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This is one of our many quizzes on Net ionic equations. Click on this to access more such competitive quizzes on Acids, Bases And Salts.

Net ionic equations Quiz 2

chemistry

 Titration is performed between oxalic acid and potassium permanganate in the presence of sulfuric acid. The ions and their stoichiometric coefficients participating in the chemical reaction will be:

 The ions and their stoichiometric coefficients participating in the reaction can be determined using the balanced net ionic equation.

 $ \text{The reaction takes place as follows.}\\ 2KMn{{O}_{4}}+3{{H}_{2}}S{{O}_{4}}+5{{(COOH)}_{2}}\to {{K}_{2}}S{{O}_{4}}+2MnS{{O}_{4}}+8{{H}_{2}}O+10C{{O}_{2}}\uparrow \\ \text{The ionic equation will be as follows}\\ 2{{K}^{+}}(aq)+2MnO_{4}^{-}(aq)+6{{H}^{+}}(aq)+3SO_{4}^{2-}(aq)+5{{C}_{2}}O_{4}^{2-}(aq)+10{{H}^{+}}(aq) \\ 2{{K}^{+}}(aq)+SO_{4}^{2-}(aq)+2M{{n}^{2+}}(aq)+2SO_{4}^{2-}(aq)+8{{H}_{2}}O(l)+10C{{O}_{2}}\uparrow  \\ \text{Hence the net ionic equation will be}\\ 2MnO_{4}^{-}(aq)+16{{H}^{+}}(aq)+5{{C}_{2}}H_{4}^{2-}(aq)\to 2M{{n}^{2+}}(aq)+10C{{O}_{2}}\uparrow +8{{H}_{2}}O(l)\\ \text{Here, the participating ions are }MnO_{4}^{-}, {{H}^{+}}, {{C}_{2}}H_{4}^{2-},\text{ and }M{{n}^{2+}}.\\ \text{Carbon dioxide gas and water are formed.}$

 What are the stoichiometric coefficients for the following net ionic equation $ \\ Zn+NO_{3}^{-}+{{H}^{+}}\to Z{{n}^{2+}}+NH_{4}^{+}+{{H}_{2}}O$ 

 The given reaction is a redox reaction. A net ionic equation must be balanced for both charge and mass.

 The balanced net ionic equation is as follows $\\ 4Zn+NO_{3}^{-}+10{{H}^{+}}\to 4Z{{n}^{2+}}+NH_{4}^{+}+3{{H}_{2}}O$ 

 The ions involved in the formation of the precipitate when silver nitrate is titrated against sodium chloride are

 The ions participating in the formation of precipitate $AgCl$ can be determined using the net ionic equation.

 $\text{The chemical equation is as follows } \\ AgN{{O}_{3}}(aq)+NaCl(aq)\to \alpha gCl(s)+NaN{{O}_{3}}(aq)\\ \text{The ionic equation is as follows}\\ A{{g}^{+}}(aq)+NO_{3}^{-}(aq)+N{{a}^{+}}(aq)+C{{l}^{-}}(aq)\to \alpha gCl(s)+N{{a}^{+}}(aq)+NO_{3}^{-}(aq)\\ \text{The net ionic equation is as follows}\\ A{{g}^{+}}(aq)+C{{l}^{-}}(aq)\to AgCl(s)$ 

 The net ionic equation at the equivalence point for the titration reaction between acetic acid and sodium hydroxide will be:

 At the equivalence point, the weak acid is completely neutralized and converted into the conjugate base. 

<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/a01a3d47-f23f-4aa6-8261-c062f05a2ce85853897454778177520.png"/> $ \\ $ The graph between pH and volume of NaOH used depicting equivalence point $ \\ $ The chemical equation will be as follows $\\ C{{H}_{3}}COOH(aq)+NaOH(aq)\to C{{H}_{3}}COONa(aq)+{{H}_{2}}O(l)\\ $ Weak acids do not easily dissociate in aqueous solutions whereas their strong conjugate base formed do. $ \\ $ The ionic equation will be as follows $ \\ C{{H}_{3}}COOH(aq)+N{{a}^{+}}(aq)+O{{H}^{-}}(aq)\to C{{H}_{3}}CO{{O}^{-}}(aq)+N{{a}^{+}}(aq)+{{H}_{2}}O(l)\\ $ The net ionic equation will be $ \\ C{{H}_{3}}COOH(aq)+O{{H}^{-}}(aq)\to C{{H}_{3}}CO{{O}^{-}}(aq)+{{H}_{2}}O(l)$ 

 The spectator ions in the following equation will be $ \\ {{H}_{2}}S{{O}_{4}}+NaOH\rightleftharpoons {{H}_{2}}O+N{{a}_{2}}S{{O}_{4}}$

 Ions that do not take part in a chemical equation but are found in the reaction solution are known as spectator ions.

 $\text{The balanced chemical equation is as follows } \\ {{H}_{2}}S{{O}_{4}}(aq)+2NaOH(aq)\rightleftharpoons 2{{H}_{2}}O(l)+N{{a}_{2}}S{{O}_{4}}(aq) \\ \text{The ionic reaction is as follows}\\ 2{{H}^{+}}(aq)+SO_{4}^{2-}(aq)+2N{{a}^{+}}(aq)+2O{{H}^{-}}(aq)\rightleftharpoons 2{{H}_{2}}O(l)+2N{{a}^{+}}(aq)+SO_{4}^{2-}(aq)\\ \text{The net ionic equation will be }\\ 2{{H}^{+}}(aq)+2O{{H}^{-}}(aq)\rightleftharpoons 2{{H}_{2}}O(l) \\ {{H}^{+}}(aq)+O{{H}^{-}}(aq)\rightleftharpoons {{H}_{2}}O(l) \\ \text{So, the spectator ions are }N{{a}^{+}}\text{ and }SO_{4}^{2-}$