Question

If $Kw$ for $H _{2} O$ at $90^{\circ} C$ is $10^{-12} M ^{2}$, then what would be the dissociation constant of water at the same temperature? [Assume $\left. d _{ H _{2} O }=1 g / mL \right]$

• A. $10^{-14}$
• B. $1.8 \times 10^{-16}$
• C. $1.8 \times 10^{-14}$
• D. $10^{-12}$

Answer 1

Answer: (C)

$K _{ H _{2} O }=\frac{\left[ H ^{+}\right]\left[ OH ^{-}\right]}{\left[ H _{2} O \right]}=\frac{10^{-12}}{55.55}$

$\therefore K _{ H _{2} O }=1.8 \times 10^{-14}$