Calculate all the ions present in the solution when 45 mL of 0.23 M mercury (II) nitrate is mixedwith 95 mL of lead (IV) nitrate.

Answer 1

mercury(II) nitrate = Hg(NO3)2

lead(IV) nitrate = Pb(NO3)4

mixing them together produces mercury(II) ions, (Hg2+), lead(IV) ions (Pb4+), nitrate ions (NO3-) and some hydrogen (H+) and hydroxide (OH-) ions from the water used.

Since the question didn't ask for the concentrations of ions, but only for what ions are present, I would say the following ions are present:

Hg²⁺

Pb⁴⁺

NO₃^-

H⁺

OH^-