[phoenixrising]

Question 3

a) How much sodium acetate powder is required to prepare a 3 M (per 200 mL) aqueous solution?

1. 3 M of C₂H₃NaO₂ = 3 mol/L of C₂H₃NaO₂
2. Convert 3 mol/L of C₂H₃NaO₂ to grams

(3 mol/L) x [(82.0343 g C₂H₃NaO₂) / 1 mol]  x 0.2 L = 49.2 g C₂H₃NaO₂

b) The lab doesn't have anhydrous sodium acetate in stock, but rather sodium acetate trihydrate. How much of the hydrate form would be required to prepare an equivalent solution?

I need help here. What exactly do they want me to convert?

[phoenixrising]

Also, what about this one? I have to fill in the table, and I already calculated the weight for both molecules. My question is...do I THEN have to convert both to mL?

[Hunter2]

The first one you have to calculate how much grmm of CH₃COONa x 3 H₂O you have to add instead of the anhydrouis Acetate. Calculation is the same but with different molecular weight.

The 2. One: Tris base is correct.  Acetic acid needs density to convert to gramm. EDTA is given as an 0.5 M solution instead of powder, you have to calculate again.

[AWK]

http://www.chemicalforums.com/index.php?topic=22089.0

[phoenixrising]

The first one you have to calculate how much grmm of CH₃COONa x 3 H₂O you have to add instead of the anhydrouis Acetate. Calculation is the same but with different molecular weight.

The 2. One: Tris base is correct.  Acetic acid needs density to convert to gramm. EDTA is given as an 0.5 M solution instead of powder, you have to calculate again.

If the molar mass of the Hydrate (H₂O) is 18.015 g/mol, the molar mass of the trihydrate is 3 x (18.015 g H₂O / mol) = 54.045 g/mol Trihydrate

(3 mol C₂H₂NaO₂ / L ) x ( 54.045 g H₂O / 1 mol C₂H₂NaO₂) x ( 0.2 L ) = 32.427 g of Trihydrate

Is that the answer?

[Hunter2]

You forgot the sodiumacetate. Molar mass of sodiumacetate + 3 water.

[phoenixrising]

You forgot the sodiumacetate. Molar mass of sodiumacetate + 3 water.

Ohhhhhhh

[phoenixrising]

You forgot the sodiumacetate. Molar mass of sodiumacetate + 3 water.

[( 3 mol C₂H₂NaO₂ / L ) x ( 82.0343 g C₂H₂NaO₂ / 1 mol C₂H₂NaO₂ ) x ( 0.2 L )] + [(3 mol H₂O / L ) x ( 18.015 g H₂O / 1 mol H₂O ) x ( 0.2 L )]

Like that?
I don't think that's right because the answer doesn't cancel out to ___ g H₂O

[phoenixrising]

You forgot the sodiumacetate. Molar mass of sodiumacetate + 3 water.

You still there?

[Hunter2]

You have to add 82 g/mol for the sodiumacetat + 3  x 18g/mol for the water what gives 136 g/mol. Use this value to calculate further
.

[phoenixrising]

You have to add 82 g/mol for the sodiumacetat + 3  x 18g/mol for the water what gives 136 g/mol. Use this value to calculate further
.

3 mol/L x (136.0793 g H₂O/mol) x (0.2 L) = 81.6 g Trihydrate?

I was looking for ____ g Trihydrate right?
That was my other question...
Somehow everything canceled out and left me with just g Trihydrate.

[Hunter2]

Trihydrate doesn't stands alone, it is everytime together with the salt. How you can weight it alone. The 81,6 g are correct of CH₃COONa x 3 H₂O