Question

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.

Answer 1

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!