Please explain throughly also explaing how to find molarity An example of a prec
ID: 940516 • Letter: P
Question
Please explain throughly
also explaing how to find molarity
An example of a precipitation titration reaction is the Mohr method, which is used to find the concentration of halide ions in solution (particularlyCl and Br). First, the sample to be analyzed is titrated with a AgNO3 solution, which results in the precipitation of a white silver solid (e.g., AgCl). Because it is difficult to tell when all the halide ions have reacted with the silver ions, a small amount of an indicator ion, CrO42, is added to the reaction. Because Ag2CrO4 is more soluble than AgCl, the Ag2CrO4precipitates out only after all of the Cl ions have precipitated out as AgCl. Thus, as soon as all of the Cl ions have reacted, the Ag2CrO4precipitation begins, marking the equivalence point of the reaction by forming a rust-colored precipitate.
As a food chemist for a major potato chip company, you are responsible for determining the salt content of new potato chip products for the packaging label. The potato chips are seasoned with table salt, NaCl. You weigh out a handful of the chips, boil them in water to extract the salt, and then filter the boiled chips to remove the soggy chip pieces. You then analyze the chip filtrate for Cl concentration using the Mohr method.
First you prepare a solution of silver nitrate, AgNO3, and titrate it against 0.500 g of KCl using the Mohr method. You find that it takes 61.8 mL of AgNO3 titrant to reach the equivalence point of the reaction.
You then use the same silver nitrate solution to analyze the chip filtrate in a Mohr reaction, finding that the solution yields a rusty brown precipitate when 49.3 mL of titrant is added.
Part A
If the sample of chips used to make the filtrate weighed 81.0 g, how much NaCl is present in one serving (155 g ) of chips?
Express your answer to three significant figures and include the appropriate units.
Explanation / Answer
61.8 mL of AgNO3 solution is equivalent to 0.500 g of KCl
therefore 49.3 mL of AgNO3 solution is equivalent to 0.500 g x 49.3 mL/ 61.8 mL of KCl = 0.399 g of KCl
Now 74.5 g of KCl is equlvalent to 58.5 g of NaCl
hence 0.399 g KCl is equivalent to 58.5 x 0.399/ 74.5 = 0.313 g of NaCl
Now 81.0 g of chips has 0.313 g of NaCl
hence 155 g of chips has 0.313 x 155/ 81.0 = 0.599 g of NaCl