Question

At $90^\circ C$ , pure water has $\left[\right.H_{3}O^{+}\left]\right.=10^{- 6}$ mol $litre^{- 1}$ . What is the value of $K_{w}$ at $90^\circ $ C?

• A. $10^{- 6}$
• B. $10^{- 12}$
• C. $10^{- 14}$
• D. $10^{- 8}$

Answer 1

Answer: (B)

At $\text{90}^{\text{o}} \text{C,} \, \left[\text{H}_{\text{3}} \text{O}^{\text{+}} \left] \, \text{=} \, \right[ \text{OH}^{-}\right]$
$= 10^{- 6} \, \text{M}$
So $K_{w}=\left[\right.H_{3}O^{+}\left]\right.\left[\right.OH^{-}\left]\right.$
$= 10^{- 6} \, \times 10^{- 6}$
$= 10^{- 12}$