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Verizon [17]
4 years ago
9

2. In Experiment SOL, you investigated the solubility of oxalic acid. Sodium oxalate, Na2C2O¬4, is the sodium salt of this acid.

Categorize it as acidic, basic, or neutral in water. Does the salt dissolve, dissociate, or both in aqueous solutions? What about oxalic acid; does it dissolve, dissociate, or both in water? Explain.
Chemistry
1 answer:
AnnZ [28]4 years ago
6 0

Answer:

Sodium oxalate is a basic salt. In water it can be dissolved and dissociated.

The oxalic acid in water has two dissociations.

Explanation:

Na2C2O4 ---> 2Na+   +  C2O4-2

Sodium oxalate is the conjugate base of a weak acid. In water this salt, dissociates completely giving rise to the sodium and oxalate ions. As Na+ comes from a strong base, in water it does not produce hydrolysis while oxalate does react in water, because it takes a proton from it and it generates a basic hydrolysis releasing OH-.

C2O4-2  + H2O ⇄  HC2O4-  +  OH-

In water the salt is basic.  The pH of an aqueous solution of this salt is basic, since OH- is generated.

The HC2O4- has a second hydrolisis, it takes another proton from water to form oxalic acid.

HC2O4-  +  H2O ⇄  H2C2O4  +  OH-

The oxalic acid acts as a weak acid, it can release 2 protons to water, to make oxalate (its conjugate base).

H2C2O4  + H2O ⇄ H3O+  + HC2O4-

HC2O4-  +  H2O ⇄  H3O+  C2O4-2

The  HC2O4-  acts as an ampholyte since it accepts and delivers protons simultaneously.

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If 507 g FeCl2 were used up in the reaction FeCl2 + 2NaOH Fe(OH)2(s) + 2NaCl, how many grams of NaCl would be made?
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6 0
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The acid-dissociation constant for benzoic acid (C6H5COOH) is 6.3×10−5. Calculate the equilibrium concentration of H3O+ in the s
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The dissociation of acid reaction is:

                       C_6H_5COOH+H_2O\rightarrow H_3O^++C_6H_5COO^-

Initial conc.        c                                 0                0

At eqm.             c-x                                 x                x

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K_a=\frac{(x)\times (x)}{(c-x)}

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6.3\times 10^{-5}=\frac{(x)\times (x)}{[(7.0\times 10^{-2})-x]}

x=2.1\times 10^{-3}M

Thus, the equilibrium concentration of H_3O^+ in the solution is, 2.1\times 10^{-3}M

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