D. 1,008 liters because you are looking for liters from a calculation of moles. Recognizing that you can do STP (22.4L) you multiply this number by 45 moles and it is 1,008 liters
Answer:
25.7 mL
Explanation:
Step 1: Given data
- Initial concentration (C₁): 0.350 M
- Final volume (V₂): 600 mL
- Final concentration (C₂): 0.150 M
Step 2: Calculate the volume of the initial solution
We have a concentrated solution and we want to prepare a diluted one. We can calculate the initial volume using the dilution rule.
C₁ × V₁ = C₂ × V₂
V₁ = C₂ × V₂ / C₁
V₁ = 0.150 M × 600 mL / 0.350 M
V₁ = 25.7 mL
Mass is often the most common and weight is its close to that but I'd go with mass
Answer: Dissociation constant of the acid is
.
Explanation: Assuming the acid to be monoprotic, the reaction follows:

pH of the solution = 6
and we know that
![pH=-log([H^+])](https://tex.z-dn.net/?f=pH%3D-log%28%5BH%5E%2B%5D%29)
![[H^+]=antilog(-pH)](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3Dantilog%28-pH%29)
![[H^+]=antilog(-6)=10^{-6}M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3Dantilog%28-6%29%3D10%5E%7B-6%7DM)
As HA ionizes into its ions in 1 : 1 ratio, hence
![[H^+]=[A^-]=10^{-6}M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D%5BA%5E-%5D%3D10%5E%7B-6%7DM)
As the reaction proceeds, the concentration of acid decreases as it ionizes into its ions, hence the decreases concentration of acid at equilibrium will be:
![[HA]=[HA]-[H^+]](https://tex.z-dn.net/?f=%5BHA%5D%3D%5BHA%5D-%5BH%5E%2B%5D)
![[HA]=0.1M-10^{-6}M](https://tex.z-dn.net/?f=%5BHA%5D%3D0.1M-10%5E%7B-6%7DM)
![[HA]=0.09999M](https://tex.z-dn.net/?f=%5BHA%5D%3D0.09999M)
Dissociation Constant of acid,
is given as:
![K_a=\frac{[A^-][H^+]}{HA}](https://tex.z-dn.net/?f=K_a%3D%5Cfrac%7B%5BA%5E-%5D%5BH%5E%2B%5D%7D%7BHA%7D)
Putting values of
in the above equation, we get


Rounding it of to one significant figure, we get
