Answer: There is no question, but we can calculate a couple of items:
Density of sea water sample = (52.987g-44.317g)/8.5ml
Inorganic content of sample (mostly salts) = (44.599g-44.317g)/(52.987g-44.317g) x 100% = percent inorganics in water sample
Explanation:
<u>Answer:</u> The volume of concentrated solution required is 9.95 mL
<u>Explanation:</u>
To calculate the pH of the solution, we use the equation:
![pH=-\log[H^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%5BH%5E%2B%5D)
We are given:
pH = 0.70
Putting values in above equation, we get:
![0.70=-\log[H^+]](https://tex.z-dn.net/?f=0.70%3D-%5Clog%5BH%5E%2B%5D)
![[H^+]=10^{-0.70}=0.199M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D10%5E%7B-0.70%7D%3D0.199M)
1 mole of nitric acid produces 1 mole of hydrogen ions and 1 mole of nitrate ions.
Molarity of nitric acid = 0.199 M
To calculate the volume of the concentrated solution, we use the equation:

where,
are the molarity and volume of the concentrated nitric acid solution
are the molarity and volume of diluted nitric acid solution
We are given:

Putting values in above equation, we get:

Hence, the volume of concentrated solution required is 9.95 mL