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
[tex]M_{CH_{3}COONa*3H_{2}O} = molar \thinspace mass = 136 g/mol[/tex]
The pH of the buffer solution is:
[tex] pH = pKa + log(\frac{[CH_{3}COONa*3H_{2}O]}{[CH_{3}COOH]}) [/tex] (1)
By solving equation (1) for [CH₃COONa*3H₂O] we have:
[tex] log [CH_{3}COONa*3H_{2}O] = pH - pKa + log [CH_{3}COOH] [/tex]
[tex] log [CH_{3}COONa*3H_{2}O] = 5.25 - (-log(1.80 \cdot 10^{-5})) + log (0.10) = -0.495 [/tex]
[tex][CH_{3}COONa*3H_{2}O] = 10^{-0.495} = 0.32 M[/tex]
Hence, the mass of the sodium acetate tri-hydrate is:
[tex]m = moles*M = [CH_{3}COONa*3H_{2}O]*V*M = 0.32 mol/L*0.250 L*136 g/mol = 10.88 g[/tex]
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!
