Answer:
Step-by-step explanation:
Given that for Part A,
the arrangement can be either AgF or NaF.
On the off chance that the arrangement is AgF ,
For this situation AgF conc. can be controlled by titration with standard choride arrangement with potassium chromate as marker.
Get ready 1 M NaCl standard arrangement by dissolving 58.5 gm of NaCl in one liter refined water. Take 25 mL of 1 N NaCl arrangement in a funnel shaped carafe include hardly any drops of potassium chromate marker.
Take the example in the burette and titrate until a perpetual ruddy earthy colored hasten is seen.
[AgF responds with NaCl as follows
AgF + NaCl \small \rightarrow AgCl ( white precipitate)+ NaF
For whatever length of time that CaCl is available in the arrangement Ag structures AgCl hasten as above. When all the chloride particles expelled this way, the overabundance silver structures Ag2CrO4 ( rosy earthy colored encourage ) with the marker.
The dissolvability of AgCl is a lot of lower than the solvency of Ag2CrO4. Consequently the later won't hasten until all chloride particles exhausted.]
Let V2 be the volume of test utilized. At that point grouping of AgF in the example is given by
C2 = V1C1/V2 = 25 x 1/V2.
Rehash the titration by making legitimate weakening if any required.
On the off chance that the arrangement is NaF (Indirect strategy)
Here the Fluoride particle in a referred to volume of test is hastened as leadchlorofluoride with NaCl and PbNO3.
NaF + NaCl + Pb(NO3)2 \small \rightarrow PbClF (hasten) + 2 NaNO3
The above hasten is separated and washed and the washed encourage is taken in a measuring utencil and re-broke up in nitric corrosive to deliver the chloride particles.
PbClF + 2 HNO3 \small \rightarrow Pb(NO3)2 + HCl + HF
Presently include a known abundance of Silver nitrate answer for the above blend to encourage the chloride as silver chloride. Channel wash and evacuate the AgCl encourage.. The Filtrate contains the overabundance silver nitrate is to be dictated by titration with Standard chloride arrangement as in the past technique.
From the first amount of silver nitrate taken and the overabundance silver nitrate decided from the titration, we can decide the amount of silver nitrate responded.
The moles of silver nitrate responded = the moles of Cl delivered = the moles of F in the example.
In this way the convergence of sodium fluoride in the obscure example might be resolved.