Question

1. Calculate the concentration of the AgNO3 solution.

2. Use your AgNO3 concentration to determine the concentration of the KSCN titrant from your standardization titrations. (See Equation 6-1)

3. Determine the amount of excess Ag+ remaining in the filtrate of your unknown solutions:

a. Equation 6-1 shows the stoichiometry for this reaction.

Ag+ + SCN- --> AgSCN

b. Use the volume of KSCN used, along with the true concentration of KSCN you determined, to find the moles of Ag+ remaining in solution after the reaction with chloride.

c. Use the concentration of AgNO3 to determine the total moles of silver you added to the reaction.

d. The difference between the values you determined in b and c equals the total number of moles of silver that reacted with the chloride in your aliquot of unknown solution.

e. Determine and report the mass% of chloride in your unknown sample.

Okay so I have my concentration of AgNO3 buthow do I use it to find the concentration of the KSCN titrant? Also How to determine the excess amount of Ag+ remaining in the filtrate of my unknown silutions?

Answer 1

Answer to your first question is already given in your question, "from standardization titrations". You have taken AgNO3 as the primary standard and KSCN as the secondary standard. Now, from the equation of standardization titration, V1S1=V2S2; V= volume, S= Strength. you can get the strength of KSCN from the strength of AgNO3 by getting a titre value of KSCN for a measured volume of AgNO3.

Answer to your second question: Now, as you've got the strength of KSCN, you can titrate now the filtrate solution. As before you'll get the concentration of Ag+ in the solution. To be more clear, volume times strength of KSCN gives the mol no of KSCN that reacts with Ag+. Now as there is one mole of KSCN for one mole of Ag+, we will get the mol no of Ag+ tgat is actually equal to the volume (in lit) times strength (in M) of KSCN solution. Thus, we get the value of the number of moles of Ag+ that is excess in the filtrate after reacting with Cl- to form the aprecipitate of AgCl.