A weak acid has a low concentration of H+ Ions and a dilute acid is a solution where acid is dissolved in a more volume of water than that of acid.
Hi,
Two sodium atoms are needed to create an ionic bond with sulfur.
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.
Q1)
firstly we need to determine the empirical formula of the compound. empirical formula is the simplest ratio of components in the compound.
percentages of the elements have been given, so lets assume we are calculating for a compound of 100g
C H O
mass 63.13 g 8.830 g 28.03 g
molar mass 12 g/mol 1 g/mol 16 g/mol
number of moles 63.13/12 8.830/1 28.03/16
5.26 8.830 1.75
divide by the smallest number of moles
5.26/1.75 8.830/1.75 1.75/1.75
= 3.01 = 5.04 =1
rounded off to the nearest whole numbers
C - 3
H - 5
O - 1
therefore empirical formula = C₃H₅O
Q2)
we have to next determine the molecular formula of the compound
molecular formula gives the actual composition of elements in the compound.
since we know the empirical formula and molecular mass, we can find how many empirical units are in the molecular formula.
mass of empirical unit = Cx3 + Hx5 + Ox1
= 12 g/mol x 3 + 1g/mol x 5 + 16 g/mol x 1
= 36 + 5 + 16 = 57 g/mol
the molecular mass = 228 g/mol
then number of empirical units in the molecular formula = 228 / 57 = 4
therefore there are 4 empirical units
then the molecular formula = 4 x empirical formula =4 (C₃H₅O)
molecular formula = C₁₂H₂₀O₄
