Question

consider these compounds:
a. ( mathrm{ca}_{3}left(mathrm{po}_{4}
ight)_{2} )
b. ( mathrm{mn}(mathrm{oh})_{2} )
c. ( mathrm{fe}_{2} mathrm{~s}_{3} )
d. ( mathrm{mg}(mathrm{oh})_{2} )
complete the following statements by entering the letter corresponding to the correct compound.
without doing any calculations it is possible to determine that barium phosphate is more soluble than ( square ), and barium phosphate is less soluble than ( square ).
it is not possible to determine whether barium phosphate is more or less soluble than ( square ) by simply comparing ( k_{s p} ) values.

Explanation:

Step1: Recall solubility product formula

For a compound \(M_pX_q\), \(K_{sp}=[M^{n+}]^p[X^{m -}]^q\). For compounds with the same \(p\) and \(q\) (same stoichiometry), we can compare \(K_{sp}\) directly. Barium phosphate \((Ba_3(PO_4)_2)\) has \(K_{sp}=[Ba^{2 +}]^3[PO_4^{3 -}]^2\).

Step2: Analyze each compound

• Compound \(A:Ca_3(PO_4)_2\), \(K_{sp}=[Ca^{2+}]^3[PO_4^{3 -}]^2\). Same stoichiometry as \(Ba_3(PO_4)_2\).
• Compound \(B:Mn(OH)_2\), \(K_{sp}=[Mn^{2+}][OH^-]^2\).
• Compound \(C:Fe_2S_3\), \(K_{sp}=[Fe^{3+}]^2[S^{2 -}]^3\).
• Compound \(D:Mg(OH)_2\), \(K_{sp}=[Mg^{2+}][OH^-]^2\).

Since \(Ba_3(PO_4)_2\) and \(Ca_3(PO_4)_2\) have the same stoichiometry (\(M_3X_2\) type), we can compare \(K_{sp}\) directly. \(Ba_3(PO_4)_2\) and \(Mn(OH)_2\) (\(MX_2\) type) and \(Mg(OH)_2\) (\(MX_2\) type) have different stoichiometries. \(Ba_3(PO_4)_2\) (\(M_3X_2\)) and \(Fe_2S_3\) (\(M_2X_3\)) have different stoichiometries.

Answer:

Barium phosphate is more soluble than \(A\). Barium phosphate is less soluble than \(B\) and \(D\). It is not possible to determine whether barium phosphate is more or less soluble than \(C\) by simply comparing \(K_{sp}\) values.