Question

To determine the concentration of SO42– ion in a sample of groundwater, 100.0 mL of the sample is titrated with 0.0250 M Ba(NO3)2, forming insoluble BaSO4. If 3.19 mL of the Ba(NO3)2 solution is required to reach the end point of the titration, what is the molarity of the SO42–? Ba(NO3)2 + SO42- → BaSO4 + NO3- in the sample?

Answer 1

Answer:

The concentration of the SO42- ion is 7.975 * 10 ^-4 M

Explanation:

Step 1: Data given

Volume of the 0.0250 M Ba(NO3)2 sample = 100 mL = 0.1 L

3.19 mL of Ba(NO3)2 solution is required to reach the end point of the titration

Step 2: The balanced equation

Ba(NO3)2 + SO42- → BaSO4 + 2NO3-

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

The number of moles of Ba(NO3)2 = Molarity of Ba(NO3)2 * Volume

The number of moles of Ba(NO3)2 = 0.025M *3.19 x 10^-3  L

 The number of moles of Ba(NO3)2 = 7.975 * 10 ^-5 moles                                          

Step 4: Calculate number of moles SO42-                                          

For 1 mole Ba'NO3)2 consumed, we need 1 mole of SO42-

For 7.975 * 10 ^-5 moles of Ba(NO3)2, we need 7.975 * 10 ^-5 moles of SO42-

Step 5: Calculate concentration of SO42-

[SO42-] = Moles SO42- / volume

[SO42-] = 7.975 * 10 ^-5 moles / 0.1L

[SO42-] = 7.975 * 10 ^-4 M

The concentration of the SO42- ion is 7.975 * 10 ^-4 M