`pH` of `0.1M` solution of `NaA` (sodium salt of weak acid `HA`) is `8.92`.Calculate `pK_(a)` of `HA`.If a drop `HPh(pK_("in")=9.52)` be added to the

Question

`pH` of `0.1M` solution of `NaA` (sodium salt of weak acid `HA`) is `8.92`.Calculate `pK_(a)` of `HA`.If a drop `HPh(pK_("in")=9.52)` be added to the

`pH` of `0.1M` solution of `NaA` (sodium salt of weak acid `HA`) is `8.92`.Calculate `pK_(a)` of `HA`.If a drop `HPh(pK_("in")=9.52)` be added to the above solution.predict whether the pink colour will visible or not the under the medical fact that our eyes can see the pink colour if the mole `%` of ionised form of indicated is `25%` or more.

Monjur Alahi

9 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

`pH=(1)/(2)(pK_(a)+pK_(w)+logc)`
`8.92=7+(1)/(2)pK_(a)+(1)/(2)(-1)`
`(1)/(2)pK_(a)=8.92-6.5=2.42`
`pK_(a)` of `HA=4.84`
`:.pK_(a)=pK_("in")+log""([ln^(-)])/([Hln])`
`=8.92=9.52+log""([ln^(-)])/([Hln])`
`=8.92=9.52+log"([ln^(-)])/([Hln])`
`([ln^(-)])/([Hln])=(ionised)/(unionised)=(1)/(4)`
Since ionised from `%=(1)/(1+4)xx100=20%` pink colour will not be visible.