Answer to Question #149887 in General Chemistry for NURUL AFIQAH BINTI MOHD HANAFI

Question

A 〖MgCO〗_3 solution was prepared by dissolving 0.500 g 〖MgCO〗_3 in sufficient water to produce 250.00 mL solution. A 20.00 mL aliquot is taken and put into an Erlenmeyer flask. A 25.00 mL of 0.1500 M 〖HNO〗_3 is added into the flask. The resulting solution is then titrated with 0.12 M NaOH. Calculate the volume of NaOH required for the back titration.

Answer 1

1) MgCO₃ + 2HNO₃ = MgNO₃ + H₂CO₃

2) HNO₃ + NaOH = NaNO₃ + H₂O

M (MgCO₃) = 84.31 g/mol

n (MgCO₃) = 0.500 / 84.31 = 0.006 mol

C_M (MgCO₃) = n/V = 0.006 / 0.25 = 0.024 M

20.00 ml of this MgCO₃ solution contains:

n₁(MgCO₃) = 20.00/1000 x 0.024 = 0.0005 mol

According to the equation 1, n(HNO₃) = 2 x n(MgCO₃) = 2 x 0.0005 = 0.001 mol

n(HNO₃)_actual = 0.1500 x 25.00/1000 = 0.0038 mol

So that present HNO₃ is in excess: 0.0038 - 0.001 = 0.0028 mol

According to the eqanion 2, n (HNO₃) = n (NaOH) = 0.0028 mol

C_M=n/V

V (NaOH) = n/C_M = 0.0028 / 0.12 = 0.03 L = 30 ml