The preparations of two aqueous solutions are described in the table below. For each solution, write the chemical formulas of the major species present at equilibrium. You can leave out water itself Write the chemical formulas of the species that will act as acids in the 'acids' row, the formulas of the species that will act as bases in the 'bases' row, and the formulas of the species that will act as neither acids nor bases in the 'other' row You will find it useful to keep in mind that NH3 is a weak base. acids ,口, 1.3 mol of HNO3 is added to 1.0 L of a 1.3 MNH, solution bases: other acids: bases: - 0.57 mol of KOH is added to 1.0 L of a solution that is 1.2 M in both NH3 and NH4CI. other:

Respuesta :

a) NaFi is formed when 1 mol of NaOH is oriented to 1.0 L of a 0.5 M HF mixture.

b) To 1.0 L of a solution that is 0.8 M in both HF & KF-producing KF, 0.3 mol of KOH is added.

An acid generates the hydronium ions H3O+ in an aqueous solution, whereas a base creates the hydroxide ions OH. Water and salt are the byproducts of the neutralization process that occurs when an acid and a base interact.

A) 1 mol of NaOH is counted to 1.0 L of a 0.5 M HF resolution.

The reaction involved in this is:

NaOH + HF → NaF + H2O

acid: HF

base: NaOH

species that neither creates an acid and neither a base nor a salt NaF

b) To 1.0 L of a mixture that seems to be 0.8 M in both HF as well as KF, 0.3 mol of KOH is added.

The reaction involved in this is:

KOH + HF → KF + H2O

acid: HF

base: KOH

the mixture that could be neither an acid nor a base or salt delivered: KF

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The question is -

The preparations of two aqueous solutions are described below. For each solution, write the chemical formulas of the major species present at equilibrium. You can leave out water itself.

For each of the questions, write the chemical formulas of the species that will act as acids, the formulas of the species that will act as bases, and the formulas of the species that will act as neither acids nor bases. Note that HF is a weak acid.

A) 1 mol of NaOH is added to 1.0 L of a 0.5 M HF solution.

B) 0.3 mol of KOH is added to 1.0 L of a solution that is 0.8 M in both HF and KF.

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