A 130.0−mL sample of a solution that is 3.0×10−3M in AgNO3 is mixed with a 225.0−mL sample of a solution that is 0.14M in NaCN.
After the solution reaches equilibrium, what concentration of Ag+(aq) remains?

Question

A 130.0−mL sample of a solution that is 3.0×10−3M in AgNO3 is mixed with a 225.0−mL sample of a solution that is 0.14M in NaCN.
After the solution reaches equilibrium, what concentration of Ag+(aq) remains?

abmon98 · Answer

Number of Moles=Concentration*Volume
130/1000*3.0*10^-3=0.00039 moles of AgNO3
Number of Moles=Concentration*Volume
225/1000*0.14=0.0315 moles of NaCN
Sodium is reactive enough to displace Ag+ to produce NaNO3
when 0.0003375 moles of Ag+ is mixed with 0.0322 moles of CN- 
we say that essentially all of the 0.00039 moles of Ag+ is converted to AgCN

the 0.00039 moles of Ag+ has been diluted to a total volume of 225+130=355 ml: 
Number of Moles=Concentration*Volume
0.00039=355/1000*Concentration
C=0.00109859154 mol/dm^3
 We are Dividing by 1000 to change the units from ml to dm^3