Calculate the number of milliliters of 0.754 M KOH required to precipitate all of the Zn2+ ions in 162 mL of 0.731 M Zn(
Posted: Tue May 17, 2022 10:35 am
Calculate the number of milliliters
of 0.754 M KOH required
to precipitate all of
the Zn2+ ions
in 162 mL
of 0.731 M Zn(NO3)2 solution
as Zn(OH)2. The equation for the
reaction is:
Zn(NO3)2(aq)
+ 2KOH(aq) Zn(OH)2(s)
+ 2KNO3(aq)
of 0.754 M KOH required
to precipitate all of
the Zn2+ ions
in 162 mL
of 0.731 M Zn(NO3)2 solution
as Zn(OH)2. The equation for the
reaction is:
Zn(NO3)2(aq)
+ 2KOH(aq) Zn(OH)2(s)
+ 2KNO3(aq)