# Why does acetate ion react with water? [closed]

When NaCl is added to water, it dissociates into $$\rm{Na^+}$$ and $$\rm{Cl^-}$$. The newly formed $$\rm{Na^+}$$ and $$\rm{Cl^-}$$ do not react with water.

When $$\rm{CH_3COONa}$$ (sodium acetate) is added to water, it dissociates into $$\rm{CH_3COO^-}$$ and $$\rm{Na^+}$$. The newly formed $$\rm{CH_3COO^-}$$ reacts with water to form $$\rm{CH_3COOH}$$ and $$\rm{OH^-}$$ ions.

Why is it that $$\rm{Cl^-}$$ does not react with water, but $$\rm{CH_3COO^-}$$ reacts with water?

• Search for concept of strong and weak acids. – Poutnik May 29 at 4:27
• Chloride ion is stable enough to exist on its own, acetate not so much and thus tries to grab $\ce{H+}$ from water. – Nisarg Bhavsar May 29 at 5:11

## 1 Answer

According to Bronsted-Lowry Acid-Base theory, stronger acids have weaker conjugate bases and vice-versa.

$$\ce{HCl}$$, being a strong acid in water, it's conjugate base i.e $$\ce{Cl-}$$ is weak or it is more stable. Although it reacts with $$\ce{H2O}$$ but to a very less extent. Whereas in the case of $$\ce{CH3COOH}$$, it being a weak acid, it's conjugate base is a strong base. So, it reacts with water and forms relatively more stable $$\ce{CH3COOH}$$.

The degree of hydrolysis, $$\mathrm{h}$$ of a salt of weak acid and strong base is approximately given by $$\mathrm{h}$$=$$\sqrt{k_w/Ck_a}$$.

If you substitute the $$\mathrm{k_a}$$ value for $$\ce{CH3COOH}$$ ( $$\mathrm{k_a}$$=$$\pu{1.8E-5}$$) for a $$\mathrm{0.1M}$$ solution of $$\ce{CH3COONa}$$, which yields $$\mathrm{h=}$$ $$\pu{7.45E-4}$$ where as for $$\mathrm{0.1M}$$ $$\ce{NaCl}$$ (although approximate) $$\mathrm{h\approx}\pu{ 10^-10}$$, which is negligible when compared to degree of hydrolysis of $$\ce{CH3COONa}$$.

• I agree that Cl- ion is stable because it has octet. Can we give a reason for the lesser stability of CH3COO- ion? – Nikhil Kumar May 29 at 5:33
• @NikhilKumar Carbon isn't enough electronegative to hold the negative charge – Jayadithya May 29 at 5:46