The dissociation constant of a weak base is 1 × 10-5 at 25°C. The pH of its 0.1 M solution at the same temperature. will be .

Question

The dissociation constant of a weak base is 1 × 10-5 at 25°C. The pH of its 0.1 M solution at the same temperature. will be . (A) 11 (B) 3 (C) 6 (D) 13

Tardigrade Admin · Accepted Answer

By Ostwald dilution law, [ OH -]in weak base =√Kb C =√1 × 10-5 × 0.1 =1 × 10-3 M pOH =- log [ OH -]=- log (1 × 10-3) pOH =3 ⇒ pH + pOH =14 ⇒ pH =14-3=11

Q. The dissociation constant of a weak base is $1 \times 1 0^{- 5}$ at $25° C$ . The pH of its $0.1 M$ solution at the same temperature. will be .

11

3

6

13

By Ostwald dilution law,
$[O H^{-}]$ in weak base $= K_{b} C$
$= 1 \times 1 0^{- 5} \times 0.1$
$= 1 \times 1 0^{- 3} M$
$pO H = - lo g [O H -] = - lo g (1 \times 1 0^{- 3})$
$pO H = 3 \Rightarrow p H + pO H = 14$
$\Rightarrow p H = 14 - 3 = 11$

Q. The dissociation constant of a weak base is 1×10−5 at 25°C. The pH of its 0.1M solution at the same temperature. will be .

11

3

6

13

By Ostwald dilution law, [OH−]in weak base =Kb​C​ =1×10−5×0.1​ =1×10−3M pOH=−log[OH−]=−log(1×10−3) pOH=3⇒pH+pOH=14 ⇒pH=14−3=11

Q. The dissociation constant of a weak base is $1 \times 1 0^{- 5}$ at $25° C$ . The pH of its $0.1 M$ solution at the same temperature. will be .

By Ostwald dilution law,
$[O H^{-}]$ in weak base $= K_{b} C$
$= 1 \times 1 0^{- 5} \times 0.1$
$= 1 \times 1 0^{- 3} M$
$pO H = - lo g [O H -] = - lo g (1 \times 1 0^{- 3})$
$pO H = 3 \Rightarrow p H + pO H = 14$
$\Rightarrow p H = 14 - 3 = 11$