Question #d34c1

1 Answer
Apr 13, 2015

I think the information you were given is incorrect. More specifically, I suspect that either the mass of the sample is too small, or the mass of the formed precipitate is too big.

Here's why.

So, you have a mixture of sodium chloride and potassium chloride which you react, in aqueous solution, with silver nitrate, #AgNO_3#.

The reaction will lead to the formation of silver chloride, a solid that will precipitate out of solution. The net ionic equation for this reaction describes what's going on

#Ag_((aq))^(+) + Cl_((aq))^(-) -> AgCl_((s))#

Notice that 1 mole of chloride ions is needed to produce 1 mole of silver chloride precipitate. This means that you can calculate the total number of moles of chloride that reacted by using the mass of the precipitate

#2.5cancel("g") * "1 mole AgCl"/(143.32cancel("g")) = "0.01744 moles AgCl"#

Since you have a #1:1# mole ratio between silver chloride and the chloride ions, you'll have

#n_(Cl^(-)) = n_(AgCl) = "0.01744 moles "# #Cl^(-)#

This represents the total number of moles of chloride present in the sample. Since both sodium chloride and potassium chloride dissociate completely in aqueous solution, you can write

#overbrace(n_(NaCl))^text(x) + underbrace(n_(KCl))_text(y) = 0.01744" "# #color(blue)((1))#

One mole of #NaCl# produces 1 mole of #Cl^(-)# ions; the same is true for #KCl#.

The second equation will use the mass of the sample, which can be written like this

#x * M_("M NaCl") + y * M_("M KCl") = "0.8870 g "#, where

#M_("M NaCl")#, #M_("M KCl")# - the molar masses of sodium chloride and potassium chloride, respectively.

#x * 58.44 + y * 74.55 = "0.8870 g "# #color(blue)((2))#,

So far so good, but here is where the problem stops making sense. Trying to solve the two equations will lead to

#x = 0.01744 - y#

#58.44 * (0.01744 - y) + 74.55 * y = 0.8870#

#1.019 - 58.44 * y + 74.55 * y = 0.8870#

#16.11 * y = -0.132 => y = color(red)(-0.008194)#

Since #y# represents number of moles, it cannot be negative. This is the problem with the data given.

The only way to get positive values for both #x# and #y# is to either

  • reduce the mass of silver chloride, which will reduce the number of moles of chloride present in solution
  • increase the mass of the sample