# Question #f395d

This is because the $O {H}^{-}$ ions from $N a O H$ ($N a O H$ $\rightarrow$ $N {a}^{+}$ + $O {H}^{_}$) neutralise the ${H}^{+}$ ions from ${H}_{3} P {O}_{4}$ (Phosphoric Acid).
The equation for the dissociation of ${H}_{3} P {O}_{4}$ is;
${H}_{3} P {O}_{4}$ $r i g h t \le f t h a r p \infty n s$ ${H}_{2} P {O}_{4}$ + ${H}^{+}$
The concentration of ${H}^{+}$ ions on the right hand side of the equation therefore decreases, and the point of equilibrium shifts to the right, in order to oppose the decrease in concentration of ${H}^{+}$ ions. The point of equilibrium shifts to the right, by more ${H}_{3} P {O}_{4}$ dissociating, regenerating ${H}^{+}$ ions, thus decreasing the concentration of Phosphoric Acid.