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
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.
When the titration of NaCl (sodium chloride) and AgNO3 (silver nitrate) is carried out at a low pH, it can have an effect on the result. This is because the low pH conditions promote the formation of silver chloride (AgCl) precipitate due to the hydrolysis of Ag+ ions.
To understand the effect, let's consider the reaction between sodium chloride and silver nitrate:
NaCl + AgNO3 → AgCl + NaNO3
In this reaction, sodium chloride (NaCl) reacts with silver nitrate (AgNO3) to produce silver chloride (AgCl) and sodium nitrate (NaNO3).
However, when the titration is carried out at a low pH, the concentration of H+ ions is high. The high concentration of H+ ions results in the reaction:
H+ + Cl- → HCl
This means that chloride ions (Cl-) react with hydrogen ions (H+) to form hydrochloric acid (HCl). As a result, the concentration of chloride ions available to react with silver ions (Ag+) is reduced.
Since the concentration of chloride ions is decreased, fewer silver chloride (AgCl) precipitates will form. This can lead to inaccurate titration results and affect the accuracy of determining the concentration of silver ions (Ag+) in the solution.