Question

Determine the concentration of strontium ions in saturated solution of strontium sulfate, SrSO4, if the Ksp for SrSO4 is 3.2x10-7.​

Answer 1

Answer:

$[Sr^{2+}]=5.66x10^{-4}M$

Explanation:

Hello there!

In this case, given the solubility equilibrium of strontium sulfate:

$SrSO_4(s)\rightleftharpoons Sr^{2+}(aq)+SO_4^{2-}(aq)$

Whose equilibrium expression is:

$Ksp=[Sr^{2+}][SO_4^{2-}]$

In such a way, we can introduce the molar solubility, s, in the equation to obtain:

$Ksp=s*s=s^2\\\\3.2x10^{-7}=s^2$

Then, we apply the square root to obtain:

$s=\sqrt{3.2x10^{-7}}\\\\s=5.66x10^{-4}M$

Which is also the concentration of strontium ions:

$[Sr^{2+}]=5.66x10^{-4}M$

Best regards!