Chapter 20, Problem 45QAP

Interpretation Introduction

Interpretation:

The balanced net ionic equation has to be written for all the reactions for the given situation.

Concept introduction:

While writing the net ionic equation, an understanding of spectator ions is very important. The spectator ions are the ions which are found in same form in both the reactant and product sides. So, while writing the net ionic equation, these ions are not involved or simply not counted.

Answer to Problem 45QAP

The net ionic equations are:

$C{r}_2{O}_7^{2-}(aq)+2O{H}^{-}(aq)\to 2Cr{O}_4^{-2}(aq)+H_{2}O(l)$

$C{r}_2{O}_4^{2-}(aq)+2A{g}^{+}(aq)\to A{g}_{2}Cr{O}_4(s)$

$A{g}_{2}Cr{O}_4(s)+4N{H}_3(aq)\to 2Ag{(N{H}_3)}_2^{+}(aq)+C{r}_2{O}_4^{2-}(aq)$

$2Ag{(N{H}_3)}_2^{+}(aq)+C{r}_2{O}_4^{2-}(aq)+4H^{+}(aq)\to A{g}_{2}Cr{O}_4(s)+4N{H}_4^{+}(aq)$

Explanation of Solution

Given Information:

(1) A solution of potassium dichromate is made basic with sodium hydroxide. The color changes from red to yellow.

(2) The addition of silver nitrate to the yellow solution gives the precipitate.

(3) The precipitate dissolves in concentrated ammonia.

(4) The dissolved precipitate again reforms when nitric acid is added.

From the given details, we can infer that four independent reactions will take place. The balanced net ionic equation will be written for each step of procedure separately.

Step1: A solution of potassium dichromate is made basic with sodium hydroxide.

The color changes from red to yellow.

$K_{2}C{r}_2{O}_7(aq)+2NaOH(aq)\to K_{2}Cr{O}_4(aq)+N{a}_{2}Cr{O}_4(aq)+H_{2}O(l)$

The ionic reaction form will be:

$2K^{+}(aq)+C{r}_2{O}_7^{2-}(aq)+2N{a}^{+}(aq)+2O{H}^{-}(aq)\to 2K^{+}(aq)+Cr{O}_4^{-2}(aq)+2N{a}^{+}(aq)+Cr{O}_4^{-2}(aq)+H_{2}O(l)$ Here, the Na+ and K+ are spectator ions and removing them will give the balanced net ionic equation as:

$C{r}_2{O}_7^{2-}(aq)+2O{H}^{-}(aq)\to 2Cr{O}_4^{-2}(aq)+H_{2}O(l)$

Step 2: The solution is added with AgNO₃ :

$C{r}_2{O}_4^{2-}(aq)+2A{g}^{+}(aq)+2N{O}_3^{-}(aq)+2H^{+}\to A{g}_{2}Cr{O}_4(s)+HN{O}_3(aq)$

The spectator ions ${\text{H}}^{\text{+}}{\text{and NO}}_{\text{3}}^{\text{-}}$ are removed.

Net ionic equation is:

$C{r}_2{O}_4^{2-}(aq)+2A{g}^{+}(aq)\to A{g}_{2}Cr{O}_4(s)$

Step3: The precipitate is dissolved in concentrated ammonia.

$A{g}_{2}Cr{O}_4(s)+4N{H}_3(aq)\to 2Ag{(N{H}_3)}_2^{+}(aq)+C{r}_2{O}_4^{2-}(aq)$

The reaction is automatically in its net ionic form.

Step 4: The dissolved precipitate again reforms when nitric acid is added.

$2Ag{(N{H}_3)}_2^{+}(aq)+C{r}_2{O}_4^{2-}(aq)+4N{O}_3^{-}(aq)+4H^{+}(aq)\to A{g}_{2}Cr{O}_4(s)+4N{H}_4^{+}(aq)+4N{O}_3^{-}(aq)$

Here, the spectator ion is ${\text{NO}}_{\text{3}}^{\text{-}}$ which is removed to give the net ionic reaction as:

$2Ag{(N{H}_3)}_2^{+}(aq)+C{r}_2{O}_4^{2-}(aq)+4H^{+}(aq)\to A{g}_{2}Cr{O}_4(s)+4N{H}_4^{+}(aq)$

Conclusion

The net ionic equations are:

$C{r}_2{O}_7^{2-}(aq)+2O{H}^{-}(aq)\to 2Cr{O}_4^{-2}(aq)+H_{2}O(l)$

$C{r}_2{O}_4^{2-}(aq)+2A{g}^{+}(aq)\to A{g}_{2}Cr{O}_4(s)$

$A{g}_{2}Cr{O}_4(s)+4N{H}_3(aq)\to 2Ag{(N{H}_3)}_2^{+}(aq)+C{r}_2{O}_4^{2-}(aq)$

$2Ag{(N{H}_3)}_2^{+}(aq)+C{r}_2{O}_4^{2-}(aq)+4H^{+}(aq)\to A{g}_{2}Cr{O}_4(s)+4N{H}_4^{+}(aq)$