Question

use the precipitation interactive to select the solution that reacts with the $pb(no_{3})_{2}$ solution to form a white s
$\bigcirc k_{2}s$
$\bigcirc ki$
$\bigcirc k_{2}co_{3}$
identify the precipitate that forms and the spectator ions that remain in the solution following this reaction.
answer bank
$s^{2 - }(aq)$ $kno_{3}(s)$
$pb^{2 + }(aq)$ $no_{3}^{-}(aq)$
$k^{+}(aq)$ $co_{3}^{2 - }(aq)$
$pbco_{3}(s)$
$pbi_{2}(s)$

Explanation:

Step1: Write the balanced chemical equation

When \(K_2CO_3\) reacts with \(Pb(NO_3)_2\), the reaction is \(K_2CO_3(aq)+Pb(NO_3)_2(aq)
ightarrow PbCO_3(s)+2KNO_3(aq)\)

Step2: Identify the precipitate

In the above reaction, \(PbCO_3\) is insoluble (forms a solid) and \(KNO_3\) is soluble. So the precipitate is \(PbCO_3(s)\)

Step3: Identify the spectator ions

Spectator ions are the ions that do not participate in the formation of the precipitate. From the dissociation of \(K_2CO_3 = 2K^{+}+CO_3^{2 -}\) and \(Pb(NO_3)_2=Pb^{2+}+2NO_3^{-}\), and the formation of \(KNO_3 = K^{+}+NO_3^{-}\). The ions \(K^{+}(aq)\) and \(NO_3^{-}(aq)\) are present in both reactant (dissociated form) and product (dissociated form of \(KNO_3\)) side. So the spectator ions are \(K^{+}(aq)\) and \(NO_3^{-}(aq)\)

Answer:

Precipitate: \(PbCO_3(s)\); Spectator ions: \(K^{+}(aq)\), \(NO_3^{-}(aq)\)