For the determination of the quantity of chloride in a physiological serum sample by the Volhard method, a sample of 25.00 ml was treated with 50.00 ml of 0.110 mol/L Silver nitrate solution and the excess of silver ions was titrated with 17.00 ml of a solution of potassium thiocyanate 0.100 mol/L using a solution containing iron III ions as an indicator. What concentration of NaCl in the sample in MOL/L and g/L
To determine the concentration of NaCl in the sample using the Volhard method, we need to understand the reaction that takes place and the stoichiometry involved.
The reaction equation for the reaction between silver nitrate (AgNO3) and potassium thiocyanate (KSCN) is:
Ag+ + SCN- → AgSCN
Based on the stoichiometry of this reaction, we know that one mole of Ag+ reacts with one mole of SCN- to form one mole of AgSCN.
In the given problem, a 25.00 ml sample of physiological serum solution was treated with 50.00 ml of 0.110 mol/L silver nitrate solution. This means that 0.110 mol of Ag+ was added to the solution. Since the stoichiometry of the reaction is 1:1, we can infer that 0.110 mol of SCN- is required to react with the 0.110 mol of Ag+ completely.
The volume of potassium thiocyanate solution required to react with the excess silver ions is 17.00 ml. Since the concentration of the potassium thiocyanate solution is given as 0.100 mol/L, we can calculate the number of moles of SCN- used in this titration:
Moles of SCN- = (Volume of KSCN solution in L) × (Concentration of KSCN solution in mol/L)
Moles of SCN- = (0.017 L) × (0.100 mol/L)
Now, subtract the number of moles of SCN- used in titration from the initial moles of SCN- added in excess:
Excess moles of SCN- = Initial moles of SCN- - Moles of SCN- used in titration
Finally, we can calculate the moles of Cl- present in the sample using the stoichiometry of the reaction:
Moles of Cl- = Excess moles of SCN-
The concentration of NaCl in the sample can be calculated using the moles of Cl- and the volume of the sample:
Concentration of NaCl (in mol/L) = Moles of Cl- / Volume of sample in L
To convert the concentration to g/L, you need to multiply the concentration by the molar mass of NaCl (which is 58.44 g/mol).
Concentration of NaCl (in g/L) = Concentration of NaCl (in mol/L) × Molar mass of NaCl (in g/mol)
Plug in the calculated values into these equations to find the concentration of NaCl in the sample in both mol/L and g/L.
To determine the concentration of NaCl in the sample in mol/L and g/L, we need to calculate the number of moles and grams of NaCl present in the 25.00 ml sample.
1. Firstly, we need to calculate the number of moles of AgNO3 used in the titration:
Moles of AgNO3 = concentration of AgNO3 x volume of AgNO3 solution
Moles of AgNO3 = 0.110 mol/L x 50.00 ml = 0.0055 mol
2. Next, we need to calculate the number of moles of KSCN used in the titration:
Moles of KSCN = concentration of KSCN x volume of KSCN solution
Moles of KSCN = 0.100 mol/L x 17.00 ml = 0.0017 mol
3. Since the AgNO3 and KSCN react in a 1:1 ratio, the number of moles of Cl- ions present in the 25.00 ml sample is also 0.0017 mol.
4. To calculate the concentration of NaCl in mol/L, we need to divide the number of moles of Cl- ions by the volume of the sample in liters:
Concentration of NaCl = 0.0017 mol / (25.00 ml / 1000)
Concentration of NaCl = 0.0017 mol / 0.025 L = 0.068 mol/L
5. Finally, to calculate the concentration of NaCl in g/L, we need to convert the concentration in mol/L to g/L using the molar mass of NaCl:
Molar mass of NaCl = 22.99 g/mol (mass of sodium) + 35.45 g/mol (mass of chlorine) = 58.44 g/mol
Concentration of NaCl = 0.068 mol/L x 58.44 g/mol = 3.97 g/L (rounded to two decimal places)
Therefore, the concentration of NaCl in the sample is approximately 0.068 mol/L and 3.97 g/L.