Question

dilu SLUHY Udses, velduse ule ujuydte udbe vid Struny dulu is too weak to establish any meaningful equilibrium in solution. Figure 2: After Neutralization, F. Re-establishes an Equilibrium. Free Free Conclusions 1. The initial reaction between a weak acid and a strong base is not an equilibrium. Each mole of base deprotonates one mole of acid in an "irreversible" (very large K value) reaction. 2. Neutralization of a weak acid with strong base produces a conjugate base that will react with water, hydrolyzing in solution 3. The resulting pH when the weak acid is completely neutralized, will be greater than 7.0. This is the result of the conjugate base that was created in solution Do You Understand? 1. In each of the following reactions between a weak acid and a strong base, all of the acid was neutralizing by adding a stoichiometric amount of base. Write the dominant equilibrium that will exist (including phase labels) in this "neutralized solution. HCOOH reacting with KOH a. HNO, reacting with UCH HNO(a) LOHE L INO)(a) HO HCOCHE KOH ) = HCOONL) HOUR HGOOD HOOORING 20 Your answer appears to be missing a charge. Parts of the formula in your answer are Parts of the formula in your answer are transposed. Your answer appears to have inconsistent states of matter. One or more of your cats appear to substitute one element for another your answer contains an incorrect or incomplete product inconsistent states of matter. Your Own contains an incorrect or incomplete reactant One or more of your products per 2. When a solution of a weak acid is completely neutralized by a strong base, the pH of the resulting solution will be: O Basic Metal Acidic

Answer 1

Both your answers are correct but I think they need the overall ionic equation maybe. So here's it.

HNO₂ (aq) + LiOH (aq) <--> LiNO₂ (aq) + H₂O (l)

A weak acid is completely neutralized by a strong base. The reaction occurring is,

HNO₂ (aq) + OH^- (aq) <---> NO₂^- (aq) + H2O (l)

2. Formic acid, a weak acid is neutralized by KOH, a strong base.

HCOOH (aq) +KOH (aq) <---> HCOOK (aq) + H₂O (l)

Which can be represented as,

HCOOH (aq) + OH^- (aq) <--->HCOO^- (aq) + H2O (l)

Please give a thumbs up if this helped and incase of any doubt ask in the comment section. Thanks