91Ó°ÊÓ

Ionic equations will be written by dissociating given compound to its cation and anion. Constant of solubility is calculated like this:

\({K_{sp}} = C{\left( {{C^ + }} \right)^c} \times C{\left( {{A^ - }} \right)^a}\)

c is for concentration\(,{C^ + }\) is for cation and $A^{-}$is for anion that are potentiated on their stoichiometry coefficient

a)\(PbC{l_2}\)

\(\begin{array}{l}PbC{l_2}\;(aq) \to P{b^{2 + }}(aq) + 2C{l^ - }(aq) \\ {K_{sp}} = \left( {P{b^{2 + }}} \right) \times 2\;\left( {{\rm{C}}{{\rm{l}}^ - }} \right)\end{array}\)

b)\(A{g_2}S\)

\(\begin{array}{l}A{g_2}S(aq) \to 2A{g^ + }(aq) + {S^{2 - }}(aq)\\ {K_{sp}} = {\left( {c\left( {A{g^ + }} \right)} \right)^2} \times c\left( {{S^{2 - }}} \right)\\\end{array}\)

C)\(S{r_3}{\left( {P{O_4}} \right)_2}\)

\(\begin{array}{l}S{r_3}{\left( {P{O_4}} \right)_2}(aq) \to 3\;S{r^{2 + }}(aq) + 2PO_4^{3 - } \\ {K_{sp}} = {\left( {3\left( {S{r^{2 + }}} \right)} \right)^3} \times {\left( {2\left( {PO_4^{3 - }} \right)} \right)^2}\end{array}\)

d)\(SrS{O_4}\)

\(\begin{array}{l}SrS{O_4}\;(aq) \to S{r^{2 + }}\;(aq) + SO_4^{2 - }\;(aq) \\ {K_{sp}} = c\left( {S{r^{2 + }}} \right) \times c\left( {SO_4^{2 - }} \right)\end{array}\)