Answer:
The molarity of barium cation in the solution is 0.173 M
Explanation:
Step 1: The balanced equation
Ba(NO3)2(aq) + Na2CrO4 (aq) → BaCrO4(s) + 2NaNO3(aq)
Step 2: Data given
Mass of Barium nitrate = 13.6 grams
Volume of 0.40M sodium chromate = 300 mL
Step 3: Calculate moles of Ba(NO3)2
Moles = mass / molar mass
Moles = 13.6 grams / 261.34 g/mol
Moles = 0.052 moles
Step 4: Calculate moles of Na2CrO4
Moles = Molarity * Volume
Moles Na2CrO4 = 0.40 * 0.3L
Moles Na2CrO4 = 0.12 moles
Step 5: Calculate limiting reactant
Na2CrO4 is in excess so all of Ba(NO3)2 will be consumed and reacts to form BaCrO4(s) in the form Ba2+
Step 6: Calculate moles of Ba2+
n(Ba2+)=n(BaCrO4) =n(Ba(NO3)2 = 0.0520 moles
Step 7: Calculate molarity of Ba2+
C=n/v so C(Ba2+)=0.0520/0.300 = 0.173 M
The molarity of barium cation in the solution is 0.173 M
