40.0 ml of 0.200 m aqueous naoh is added to 200.0 ml of 0.100 m aqueous nahco3 in a flask maintained at 25 ? c. neglecting the effects of dilution, what is q for this reaction?

Answer : The 'q' for this reaction is, 328 J

Explanation :

First we have to calculate the moles of and .

From this we conclude that, the moles of are less than moles of . So, the limiting reactant is, NaOH.

Now we have to calculate the 'q' for this reaction.

The balanced chemical reaction will be,

The expression used for 'q' of this reaction is:

where,

n = number of moles of limiting reactant = 0.008 mole

At room temperature,

= -677 kJ/mole

= -286 kJ/mole

= -692 kJ/mole

= -230 kJ/mole

Now put all the given values in the above expression, we get:

     (conversion used : 1 kJ = 1000 J)

Therefore, the 'q' for this reaction is, 328 J

the ratio of the conjugate base to the acid required to attain the desired ph.