Question

At $90^{\circ }C$ , pure water has $[H_3{O}^{+}]=10^{-6}$ mol $litr{e}^{-1}$ . 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 $K_w=[H_3{O}^{+}][O{H}^{-}]$
$=10^{-6}\times 10^{-6}$
$=10^{-12}$