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

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(([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!