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# Question and Answer

## QuestionChemistryClass 11

Is it possible for a weak acid, say acetic acid $${\left({K}_{{{a}}}={1.8}\times{10}^{{-{5}}}\right)}$$ to have a pH = 7 by adding any other chemical and if so how much amount of it should be added ?

If acid is completely neturalised by a base (say NaOH), the salt fromed will have $${p}{H}\lt{7}$$ . The only possibility to have $${p}{H}={7}$$ for $${C}{H}_{{{3}}}{C}{O}{O}{H}$$ can be obtained by addition of $${C}{H}_{{{3}}}{C}{O}{O}^{{-}}$$ (a conjugate base say $${C}{H}_{{{3}}}{C}{O}{O}{N}{a}$$) in it.
$${C}{H}_{{{3}}}{C}{O}{O}{H}\Leftrightarrow{C}{H}_{{{3}}}{C}{O}{O}^{{-}}+{H}^{{+}}$$
$${K}_{{{a}}}=\frac{{{\left[{C}{H}_{{{3}}}{C}{O}{O}^{{-}}\right]}{\left[{H}^{{+}}\right]}}}{{{\left[{C}{H}_{{{3}}}{C}{O}{O}{H}\right]}}}$$
$$\therefore\frac{{{\left[{C}{H}_{{{3}}}{C}{O}{O}^{{-}}\right]}}}{{{\left[{C}{H}_{{{3}}}{C}{O}{O}{H}\right]}}}=\frac{{{1.8}\times{10}^{{-{5}}}}}{{{10}-{7}}}={1.8}\times{10}^{{{2}}}$$
Thus addition of 180 times concentratgikon of $${C}{H}_{{{3}}}{C}{O}{O}{N}{a}$$ to $${C}{H}_{{{3}}}{C}{O}{O}{H}$$ will give a solution of $${p}{H}={7}$$