Question #4359c

1 Answer
May 18, 2017

Answer:

#[Cl^-]=0.667*mol*L^-1#

Explanation:

This question has a sting in the tail. It expects you (i) to do the tedious arithmetic involved in the percentage mass; and (ii) it tries to distract students who know that #BaSO_4# is insoluble in aqueous solution. Of course, the insolubility of #BaSO_4# is a distractor.

#Ba^(2+) + SO_4^(2-) rarr BaSO_4(s)darr#

But we only need to assess the concentration of chloride ion WHICH REMAINS IN SOLUTION.

#"Moles of"# #BaCl_2=(50*mLxx20.8%*g*mL^-1)/(208.23*g*mol^-1)# #=0.050*mol#.

And thus there are #0.100*mol# with respect to #"chloride ion"#. And of course this is present in a COMBINED VOLUME of #100*mL+50*mL=150*mL=0.150*L#

Now concentration is simply #"moles of solute"/"volume of solution"=(0.100*mol)/(150xx10^-3L)#

#[Cl^-]=0.667*mol*L^-1#