**Please answer all parts of the question. Otherwise don't answer. **Net Ionic Equation for Acid-Base Reaction**When aqueous solutions of hydrobromic acid (HBr) and potassium hydroxide (KOH) are combined, a neutralization reaction occurs. The hydrogen ions from the hydrobromic acid react with the hydroxide ions from the potassium hydroxide to form water. The potassium ions (K⁺) and bromide ions (Br⁻) do not participate in the chemical reaction and remain in the solution as spectator ions.**Net Ionic Equation:**\[ \text{H}^+ (aq) + \text{OH}^- (aq) \rightarrow \text{H}_2\text{O} (l) \]This equation shows the formation of water, which is the main product of the reaction. The spectator ions are omitted from the net ionic equation as they do not affect the outcome of the reaction. An aqueous solution of hydroiodic acid is standardized by titration with a 0.177 M solution of barium hydroxide.If 12.3 mL of base are required to neutralize 21.2 mL of the acid, what is the molarity of the hydroiodic acid solution?[Answer Box] M hydroiodic acid

Ionic equation is an equation in which ions are explicitly shown. Net ionic equation is the chemical equation for a reaction which lists only those species participating in the reaction. Spectator ions are the ion that appears unchanged on both sides of a reaction arrow. To write the net ionic equation, steps mentioned below must be followed. Write the balanced molecular equation Write the complete ionic equation showing the strong electrolytes completely dissociated into cations and anions. Cancel the spectator ions on both sides of the ionic equation. Check that charges and number of atoms are balanced in the net ionic equation. The net ionic equation for the reaction between Hydrogen bromide and Potassium hyroxide is given below: Balanced Chemical equation:HBr (aq)+KOH(aq) →KBr(aq)+H2O(l)Complete ionic equation is as followsH+ (aq)+Br-(aq)+K+ (aq)+OH-(aq) → K+ (aq)+Br-(aq) +H2O(l)Net ionic equation:H+ (aq)+OH-(aq) →H2O(l) The reaction is shown below: 2 HI +Ba(OH)2 →BaI2 +2 H2O Number of moles of HI can be calculated as follows: No.of moles=Molarity×VolumeNo.of moles of Ba(OH)2=0.177 M×0.0123 LNo.of moles of Ba(OH)2=0.00218 mol From the reaction, it is clear that one mole of Barium hydroxide requires two moles of HI. Thus, number of moles of HI will be double that of Barium hydroxide. Molarity of HI can be determined as follows: No. of moles=0.00218 mol Ba(OH)2×2 mol HI1 mol Ba(OH)2=0.00436 mol HIMolarity=No. of molesVolume(L)Molarity=0.00436 mol HI0.0212 LMolarity=0.0206 M