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. 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 HCH,CO2 is a weak acid. 0.19 mol of NaOH is added to 1.0 L of a 1.2M HCH3CO2 solution. acids: 0 bases: 0 other: acids: 0 x s ? 0.1 mol of HNO3 is added to 1.0L of a solution that is 0.4 M in both HCH,CO2 and KCH3CO2- other: 0

Answer 1

1-

Given NaOH is added to HCH3CO2 solution.

Now we know HCH3CO2 is a weak acicd where as NaOH is a strong base. So the neutralization reaction betwen them is-

NaOH + HCH3CO2 ----------> NaCH3CO2 + H2O

i.e 1 mole of NaOH will neutralize 1 mole of NaCH3CO2

Now given

Initial mols of NaoH taken = 0.19 mols

Concentration of HCH3CO2 taken = 1.2 M

Volume of HCH3CO2 taken = 1L

That means mols of HCH3CO2 taken = conc * vol

= 1.2 M * 1L

= 1.2 mols

So the reacation between them will be-

Reaction	NaOH +	HCH3CO2 ---------->	NaCH3CO2
Initial	0.19 mols	1.2 mols	0
Change	-0.19 mols	-0.19 mols	+0.19 mols
Equilibrium	0	1.01‬ mols	0.19 mols

That means finally in our solution, we have-

Weak acid (HCH3CO2) and its salt (NaCH3CO2) is present

Again NaCH3CO2 dissociate as- NaCH3CO2 ----------> Na+ + CH3CO2-

Now CH3CO2- is the conugate base of HCH3CO2

So we can say, at equilibrirum, we have

weak acid : HCH3CO2

Conjugate base : CH3CO2-

Others : Na+  

2- Similalrly-

Given HNO3 is added to a solution of HCH3CO2 and KCH3CO2

Now KCH3CO2 dissociate as- KCH3CO2 ----------> K+ + CH3CO2-

And the strong acid HNO3 will react with the conjugate base CH3CO2- as-

HNO3 + CH3CO2- ----------> HCH3CO2 + NO3-  

i.e 1 mole of HNO3 will neutralize 1 mole of CH3CO2-

Now given

Initial mols of HNO3 taken = 0.1 mols

Concentration of CH3CO2- taken = 0.4 M

Volume of CH3CO2- taken = 1L

That means mols of CH3CO2- taken = conc * vol

= 0.4 M * 1L

= 0.4 mols

So the reacation between them will be-

Reaction	HNO3 +	CH3CO2- ---------->	HCH3CO2
Initial	0.1 mols	0.4 mols	0
Change	-0.1 mols	-0.1 mols	+0.1 mols
Equilibrium	0	0.3‬ mols	0.1 mols

Agian initially we have HCH3CO2 present = 0.4 moles

So we can say, at equilibrirum, we have

weak acid : HCH3CO2

Conjugate base : CH3CO2-

Others : K+ and NO3-