At 80° C, distilled water has concentration equal to 1 × 10-6 mole/litre. The value of Kw at this temperature will be

Question

At 80° C, distilled water has concentration equal to 1 × 10-6 mole/litre. The value of Kw at this temperature will be (A) 1 × 10-6 (B) 1 × 10-12 (C) 1 × 10-9 (D) 1 × 10-15

Tardigrade Admin · Accepted Answer

[ H 3 O +]=1 × 10-6 mol L -1 or [ H +]=1 × 10-6 mol L -1 As for distilled water [ H +]=[ OH -] [ H +]=[ OH -]=1 × 10-6 mol L -1 K w =[ H +][ OH -] =10-12 mol 2 L -2

Q. At $8 0^{\circ} C$ , distilled water has concentration equal to $1 \times 1 0^{- 6}$ mole/litre. The value of $K_{w}$ at this temperature will be

$1 \times 1 0^{- 6}$

$1 \times 1 0^{- 12}$

$1 \times 1 0^{- 9}$

$1 \times 1 0^{- 15}$

$[H_{3} O^{+}] = 1 \times 1 0^{- 6} m o l L^{- 1}$

or $[H^{+}] = 1 \times 1 0^{- 6} m o l L^{- 1}$

As for distilled water $[H^{+}] = [O H^{-}]$

$[H^{+}] = [O H^{-}] = 1 \times 1 0^{- 6} m o l L^{- 1}$

$K_{w} = [H^{+}] [O H^{-}]$

$= 1 0^{- 12} m o l^{2} L^{- 2}$

Q. At 80∘C, distilled water has concentration equal to 1×10−6 mole/litre. The value of Kw​ at this temperature will be

1×10−6

1×10−12

1×10−9

1×10−15