The pH of 0.1, M monobasic acid is 4.50 Calcu | vedclass

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Question

$pH =-\log \left[ H ^{+}ight]$

Therefore, $\left[ H ^{+}ight]=10^{- pH } =10^{-4.50} $

$=3.16 	imes 10^{-5} $

$\left[ H ^{+}ight]=\lef

Vedclass · Accepted Answer

Vedclass · Answer

$pH =-\log \left[ H ^{+}ight]$

Therefore, $\left[ H ^{+}ight]=10^{- pH } =10^{-4.50} $

$=3.16 	imes 10^{-5} $

$\left[ H ^{+}ight]=\left[ A ^{-}ight]=3.16 	imes 10^{-5}$

Thus, $K_{ a }=\left[ H ^{+}ight]\left[ A ^{-}ight] /[ HA ]$

${[HA]_{eqlbm}} = 0.1 - \left( {3.16 	imes {{10}^{ - 5}}} ight) \simeq 0.1$

$K_{ a }=\left(3.16 	imes 10^{-5}ight)^{2} / 0.1=1.0 	imes 10^{-8}$

$p K_{ a }=-\log \left(10^{-8}ight)=8$

Alternatively, "Percent dissociation" is another useful method for measure of strength of a weak acid and is given as:

Percent dissociation

$ = {[HA]_{{m{dissociated }}}}/{[HA]_{{m{initial }}}} 	imes 100\% \,\,\,\,\,\,\left( {7.32} ight)$

The $pH$ of $0.1\, M$ monobasic acid is $4.50$ Calculate the concentration of species $H ^{+},$ $A^{-}$ and $HA$ at equilibrium. Also, determine the value of $K_{a}$ and $pK _{a}$ of the monobasic acid.

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