Question

Write the equations describing how sodium oxalate and sodium carbonate
dissolve in water, and obtain the molar ratios of the ions to their source compounds

Answer 1

Answer on question #73917, chemistry

Sodium oxalate react with water (Hydrolysis of sodium oxalate)

$$\mathrm{Na_2C_2O_4} + \mathrm{H_2O} \rightleftharpoons \mathrm{NaHCO_3} + \mathrm{NaOH}$$

which may be expressed as follows in terms of ions:

$$\begin{array}{l} \mathrm{Na_2C_2O_4} \rightleftharpoons 2\mathrm{Na^+} + \mathrm{C_2O_4^{2-}} \\ \mathrm{C_2O_4^{2-}} + \mathrm{H_2O} \rightleftharpoons \mathrm{HCO_3^-} + \mathrm{OH^-}. \end{array}$$

sodium oxalate consistent of $2\mathrm{Na}^+$ ion and 1 oxalate ion present, molar ratio is (2:1)

The hydrolysis may be considered as due to entirely to the small value of the ionization constant of the second hydrogen of oxalic acid, which is only about one thousandth of that of the first hydrogen of this acid.

Sodium carbonate react with water (hydrolysis of sodium carbonate)

Na2CO3 ---> 2Na+ + CO3--

CO3-- + H2O ---> HCO3- + OH-

and

HCO3- + H2O ---> H2CO3 + OH-

Sodium carbonate consistent of $2\mathrm{Na}^+$ ion and 1 CO3^-2 ion present. Molar ratio is (2:1)

Sodium carbonate is a weak basic compound. After hydrolysis it form carbonic acid H2CO3.

Answer provided by AssignmentExpert.com