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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 ?

Answer

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 \({p}{H}={7}\)
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