3 [easy] Eqlibrium qstns

[lanvins]

1. Dissolved CO2 acts as a weak acid in water according to the equation
CO2(aq) + H2O(l)---> H+(aq) + HCO3–(aq)
and the acidity constant of CO2 in water at 25°C is given by 4.5 × 10–7M2
Some CO2 is added to a solution of NaHCO3 at 25°C. In the solution, the concentration of the hydrogen carbonate ions (HCO3–) is 0.050 M and the CO2 concentration is 0.0020 M. Calculate the pH of the solution.

In this question isn't the concentration of  HCO3–(aq) equal to H+(aq) because their being produced at the same rate? Therefore why can't you just put the .05M straight into -log10 when calculating the pH?

2. Methanoic acid and ethanoic acid are both weak acids with the following acidity constants.
Ka in M at 25oC
methanoic acid (HCOOH) 1.82 × 10–4
ethanoic acid (CH3COOH) 1.74 × 10–5
Two separate solutions were prepared, one of 0.1 M methanoic acid and the other of 0.1 M ethanoic acid.
Which one of the following would be present in the highest concentration at 25oC?
A. CH3COOH in the ethanoic acid solution
B. CH3COO– in the ethanoic acid solution
C. HCOOH in the methanoic acid solution
D. HCOO– in the methanoic acid solution

i just don't understand the answer given(it's A by the way)

thanks

[Mao]

1. Dissolved CO2 acts as a weak acid in water according to the equation
CO2(aq) + H2O(l)---> H+(aq) + HCO3–(aq)
and the acidity constant of CO2 in water at 25°C is given by 4.5 × 10–7M2
Some CO2 is added to a solution of NaHCO3 at 25°C. In the solution, the concentration of the hydrogen carbonate ions (HCO3–) is 0.050 M and the CO2 concentration is 0.0020 M. Calculate the pH of the solution.

In this question isn't the concentration of  HCO3–(aq) equal to H+(aq) because their being produced at the same rate? Therefore why can't you just put the .05M straight into -log10 when calculating the pH?

not quite.

if CO₂ was put in water, then will approximate , as the only (significant) sources of these two chemicals are from the above reaction.

in this case, however, was added to . which means, will not approximate as we have started off with some , it is not solely produced from the above reaction.

even though they are being produced at the same rate, the amounts [hence concentration] do not necessarily need to be the same. the change in number of moles from initial to equilibrium, however, will be in stoichiometrical ratio (in this case, 1:1)

2. Methanoic acid and ethanoic acid are both weak acids with the following acidity constants.
Ka in M at 25oC
methanoic acid (HCOOH) 1.82 × 10–4
ethanoic acid (CH3COOH) 1.74 × 10–5
Two separate solutions were prepared, one of 0.1 M methanoic acid and the other of 0.1 M ethanoic acid.
Which one of the following would be present in the highest concentration at 25oC?
A. CH3COOH in the ethanoic acid solution
B. CH3COO– in the ethanoic acid solution
C. HCOOH in the methanoic acid solution
D. HCOO– in the methanoic acid solution

i just don't understand the answer given(it's A by the way)

looking at the , ethanoic acid dissociates less than methanoic acid [as indicated by the s]
and both are weak acids [as indicated by the s again], hence

looking at the options, we are to pick the species of highest concentration. obviously it will not be the methanoate ion nor the ethanoate ion as little dissociation occurs. also since ethanoic acid dissociates to a lesser extent than methanoic acid, there would be more present, hence A

[lanvins]

so, for the first question you'll always be told if its not equal, otherwise can you just assume so?

[Mao]

so, for the first question you'll always be told if its not equal, otherwise can you just assume so?

you'll be told if it is being added to water, in which case you can make that assumption

the key is, don't assume first make assumptions last =]