Question

5. Write a net ionic equation that occurs in a Na2HPO4/NaH2PO4 buffer solution when: A) a small amount of HCl is added (2 points) B) a small amount of sodium hydroxide is added (2 points)

Answer 1

Answer: (A) $H_{3}O^{+}(aq) + HPO^{2-}_{4}(aq) \rightarrow H_{2}PO^{2-}_{4}(aq) + H_{2}O(l)$

(B) $H_{2}PO^{-}_{4}(aq) + OH^{-}(aq) \rightarrow HPO^{2-}_{4}(aq) + H_{2}O(l)$

Explanation:

(A) As we know that HCl is a strong acid and when it is added to an aqueous solution then it leads to increase in the concentration of hydrogen ions. And, when an acid or base is added to a solution then any resistance by the solution in changing the pH of the solution is known as a buffer.

This means that addition of buffer into the given solution will not cause much change in the concentration of $H_{3}O^{+}$ in large amount.

As both the buffer components are salt then they will remain dissociated as follows.

       $Na_{2}HPO_{4}(aq) \rightarrow 2Na^{+}(aq) + HPO^{2-}_{4}(aq)$

 $NaH_{2}PO_{4}(aq) \rightarrow Na^{+}(aq) + H_{2}PO^{-}_{4}(aq)$

Hence, net ionic equation will be as follows.

       $H_{3}O^{+}(aq) + HPO^{2-}_{4}(aq) \rightarrow H_{2}PO^{2-}_{4}(aq) + H_{2}O(l)$

(B)  When we add small amount of sodium hydroxide into the solution then there will occur an increase in concentration of hydroxide ions into the solution. But then due to the presence of buffer there will occur not much change in concentration and the acid will get converted into salt.

     $NaOH(aq) \rightarrow Na^{+}(aq) + OH^{-}(aq)$

The net ionic equation is as follows.

        $H_{2}PO^{-}_{4}(aq) + OH^{-}(aq) \rightarrow HPO^{2-}_{4}(aq) + H_{2}O(l)$