Please can you answer #8
1 answer:
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Answer:
10.88 g
Explanation:
We have:
[CH₃COOH] = 0.10 M
pH = 5.25
Ka = 1.80x10⁻⁵
V = 250.0 mL = 0.250 L
The pH of the buffer solution is:
(1)
By solving equation (1) for [CH₃COONa*3H₂O] we have:
Hence, the mass of the sodium acetate tri-hydrate is:
Therefore, the number of grams of CH₃COONa*3H₂O needed to make an acetic acid/sodium acetate tri-hydrate buffer solution is 10.88 g.
I hope it helps you!
Answer : The correct option is, 0.21 moles
Solution : Given,
Molarity of the solution = 0.85 M = 0.85 mole/L
Volume of solution = 250 ml = 0.25 L
Molarity : It is defined as the number of moles of solute present in one liter of the solution.
Formula used :
Now put all the given values in this formula, we get the moles of solute of the solution.
Therefore, the moles of solute is, 0.21 moles