Answer:
30 Liters of 40% acid solution and 10 L of 60% acid solution is needed.
Explanation:
Let volume of the 40% acid solution be x.
Let volume of the 60% acid solution be y.
Volume of solution formed after mixing both solution = 40 L
x + y = 40 L..[1]
Volume of acid 40% solution = 40% of x= 0.4x
Volume of acid 60% solution = 60% of y= 0.6y
Volume of acid formed = 45% of 40 L = 
..[2]
Solving [1] and [2]
x = 30 L , y = 10 L
30 Liters of 40% acid solution and 10 L of 60% acid solution is needed.
Answer:
The final and initial concentration of the acid and it's conjugate base are approximately equal, that is we use the weak acid approximation.
Explanation:
The Henderson-Hasselbalch is used to calculate the pH of a buffer solution. It depends on the weak acid approximation.
Since the weak acid ionizes only to a small extent, then we can say that [HA] ≈ [HA]i
Where [HA] = final concentration of the acid and [HA]i = initial concentration of the acid.
It also follows that [A^-] ≈ [A^-]i where [A^-] and[A^-]i refer to final and initial concentrations of the conjugate base hence the answer above.
Answer:
The concentration of species in 500 mL of a 2.104 M solution of sodium sulfate is 4.208 M sodium ion and 2.104 M sulfate ion. (option E)
Explanation:
Step 1: Data given
Volume = 500 mL = 0.500 L
The concentration sodium sulfate = 2.104 M
Step 2: The equation
Na2SO4 → 2Na+ + SO4^2-
For 1 mol Na2SO4 we have 2 moles sodium ion (Na+) and 1 mol sulfate ion (SO4^2-)
Step 3: Calculate the concentration of the ions
[Na+] = 2*2.104 M = 4.208 M
[SO4^2-] = 1*2.104 M = 2.104 M
The concentration of species in 500 mL of a 2.104 M solution of sodium sulfate is 4.208 M sodium ion and 2.104 M sulfate ion. (option E)
Answer:
tin (IV) oxide octahydrate contains 8 part water
What is the percent composition of water found in tin (IV) oxide octahydrate?
