how many total moles of ions are released when 0.0822 moles of Na2HPO4 dissolves completely in water?

1 mole of Na₂HPO₄ dissociates in water to give 3 moles of ions (2 moles of Na⁺ ions and 1 mole of HPO₄²⁻ ions.

Total no. of moles of ions released = (0.0822 mol) × 3 = 0.2466 mol ≈ 0.247 mol (to 3 sig. fig.)

Disodium hydrogenphosphate.....

Your teacher probably expects you to say that you would get 0.1644 moles of Na+, 0.0822 moles of H+ and 0.0822 moles of PO4^2-, but that would be incorrect.

The salt will ionize to produce...
Na2HPO4 --> 2Na+ + HPO4^2-
0.0822 mol ...... 0 .......... 0
....... 0 ......0.1644 mol .. 0.0822 mol

HPO4^2- is a weak acid and essentially does not ionize to make H+ and PO4^3-. (PO4^3- is found only in very, very basic solutions.) But, HPO4^2- will react with water to make H2PO4^- and OH-
HPO4^2- + HOH(l) <==> H2PO4^- + OH- ............ Kb = 0.021
Therefore, some of the HPO4^2- will be converted to H2PO4^2- and OH-

0.0822 mol Na2HPO4 produces ....
0.1644 mol Na+ ions
some HPO4^2- ions
less than 0.0822 moles H2PO4^- ions
less than 0.0822 moles OH-

Assume a 1L solution
HPO4^2- + HOH(l) <==> H2PO4^- + OH- ............ Kb = 0.021
0.0822M .............................0 ............. 0 ............... initial
-x ...................................... +x ........... +x ............... change
0.0822-x ............................ x .............. x ............... equilibrium

Kb = [H2PO4^-][OH-] / [HPO4^2-]
0.021 = x² / (0.0822-x)
--- some algebra goes here ---
x = 0.0324
[HPO4^2-] = 0.0822-0.0324 = 0.0498M
[H2PO4^-] = 0.0324M
[OH-] = 0.0324M

The bottom line:
0.1644 moles Na+
0.0498 moles HPO4^2-
0.0324 moles H2PO4^-
0.0324 moles OH-