Question

Suppose 0.275g of barium nitrate is dissolved in 100.mL of a 17.0mM aqueous solution of sodium chromate. Calculate the final molarity of nitrate anion in the solution. You can assume the volume of the solution doesn't change when the barium nitrate is dissolved in it. Be sure your answer has the correct number of significant digits.

Answer 1

mass of Ba(NO3)2 = 0.275 g

moles of Ba(NO3)2 = 0.275 / 261.34 = 1.05 x 10^-3

moles of Na2CrO4 = 17 x 10^-3 x 0.1 = 1.7 x 10^-3 mol

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

      1                  1

1.05 x 10^-3    1.7 x 10^-3

here Ba(NO3)2 is limiting reagent .

moles of NO3- = 2 x 1.05 x 10^-3 = 2.104 x 10^-3 mol

Molarity of NO3- = 2.104 x 10^-3 / 0.1 = 0.0210 M

final molarity of nitrate anion = 0.0210 M