Help please. Am doing a past exam paper and I hit a road block with this question:
Chloride in a brine solution is determined by a precipitation titration. A 10.00 mL
aliquot of the solution is titrated with 15.00mL of standard 0.1182 M AgNO3
solution. The excess silver is titrated with standard 0.1010 M KSCN solution,
requiring 2.38 mL to reach the red Fe(SCN)2+ endpoint. Calculate the
concentration of chloride in the brine solution, in g/1000mL.
Thank you! :)
Do I calculate the moles of KSCN and the moles of AgNO3
.
Moles reacted: moles AgNO3 - moles KSCN
Then say : C =mols reacted/volume(the 10 ml volume x 10 to the minus 3)
Then multiple mol/L by g/mol then convert to g/ml
To tackle this question, we will use the concept of stoichiometry and titration calculations. Here's how you can go about solving it step by step:
Step 1: Write the balanced chemical equation
AgNO3 + KSCN → AgSCN + KNO3
The equation shows that 1 mole of AgNO3 reacts with 1 mole of KSCN to form 1 mole of AgSCN.
Step 2: Calculate the number of moles of AgNO3 used in the titration
Given that the volume of the AgNO3 solution is 15.00 mL and its concentration is 0.1182 M:
Moles of AgNO3 = concentration × volume
Moles of AgNO3 = 0.1182 M × 0.01500 L
Moles of AgNO3 = 0.001773 mol
Step 3: Determine the number of moles of KSCN used to titrate the excess silver
Given that the volume of KSCN solution used is 2.38 mL and its concentration is 0.1010 M:
Moles of KSCN = concentration × volume
Moles of KSCN = 0.1010 M × 0.00238 L
Moles of KSCN = 0.00024078 mol
Step 4: Calculate the moles of AgNO3 that reacted with KSCN
From the balanced equation, we know that 1 mole of AgNO3 reacts with 1 mole of KSCN. Therefore, the number of moles of AgNO3 that reacted with KSCN is equal to the number of moles of KSCN used:
Moles of AgNO3 reacted = Moles of KSCN = 0.00024078 mol
Step 5: Determine the moles of AgNO3 that reacted with chloride ions
Since AgNO3 reacts with KSCN on a one-to-one basis, the moles of AgNO3 that reacted with KSCN are also equal to the moles of AgNO3 that reacted with chloride ions present in the 10.00 mL aliquot of the brine solution.
Moles of AgNO3 reacted with chloride ions = Moles of AgNO3 reacted = 0.00024078 mol
Step 6: Calculate the concentration of chloride ions in the brine solution
Since the aliquot taken was 10.00 mL, we need to calculate the concentration of chloride ions in grams per 1000 mL (g/1000mL):
Concentration of chloride ions = (Moles of AgNO3 reacted with chloride ions × Molar mass of chloride ions) / Volume of aliquot
Concentration of chloride ions = (0.00024078 mol × 35.45 g/mol) / 0.01000 L
Concentration of chloride ions = 0.8505 g/1000mL
So, the concentration of chloride in the brine solution is 0.8505 g/1000mL.