KOH is the base thus, it gives OH^– ions thus it cannot remove OH^– ions from reaction mixture but it adds on the concentration of OH^– ions.

So, an acid must be added but if a strong acid is added to the reaction mixture then in acidic condition the MnO₄ ^– formed reduces to give Mn²⁺ thus, HCl which is a strong acid and SO₂ which on treating with water forms a strong H₂SO₄ cannot be used for this purpose.

Thus, CO₂ which forms H₂CO₃ a weak acid reacts to remove OH^– but not that much acidic that MnO₄ ^– undergo reduction. Thus, CO₂ is used for this reaction for completion.

For this example,

CH₃COOH ⇌ CH₃COO^– + H⁺ ;

CH₃COONa ⇌ CH₃COO^– + Na⁺

when few drops of HCl are added to this buffer, the H⁺ of HCl immediatly combine with CH₃COO^– ions to form undissociated acetic acid molecules. Thus there will be no H⁺ ions to combine with CH₃COO^– ions to form undissociated acetic acid molecules. Thus there will be no appreciable change in its pH value. Like wise if few drops of NaOH are added, the OH^– ions will combine with H⁺ ions to form unionised water molecule. Thus pH of solution will remain constant.

2SO_2(g) + O_2(g) $$\rightleftharpoons$$ 2SO_3(g),    K

SO_3(g) $$\rightleftharpoons$$ SO_2(g) + $${1 \over 2}$$ O_2(g),     K' = $$\sqrt {{1 \over K}} $$

$$ \therefore $$ K' = $$\sqrt {{1 \over {278}}} = $$ 6.0 $$ \times $$ 10^$$-$$2

A₂(g) + B_2(g) $$\rightleftharpoons$$ 2AB_(g);

K_c = $${{{{\left[ {AB} \right]}^2}} \over {\left[ {{A_2}} \right]\left[ {{B_2}} \right]}}$$

= $${{{{\left( {2.8 \times {{10}^{ - 3}}} \right)}^2}} \over {3 \times {{10}^{ - 3}} \times 4.2 \times {{10}^{ - 3}}}}$$

= 0.62