A 60.00 mL solution of 0.3600 M AgNO3. was added to a solution of AsO43. Ag3AsO4

Question

A 60.00 mL solution of 0.3600 M AgNO3. was added to a solution of AsO43. Ag3AsO4 precipitated and was filtered off. Fe3 was added and the solution was titrated with 0.2050 M KSCN. After 31.60 mL of KSCN solution had been added, the solution turned red. What mass of AsO43 was in the original solution?

For this Volhard titration, an excess amount of Ag is added to a solution containing AsO43. Ag3AsO4 precipitates, and the remaining solution contains Ag . After the addition of Fe3 , KSCN is added, resulting in the precipitation of AgSCN. Once all the Ag present is consumed, SCN reacts with Fe3 , causing the formation of FeSCN2 , a red complex.

To find the amount of AsO43 initially present, first calculate the amount of Ag added, then the amount of Ag that reacts with KSCN. The difference is the amount of Ag that reacted with AsO43. Using reaction stoichiometry, calculate the amount of AsO43 originally present.

Explanation / Answer

3AgNO3 + AsO43- ----> Ag3AsO4 + 3 NO3-

moles of AgNO3 = 0.06*0.36 = 0.0216 moles

moles of KSCN = 0.0316*0.205 = 6.478*10^-3 moles

AgNO3 + KSCN ---> Ag(SCN) + KNO3

so moles of AgNO3 left = 6.478*10^-3 moles

so moles of AgNO3 that reacted with AsO43- = 0.0216 - 6.478*10^-3 = 0.015122 moles

so moles of AsO43- = 0.015122/3 = 5.041*10^-3 moles

so mass of = 5.041*10^-3*139 = 0.70069 gm = 700.699mg

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