I need help with this question (1 Viewer)

[An0nyM0usUseR455]

Can someone please send their working for this please?

I dont get it

 

[An0nyM0usUseR455]

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[An0nyM0usUseR455]

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[indeed]

Idk if this is right, it's based on the assumption that the chemical formula of the salt is X+ and Y- (1 charged)

 

[Luukas.2]

I think the question means salt in the specific, non-chemical usage... ie. table salt, sodium chloride.

The silver nitrate then precipitates the chloride ions as silver chloride solid, which is removed by filtration. The excess (remaining) silver cations are quantified by titration with thiocyanate (precipitated as silver thiocyanate, likely using an iron(III) salt as the indicator).

 

[qldbulls]

Yeah it's literally just a back titration for NaCl then x10

 

[An0nyM0usUseR455]

View attachment 40875

Idk if this is right, it's based on the assumption that the chemical formula of the salt is X+ and Y- (1 charged)

your working makes sense but the answer is 0.12 i still dont get how

 

[An0nyM0usUseR455]

I think the question means salt in the specific, non-chemical usage... ie. table salt, sodium chloride.

The silver nitrate then precipitates the chloride ions as silver chloride solid, which is removed by filtration. The excess (remaining) silver cations are quantified by titration with thiocyanate (precipitated as silver thiocyanate, likely using an iron(III) salt as the indicator).

so do I find the number of moles of thiocyanate that reacted with the excess Ag and subtract from the total Ag used. Then do I make use of the fact that everything is 1:1 so no. of mol of salt = no. of mol Ag used and divide by volume and then multiply by 10 cuz its diluted?

thats what i did in the first place and the answer is 0.12

 

[qldbulls]

so do I find the number of moles of thiocyanate that reacted with the excess Ag and subtract from the total Ag used. Then do I make use of the fact that everything is 1:1 so no. of mol of salt = no. of mol Ag used and divide by volume and then multiply by 10 cuz its diluted?

thats what i did in the first place and the answer is 0.12

Moles Ag original = (0.12x0.025)
Molarity of Ag remaining in the 25mL of filtrate = (0.11x0.0123) given stoichiometry with the thiocyanate
Filtrate overall is 50mL therefore actual moles of Ag remaining = 2x(0.11x0.0123)
Take that away from the original moles of Ag (0.12x0.025) - 2x(0.11x0.0123) = 0.294x10^-3 moles used in reaction with NaCl
Therefore molarity of NaCl solution = (0.294x10^-3)/(0.025) = 0.01176
Multiply by 10 for dilution = 0.1176 = 0.12M (2sf)

Chem in focus is a decent textbook eh?