The molarity of formic acid is 100 mM or
. The dissociation reaction of formic acid is as follows:

The expression for dissociation constant of the reaction will be:
![K_{a}=\frac{[HCOO^{-}][H^{+}]}{[HCOOH]}](https://tex.z-dn.net/?f=K_%7Ba%7D%3D%5Cfrac%7B%5BHCOO%5E%7B-%7D%5D%5BH%5E%7B%2B%7D%5D%7D%7B%5BHCOOH%5D%7D)
Rearranging,
![[HCOO^{-}]=\frac{K_{a}[HCOOH]}{[H^{+}]}](https://tex.z-dn.net/?f=%5BHCOO%5E%7B-%7D%5D%3D%5Cfrac%7BK_%7Ba%7D%5BHCOOH%5D%7D%7B%5BH%5E%7B%2B%7D%5D%7D)
Here, pH of solution is 4.15 thus, concentration of hydrogen ion will be:
![[H^{+}]=10^{-pH}=10^{-4.15}=7.08\times 10^{-5}M](https://tex.z-dn.net/?f=%5BH%5E%7B%2B%7D%5D%3D10%5E%7B-pH%7D%3D10%5E%7B-4.15%7D%3D7.08%5Ctimes%2010%5E%7B-5%7DM)
Similarly,
thus,

Putting the values,
![[HCOO^{-}]=\frac{(1.78\times 10^{-4}M)(100\times 10^{-3}M)}{(7.08\times 10^{-5}M}=0.2511 M](https://tex.z-dn.net/?f=%5BHCOO%5E%7B-%7D%5D%3D%5Cfrac%7B%281.78%5Ctimes%2010%5E%7B-4%7DM%29%28100%5Ctimes%2010%5E%7B-3%7DM%29%7D%7B%287.08%5Ctimes%2010%5E%7B-5%7DM%7D%3D0.2511%20M)
Therefore, the concentration of formate will be 0.2511 M.
D as the two substances could have close boiling points, so if the temperature is not controlled well, both substances will evaporate.
The answer to this question is: The pressure will be increased two times
The gas law includes an interaction of pressure(p), volume(V) and temperature(T). In this case, the gas has same amount and volume but then the temperature is doubled. The calculation would be:
p1 * V1 / T1= p2*V2 /T2
p1 * 1 / 1 = p2 * 1 / 2
p1 / (1/2)=p2
p2= 2* p1
I think it is because it has an even number of valence electrons.