Question

Suppose 0.248g of barium nitrate is dissolved in 50.mL of a 18.0mM aqueous solution of sodium chromate. Calculate the final molarity of barium cation in the solution. You can assume the volume of the solution doesn't change when the barium nitrate is dissolved in it.

Answer 1

the balanced equation is as follows

Ba(NO3)2 (aq) + Na2CrO4 (aq) --------------------> BaCrO4 (s) + 2 NaNO3 (aq)

no of moles = mass / molar mass

for Ba(NO3)2 = 0.248 g / 261.337 g/mol => 0.0009489 moles

sodium chromate moles = 0.018 M * 0.050 L => 0.00090 moles

mole ratio of both reactants are same 1:1

1 mol Na2CrO4 -------------------> 1 mol Ba(NO3)2

0.00090 mol ------------------> 0.00090 moles

excess reactant is Ba(NO3)2

excess moles after the reaction => 0.0009489 - 0.00090 => 0.00004896 moles

the precipitate is BaCrO4 so the remaining excess moles of Ba2+ is 0.00004896 moles

concentration of Ba2+ ion = 0.00004896 moles / 0.050 L => 0.0009793 M

answer => 9.793 * 10^-4 M