Problem Page Suppose 25.1g of barium nitrate is dissolved in 200. mL of a 0.50 M aqueous solution of ammonium sulfate. 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. Round your answer to 3 significant digits.

1.26 M

Explanation:

The ion nitrate is NO₃⁻ and the Barium is from group 2 so it forms the ion Ba²⁺, so the barium nitrate has the formula: Ba (NO₃) ₂. The molar masses are: Ba: 137 g/mol, N = 14 g/mol, O = 16 g/mol, so the molar mass of barium nitrate is:

137 + 2x (14 + 3x16) = 199 g/mol

The number of moles is the mass divided by the molar mass, so:

n = 25.1/199 = 0.126 mol of Ba (NO₃) ₂

In 1 mol of the salt, there are 2 moles of NO₃⁻, so the number of moles of nitrate is 0.252 mol. Nitrates formed with ammonium (that can react when the solid dissolves) and with elements from group 1 and 2 are completely soluble in water. So, the moles of nitrate will remain 0.252 mol.

The molarity is the number of moles divided by the volume (0.2 L):

[NO₃⁻] = 0.252/0.2 = 1.26 M