Equivalent Mass of Sodium Acetate: 20.25g

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The equivalent mass of sodium acetate (NaC2H3O2) is not 20.25g, as sodium acetate does not donate protons due to the presence of Na+ which is not a proton. The acetate ion (CH3COO-) is a weak base and does not contribute to acidity, as the hydrogen atoms in the acetate structure are not acidic. The dissociation of sodium acetate in water results in the formation of hydroxide ions (OH-) rather than protons, leading to a basic solution. Additionally, the equivalent weight of any compound must be defined in the context of the specific reaction involved, as it varies based on the number of electrons transferred.

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bomba923
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Would the equivalent mass for sodium acetate NaC2H3O2 be 20.25g ?

Sodium acetate has four protons it can donate, right? The 3H+ and the Na+, right??
 
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bomba923 said:
Sodium acetate has four protons it can donate, right? The 3H+ and the Na+, right??
:eek: No!
1) Na+ is not a proton, so sodium acetate cannot donate a corresponding proton (acetic acid, however, can).
2) As far as I know none of the three hydrogen atoms on the acetate ion are acidic. You cannot just look at the overall formula NaC2H3O2 and think that you can transform it into "C2O2(4-)". You must look at how the hydrogen atoms are bound in the acetate ion: CH3COO-. The three hydrogen atoms are sitting on a methyl group and thus at most very very weakly acidic. A good help is a table of acid/base constants.

I don't know what the equivalent mass is that you are talking about, but sodium acetate will dissociate in as many Na+ as CH3COO- (or OAc-) ions (for charge balance). But since acetic acid (CH3COOH) is only a moderate acid (pKa=4.76), actually the acetate ion will act as a weak base and take up protons from water molecules.

So:
:cool: Acetic acid, CH3COOH is a monoprotic acid and will dissociate partly into H+ and CH3COO-, the extent depending on the original pH of the solution and on the concentration of acetid acid. When dissolved in water, at a concentration below 10^-4.76 CH3COO- will dominate (and hence most protons will have been given off, although they're not so many) and above 10^-4.76 undissociated CH3COOH will dominate over CH3COO- (although the total amount of protons given off of course is higher than at the lower concentration). This can be easily visualised with a log diagram (log konc. versus pH), where one assumes that all the protons existing in the solution come from the acid, i.e. [H+]=[CH3COO-], and then one just locates the pH at which these curves intersect. In the diagram one also sees that the assumption breaks down below concentrations around 10^-7, since then the amount of H+ ions is to a large extent determined by the dissociation constant of water.
:cool: Sodium acetate is highly soluble in distilled water and will dissociate completely. However, this will not generate any protons. Instead, being the conjugate base of acetic acid, the acetate ion will take up protons fom water, leaving OH- ions behind. Now one thus assumes that the amount of OH- ions is equal to the amount of CH3COOH molecules produced, but now one can see in the diagram that this assumption breaks down already below 10^-4.76 since we just stated that OH- ions are generated, implying that the pH is raised to above 7 in originally neutral water. But at e.g. [CH3COO-]=0.001M, one finds the intersection of [CH3COOH] and [OH-] (versus pH): [OH-] of course has a straight line, passing through (7,-7) and e.g. (8,-6), while the [CH3COOH] curve in this portion also is a straight line going through (7,-7+4.76-3) and (8,-8+4.76-3), thus they intersect at (7.88,-6.12). So the pH is increased with almost one unit. On the other hand, of our total sodium acetate concentration of 10^-3 M still only 10^-6.12 M is acetic acid! But at least I demonstrated that it is (weakly) basic and not at all acidic.
 
...and one more little thing : bomba, you CAN NOT specify the equivalent weight of a compound without specifying the reaction involved. The number of electrons transferred per mole depends on the reaction. With some compounds, there may not be more than one choice, but in general it is good practice to also specify the reaction.
 

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