The silver ions in a 25.00 ml of aq solution were converted to dicyanoargentate by the addition of an excess soultion containing Ni((CN)4)2-.
Ni((CN)4)2- + 2Ag+ = 2Ag((CN)2)- + NI2+
Liberated nickle required 43.77ml of 0.02408M EDTA for tritation.
Now i need to work out the concentration of the silver in the example.
I am at a complete stump because its in a sense titration of something that has the silver in but requires a double calculation. Please help and push me in the right direction.
Ni((CN)4)2- + 2Ag+ = 2Ag((CN)2)- + NI2+
Liberated nickle required 43.77ml of 0.02408M EDTA for tritation.
Now i need to work out the concentration of the silver in the example.
I am at a complete stump because its in a sense titration of something that has the silver in but requires a double calculation. Please help and push me in the right direction.
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EDTA will bind with metals in a 1:1 ratio
according to the balanced equation, for every 1 mole of Ni liberated, 2 moles of Ag are bound to the CN-
so, if 0.04377L an 0.02408M EDTA is used, this is 1.054x10^-3moles EDTA used.
this means that 1.054x10^-3moles Ni was liberated and bound by EDTA
this also means that 2.108x10^-3moles Ag was bound to the CN-
according to the balanced equation, for every 1 mole of Ni liberated, 2 moles of Ag are bound to the CN-
so, if 0.04377L an 0.02408M EDTA is used, this is 1.054x10^-3moles EDTA used.
this means that 1.054x10^-3moles Ni was liberated and bound by EDTA
this also means that 2.108x10^-3moles Ag was bound to the CN-