and the subshell, which has 5 orbitals with 10 electrons, for a total of 9 orbitals and 18 electrons.
We assume that we have Ka= 4.2x10^-13 (missing in the question)
and when we have this equation:
H2PO4 (-) → H+ + HPO4-
and form the Ka equation we can get [H+]:
Ka= [H+] [HPO4-] / [H2PO4] and we have Ka= 4.2x10^-13 & [H2PO4-] = 0.55m
by substitution:
4.2x10^-13 = (z)(z)/ 0.55
z^2 = 2.31x 10^-13
z= 4.81x10^-7
∴[H+] = 4.81x10^-7
when PH equation is:
PH= -㏒[H+]
= -㏒(4.81x10^-7) = 6.32
