Answer :
2AgNO3+K2CrO4⇒Ag2CrO4(s)+2KNO3
Hence by mixing 0.0024M AgNO3 and 0.004M
K2CrO4, we will have Ag2CrO4 which is precipitated out and leave us with
0.0024M KN03 which is mixed with (0.004-0.0024/2)M, it can be 0.0028M, of K2Cr04
Hence by mixing 0.0024M AgNO3 and 0.004M
K2CrO4, we will have Ag2CrO4 which is precipitated out and leave us with
0.0024M KN03 which is mixed with (0.004-0.0024/2)M, it can be 0.0028M, of K2Cr04
If the standard solutions had unknowingly been made up to be 0.0024 M AgNO₃ and 0.0040 M K₂CrO₄ the chemical equation would be:
2AgNO₃ + K₂CrO₄ → Ag₂CrO₄ (s) + 2KNO₃
By mixing 0.0024 M AgNO₃ and 0.004 M K₂CrO₄ one would have Ag₂CrO₄ precipitated out.
Thus, one is left with 0.0024 M KNO₃ mixed with (0.004 -0.0024 / 2) M = 0.0028 M of K₂CrO₄.