Question

equal volumes of 0.2 m solutions of lead(ii) nitrate and potassium bromide are combined to form lead (ii) bromide as a yellow precipitate. which of the following is the correct net ionic equation for the reaction?
a ( mathrm{pb}^{2+}(a q)+2 mathrm{br}^{-}(a q)
ightarrow mathrm{pbbr}_{2}(s) )
b ( mathrm{k}^{+}(a q)+mathrm{no}_{3}^{-}(a q)
ightarrow mathrm{kno}_{3}(a q) )
c ( mathrm{pb}^{2+}(a q)+2 mathrm{no}_{3}^{-}(a q)+2 mathrm{~k}^{+}(a q)+2 mathrm{br}^{-}(a q)
ightarrow 2 mathrm{~k}^{+}(a q)+2 mathrm{no}_{3}^{-}(a q)+mathrm{pbbr}_{2}(s) )
d ( mathrm{pb}left(mathrm{no}_{3}
ight)_{2}(a q)+2 mathrm{kbr}(a q)
ightarrow mathrm{pbbr}_{2}(s)+2 mathrm{kno}_{3}(a q) )

Explanation:

Step1: Write the complete ionic equation

Lead(II) nitrate \(Pb(NO_3)_2\) dissociates as \(Pb^{2 +}(aq)+2NO_3^{-}(aq)\), potassium bromide \(KBr\) dissociates as \(K^{+}(aq)+Br^{-}(aq)\). The complete ionic equation is \(Pb^{2 +}(aq)+2NO_3^{-}(aq)+2K^{+}(aq)+2Br^{-}(aq)\to PbBr_2(s)+2K^{+}(aq)+2NO_3^{-}(aq)\)

Step2: Cancel out spectator ions

Spectator ions (\(K^{+}\) and \(NO_3^{-}\)) do not participate in the reaction. After canceling them, we get \(Pb^{2 +}(aq)+2Br^{-}(aq)\to PbBr_2(s)\)

Answer:

A. \(Pb^{2 +}(aq)+2Br^{-}(aq)\to PbBr_2(s)\)