Asked by Mina
assuming in the titration of NaCl and AgNO3, a student erroneously carried it out at a low PH,what effect will it be on the result. explain with an ionic equation
Answers
Answered by
DrBob222
There is no problem with the NaCl or the AgNO3; the problem exists because of the K2CrO4 indicator solution. In practice the pH is adjusted to between 7 and 10; NaHCO3 is usually used. At lower pH values, the CrO4^2- is converted to HCrO4^-.
CrO4^2- + H^+ ==> HCrO4^-
Therefore, at the equivalence point, the (CrO4^2-) is too low to form Ag2CrO4 in enough quantity to see the color and you must add more AgNO3 to achieve that end point. That means the %Cl is too high.
CrO4^2- + H^+ ==> HCrO4^-
Therefore, at the equivalence point, the (CrO4^2-) is too low to form Ag2CrO4 in enough quantity to see the color and you must add more AgNO3 to achieve that end point. That means the %Cl is too high.
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