The value of #K_(sp)# for #SrSO_4# is #2.8 times 10^-7#. What is the solubility of #SrSO_4# in moles per liter?

1 Answer
Sep 5, 2017

Answer:

#"Solubility"# #~=# #100*mg*L^-1?#

Explanation:

Now we interrogate the solubility equilibrium.....

#SrSO_4(s) rightleftharpoons Sr^(2+) + SO_4^(2-)#

And #K_"sp"=[Sr^(2+)][SO_4^(2-)]#

We write #S="solubility of strontium sulfate."# And thus #S=[Sr^(2+)]#, and #S=[SO_4^(2-)]#...and we substitute these values into the solubility expression......

#K_"sp"=[Sr^(2+)][SO_4^(2-)]=SxxS=S^2#

And thus..............................

#S=sqrt(K_"sp")=sqrt(2.8xx10^-7)=5.29xx10^-4*mol*L^-1#.

And the gram solubility is thus.....

#5.29xx10^-4*mol*L^-1xx183.68*mol*L^-1=??#

If we attempted to dissolve the slat up in say #0.100*mol*L^-1# #"sodium sulfate"#, do you think the solubility would increase or decrease? Why?