When NaOH is added to the buffer solution containing [tex]\(H_3PO_4\)[/tex] and [tex]\(NaH_2PO_4\)[/tex], it will affect the equilibrium of the reaction:
[tex]\[ H_3PO_4(aq) \rightleftharpoons H^+(aq) + H_2PO_4^-(aq) \][/tex]
Here’s a detailed, step-by-step explanation of what happens:
1. Initial Conditions: We start with a buffer solution made from [tex]\(H_3PO_4\)[/tex] and [tex]\(NaH_2PO_4\)[/tex]. The equilibrium in this solution follows the reaction:
[tex]\[ H_3PO_4(aq) \rightleftharpoons H^+(aq) + H_2PO_4^-(aq) \][/tex]
2. Addition of NaOH: Sodium hydroxide ([tex]\(NaOH\)[/tex]) is a strong base and dissociates completely in water to give [tex]\(Na^+\)[/tex] and [tex]\(OH^-\)[/tex] ions.
3. Reaction of [tex]\(OH^-\)[/tex] with [tex]\(H^+\)[/tex]: The hydroxide ions ([tex]\(OH^-\)[/tex]) from [tex]\(NaOH\)[/tex] will react with the hydrogen ions ([tex]\(H^+\)[/tex]) in the solution to form water:
[tex]\[ OH^-(aq) + H^+(aq) \rightarrow H_2O(l) \][/tex]
4. Decrease in [tex]\(H^+\)[/tex] Concentration: As [tex]\(OH^-\)[/tex] ions react with [tex]\(H^+\)[/tex] ions, the concentration of [tex]\(H^+\)[/tex] ions in the solution decreases.
5. Shift in Equilibrium: According to Le Chatelier’s principle, any change in the concentration of reactants or products in a chemical equilibrium will cause the equilibrium to shift in a direction that counteracts the change. In this case, the decrease in [tex]\(H^+\)[/tex] ions will cause the equilibrium to shift to the right (towards the products) to produce more [tex]\(H^+\)[/tex] ions.
[tex]\[ H_3PO_4(aq) \rightleftharpoons H^+(aq) + H_2PO_4^-(aq) \][/tex]
By shifting to the right, more [tex]\(H_3PO_4\)[/tex] will dissociate to make up for the loss of [tex]\(H^+\)[/tex] ions. Therefore, the equilibrium will shift towards the products.
Based on this analysis, when NaOH is added to the buffer solution, the equilibrium shifts to products.
