Question

The net ionic equation for the reaction between aqueous solutions of HF and KOH is.... a) HF + KOH ---> H2O + K^+ + F^- b) HF + OH^- ---> H2O + F^- c) HF + K^+ ----> H2O + KF d) H^+ + OH^- ----> H2O e) H^+ + F^- + K^+ + OH^- ---> H2O + K^+ + F^- can you please explain how you got to the answer, thank you!

Answer 1

First write the molecular equation

HF(aq) + KOH(aq) --------> KF(aq) + H₂ O(l)

KOH is a strong base, so it dissociated completely in water.

KOH(aq) --------> K⁺(aq) + OH^-(aq)

HF is a weak acid, dissociated only partially in water,

HF(aq) ----------> H⁺(aq) + F^-(aq)

Complete ionic equation will be

HF(aq) + (aq) +K⁺(aq) + OH-(aq) --------> K⁺ (aq) + F-(aq) + H₂O(l)            

K⁺ (aq) is on both sides, so it cancels out. The net ionic equation will be

HF(aq) + OH^- (aq) -------> F^-(aq) + H₂O(l)

b) HF + OH^- ---> H₂O + F^- is correct option.