If the pH of 0.10 M monobasic acid at 298 K is 5.0 the value the value of `PK_(a)` at the same temperature is

Options

A. `5.0`
B. `8.0`
C. `9.0`
D. `6.0`

Answer

C

Explanation

Given : pH for monobasic acid = 50 The reaction for dissociation of monobasic acid is given by
`HA+(aq) rarr H^(+)(aq) + A^(-)(aq)`
Initial conc. `" " `0.10 `" " 0" "0`
Conc of `" "(0.10-x) " "x" "x`
equilibrium
Since pH =- log `[H^(+)]`
`:. [H^(+)]=10^(-5) ` M
Thus x = `10^(-5)M`
Now `K_(a)= ([H^(+)][A^(-)])/([HA])`
`:. K_(a)= ([10^(-5)][10^(-5)])/[[0.10])`
`K_(a)= 10^(-9)`
`implies pK_(a)=-"log"[K_(a)]=-"log"[10^(-9)]`
`:. pK_(a)=9`