The preparations of two aqueous solutions are described in the table below. For

Question

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 table You can leave out 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 HCH3CO2 is a weak acid. O acids: 1.6 mol of KOH is added to 1.0 L of a 1.0 MHCHCO2bases: solution. O other: 0 0.05 mol of HI is added toacids: 1.0 L of a solution that is 0.6 M in both HCH3CO2 and NaCH3CO2 bases: Oother:[

Explanation / Answer

1) HCH3CO2 is a weak acid.

When a strong base KOH is added to it, the reaction is

HCH3CO2 + KOH --------------> K CH3CO2 + H2O

1.0 x1.0 0 0 0 Initial moles

--- 1.6 ------- -- change

0 0.6 1.0 - at equilibrium

Thus the major species at equilibrium are KOH (base) and KCH3CO2 (conjugate base)

2)the solution has weak acid HCH3CO2 and its conjugate base NaCH3CO2 .

When 0.05 mol of strong acid HI is added the reaction is

NaCH3CO2 + HI -------------->   HCH3CO2 + NaI

1.0x0.6 0 1.0x0.6 0 initial moles

--- 0.05 --- - change

0.55 0 0.65 0.05 at equilibrium

Thus the major species present at equilibrium are

acids  HCH3CO2

bases NaCH3CO2 [conjugate base]

and others NaI

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