If Kw for H 2 O at 90° C is 10-12 M 2, then what would be the dissociation constant of water at the same temperature? [Assume . d H 2 O =1 g / mL ]

Question

If Kw for H 2 O at 90° C is 10-12 M 2, then what would be the dissociation constant of water at the same temperature? [Assume . d H 2 O =1 g / mL ] (A) 10-14 (B) 1.8 × 10-16 (C) 1.8 × 10-14 (D) 10-12

Tardigrade Admin · Accepted Answer

K H 2 O =([ H +][ OH -]/[ H 2 O ])=(10-12/55.55) ∴ K H 2 O =1.8 × 10-14

Q. If $K w$ for $H_{2} O$ at $9 0^{\circ} C$ is $1 0^{- 12} M^{2}$ , then what would be the dissociation constant of water at the same temperature? [Assume $d_{H_{2} O} = 1 g / m L]$

$1 0^{- 14}$

$1.8 \times 1 0^{- 16}$

$1.8 \times 1 0^{- 14}$

$1 0^{- 12}$

$K_{H_{2} O} = \frac{[ H ^{+} ] [ O H ^{-} ]}{[ H _{2} O ]} = \frac{1 0 ^{- 12}}{55.55}$

$∴ K_{H_{2} O} = 1.8 \times 1 0^{- 14}$

Q. If Kw for H2​O at 90∘C is 10−12M2, then what would be the dissociation constant of water at the same temperature? [Assume dH2​O​=1g/mL]

10−14

1.8×10−16

1.8×10−14

10−12