Respuesta :
the concentration of Na after the two solutions are mixed together is 0.662 M.
1. Volume of Na2CrO4 = 132.4 ml = 0.1324 L.
2. Concentration of Na2CrO4 = 2.50 M
3. Volume of AgNO3 = 125 ml = 0.125 L
4. Concentration of AgNO3 = 2.50 M.
Calculation :-
1. Moles of Na2CrO4 = Molarity * Volume in L.
Moles of Na2CrO4= 2.50 * 0.1324 = 0.331 mol
2. Moles of AgNO3 = Molarity * volume in L
Moles of AgNO3 = 2.50 * 0.125 = 0.3125 mol
Solution :
the reaction of aqueous silver nitrate (AgNO3) with aqueous sodium chromate (Na2CrO4) to form solid silver chromate (Ag2CrO4) and aqueous sodium nitrate (NaNO3).
2 AgNO3(aq) + Na2CrO4 (aq) \rightarrow Ag2CrO4(s) + 2 NaNO3(aq).
Above balanced reaction shows that,
2 moles of AgNO3 reacts with 1 mol Na2CrO4.
Let, 0.3125 mol AgNO3 reacts with x mol Na2CrO4.
x = 0.3125 / 2
x = 0.1563 mol Na2CrO4.
Now,
1. Balanced reaction shows that,
2 mol AgNO3 forms 2 mol NaNO3.
So, 0.3125 mol AgNO3 forms 0.3125 mol NaNO3.
So, moles of NaNO3 = 0.3125 mol.
As 1 molecule of NaNO3 gives 1 Na^+ ion.
0.3125 mol NaNO3 gives 0.3125 mol Na^+ ions...... A
Aslo,
2. remaining moles of Na2CrO4 = total moles - used moles
Remaining moles of Na2CrO4 = 0.331 - 0.1563 = 0.1748 mol.
1 molecule of Na2CrO4 gives 2 Na^+ ions.
So, 0.1748 moles of Na^2+ ions gives 0.3496 mol Na^+ ions............ B
3.So, total moles of Na^+ ions= 0.3496 + 0.3125......(from A and B)
Total moles of Na^+ ions = 0.662 mol.
So, concentration of Na^+ ions = 0.662 M.
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