At 25°C, pH of a 10-8 M aqueous KOH solution will be

Question

At 25°C, pH of a 10-8 M aqueous KOH solution will be (A) 6.0 (B) 7.02 (C) 8.02 (D) 9.02

Tardigrade Admin · Accepted Answer

NaOH leftharpoons Na ++ OH - [ OH -]=10-8 M H 2 O leftharpoons H ++ OH - [ OH -]=10-7 M ∴ [ OH -] text total =(10-8+10-7) M =10-7(1.1) =1.1 × 10-7 ∴ pOH = log 1.1 × 10-7 ≃ 6.98 ∴ pH =14-6.98 =7.02

Q. At $2 5^{\circ}$ C, pH of a $1 0^{- 8}$ M aqueous $K O H$ solution will be

6.0

7.02

8.02

9.02

$N a O H ⇌ N a^{+} + O H^{-} [O H^{-}] = 1 0^{- 8} M$
$H_{2} O ⇌ H^{+} + O H^{-} [O H^{-}] = 1 0^{- 7} M$
$∴ [O H^{-}]_{total} = (1 0^{- 8} + 1 0^{- 7}) M$
$= 1 0^{- 7} (1.1)$
$= 1.1 \times 1 0^{- 7}$
$∴ pO H = lo g 1.1 \times 1 0^{- 7} ≃ 6.98$
$∴ p H = 14 - 6.98$
$= 7.02$

Q. At 25∘C, pH of a 10−8 M aqueous KOH solution will be

6.0

7.02

8.02

9.02

NaOH⇌Na++OH−[OH−]=10−8M H2​O⇌H++OH−[OH−]=10−7M ∴[OH−]total ​=(10−8+10−7)M =10−7(1.1) =1.1×10−7 ∴pOH=log1.1×10−7≃6.98 ∴pH=14−6.98 =7.02

Q. At $2 5^{\circ}$ C, pH of a $1 0^{- 8}$ M aqueous $K O H$ solution will be

$N a O H ⇌ N a^{+} + O H^{-} [O H^{-}] = 1 0^{- 8} M$
$H_{2} O ⇌ H^{+} + O H^{-} [O H^{-}] = 1 0^{- 7} M$
$∴ [O H^{-}]_{total} = (1 0^{- 8} + 1 0^{- 7}) M$
$= 1 0^{- 7} (1.1)$
$= 1.1 \times 1 0^{- 7}$
$∴ pO H = lo g 1.1 \times 1 0^{- 7} ≃ 6.98$
$∴ p H = 14 - 6.98$
$= 7.02$