Question

Suppose 3.40g of copper(II) nitrate is dissolved in 150.mL of a 0.20M 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 copper(II) nitrate is dissolved in it.Round your answer to 3 significant digits.

Answer 1

Answer:

The molarity of the nitrate ion is 0.241 M

Explanation:

Step 1: Data given

Mass of copper(II) nitrate = 3.40 grams

volume of the 0.20 M aqueous solution of sodium chromate = 150 mL =0.15 L

Step 2: The balanced equation

Cu(NO3)2 + Na2CrO4 → CuCrO4 + 2NaNO3 → 2Na+ + 2NO3-

For 1 mole Cu(NO3)2 consumed, we produce 2 moles of NO3-

Step 3: Calculate number of moles of Cu(NO3)2

Number of moles = Mass of Cu(NO3)2/ Molar mass of Cu(NO3)2

Moles Cu(NO3)2 = 3.40 grams /187.56 g/mol

Moles Cu(NO3)2 = 0.0181 moles

Step 4: Calculate moles of nitrate ion

For 1 mole Cu(NO3)2 consumed, we produce 2 moles of NO3-

For 0.0181 mole of Cu(NO3)2, we have 2* 0.0181 = 0.0362 moles of NO3-

Step 5: Calculate molarity of the nitrate ion

Molarity : Moles / volume

Molarity NO3- = 0.0362 moles / 0.150 L

Molarity NO3- = 0.241 M

The molarity of the nitrate ion is 0.241 M