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?

Question

Owen Ambrose · Accepted Answer

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

We are now questioning the equilibrium of solubility.

#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#

Thus,...

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

And in this way is the gram solubility...

5.29x10^-4molL^-1x183.68molL^-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?

Owen Ambrose · Answer

undefined To find the solubility of SrSO4 in moles per liter (M), use the equation:

$$ K_{sp} = [Sr^{2+}][SO_{4}^{2-}] $$

Given that $ K_{sp} = 2.8 \times 10^{-7} $, and assuming SrSO4 dissociates completely into Sr2+ and SO42- ions:

$$ K_{sp} = (x)(x) = x^2 $$

$$ 2.8 \times 10^{-7} = x^2 $$

$$ x = \sqrt{2.8 \times 10^{-7}} $$

$$ x \approx 5.3 \times 10^{-4} \, \text{M} $$

Therefore, the solubility of SrSO4 in moles per liter is approximately $5.3 \times 10^{-4}$ M.