A) \[100\sqrt{\frac{{{K}_{a}}}{C}}\]
B) \[\frac{100}{1+{{10}^{(p{{K}_{a}}-pH)}}}\]
C) Both [a] and [b] are correct
D) None of these
Correct Answer: C
Solution :
\[HA\xrightarrow{\,}\,{{H}^{+}}+{{A}^{-}}\] \[\therefore \] \[x=\sqrt{\frac{{{K}_{a}}}{C}}\] (we know) Also, \[HA\xrightarrow{\,}\,{{H}^{+}}+{{A}^{-}}\] Intial conc. C 0 0 Equil conc. \[C-Cx\] \[{{H}^{+}}\] \[Cx\] \[\therefore \] \[{{K}_{a}}=\frac{[{{H}^{+}}]\,[Cx]}{C(1-x)}=\frac{[{{H}^{+}}]x}{1-x}\] \[\therefore \] \[\log \,{{K}_{a}}=\log \,[{{H}^{+}}]\,+\log \,\frac{x}{1-x}\] \[\therefore \] \[p{{K}_{a}}=pH+\log \,\frac{1-x}{x}\] \[\Rightarrow \] \[x=\frac{1}{1+{{10}^{(p{{K}_{a}}-pH)}}}\]
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