Question

which of the following is the net ionic equation for the reaction of calcium bromide and sodium phosphate?
a (ca^{2 +}(aq)+2br^{-}(aq)\to cabr_{2}(aq))
b (3cabr_{2}(aq)+2na_{3}po_{4}(aq)\to ca_{3}(po_{4})_{2}(s)+6nabr(aq))
c (3ca^{2 +}(aq)+2po_{4}^{3 -}(aq)\to ca_{3}(po_{4})_{2}(s))
d (na^{+}(aq)+br^{-}(aq)\to nabr(aq))

Explanation:

Step1: Write the balanced molecular equation

The reaction between calcium bromide ($CaBr_2$) and sodium phosphate ($Na_3PO_4$) is a double - displacement reaction. The balanced molecular equation is $3CaBr_2(aq)+2Na_3PO_4(aq)
ightarrow Ca_3(PO_4)_2(s) + 6NaBr(aq)$.

Step2: Write the complete ionic equation

Break all the soluble ionic compounds into their ions. $3Ca^{2 +}(aq)+6Br^{-}(aq)+6Na^{+}(aq)+2PO_4^{3 -}(aq)
ightarrow Ca_3(PO_4)_2(s)+6Na^{+}(aq)+6Br^{-}(aq)$.

Step3: Write the net ionic equation

Cancel out the spectator ions (ions that appear on both sides of the equation). The spectator ions are $Na^{+}$ and $Br^{-}$. The net ionic equation is $3Ca^{2+}(aq)+2PO_4^{3 -}(aq)
ightarrow Ca_3(PO_4)_2(s)$.

Answer:

C. $3Ca^{2+}(aq)+2PO_4^{3 -}(aq)
ightarrow Ca_3(PO_4)_2(s)$