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Charra [1.4K]
3 years ago
12

calculate the number of grams of CH3COONa * 3H2O (sodium acetate tri-hydrate) needed to make 250.0 mL of a CH3COOH (acetic acid)

/ CH3COONa * 3H2O buffer. The target pH of the buffer is 5.25. The given concentration of [CH3COOH] is equal to 0.10 M. Ka = 1.80 x 10-5 for acetic acid.
Chemistry
1 answer:
anastassius [24]3 years ago
4 0

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

M_{CH_{3}COONa*3H_{2}O} = molar \thinspace mass = 136 g/mol

The pH of the buffer solution is:  

pH = pKa + log(\frac{[CH_{3}COONa*3H_{2}O]}{[CH_{3}COOH]})    (1)

By solving equation (1) for [CH₃COONa*3H₂O] we have:  

log [CH_{3}COONa*3H_{2}O] = pH - pKa + log [CH_{3}COOH]  

log [CH_{3}COONa*3H_{2}O] = 5.25 - (-log(1.80 \cdot 10^{-5})) + log (0.10) = -0.495                                

[CH_{3}COONa*3H_{2}O] = 10^{-0.495} = 0.32 M

Hence, the mass of the sodium acetate tri-hydrate is:

m = moles*M = [CH_{3}COONa*3H_{2}O]*V*M = 0.32 mol/L*0.250 L*136 g/mol = 10.88 g

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!      

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