Question

On hydrolysis of sodium carbonate, the reaction takes place between ?

• A. $N{a}^{+}andwater$
• B. $N{a}^{+}andO{H}^{-}$
• C. $C{O}_3^{2-}andwater$
• D. $C{O}_3^{2-}and{H}^{+}$

Answer 1

Answer: (C)

$C{O}_3^{2-}andwater$

In wa­ter, sodi­um car­bon­ate will hy­drolyze, caus­ing the neu­tral medi­um to be­come al­ka­line. In the pres­ence of al­ka­lis, hy­drol­y­sis of sodi­um car­bon­ate is sup­pressed, but in an acid medi­um hy­drox­ide ions bond, and an al­ka­li re­ac­tion of the medi­um takes place. The equa­tion of hy­drol­y­sis (in ion­ic form):

$C{O}_3^{2-}+H_{2}O\leftrightarrow HC{O}_3^{-}+O{H}^{-}$ hy­drol­y­sis of sodi­um car­bon­ate $N{a}_{2}C{O}_3$

$N{a}_{2}C{O}_3+HOH\leftrightarrow NaOH+NaH­C{O}_3$

$2N{a}^{+}+C{O}_3^{2-}+HOH\leftrightarrow N{a}^{+}+OH{⁻}^{-}+N{a}^{+}+HC{O}_3^{-}$

Here sodi­um runs out, leav­ing:

$C{O}_3^{2-}+HOH\leftrightarrow O{H}^{-}+HC{O}_3^{-}$

Hence option A is corrrect answer.