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Chemistry
At 90° C, pure water has H 3 O +ion concentration of 10-6 mol / L -1. The Kw at 90° C is :
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Q. At $90^{\circ} C$, pure water has $H _{3} O ^{+}$ion concentration of $10^{-6} mol / L ^{-1}$. The $K_{w}$ at $90^{\circ} C$ is :
Manipal
Manipal 2005
Equilibrium
A
$ {{10}^{-6}} $
13%
B
$ {{10}^{-14}} $
20%
C
$ {{10}^{-12}} $
47%
D
$ {{10}^{-8}} $
20%
Solution:
$ K_{w}=\left[ H ^{+}\right]\left[ OH ^{-}\right] $
$=\left(10^{-6}\right) \times\left(10^{-6}\right)$
$\left(\because\left[ H ^{+}\right]=\left[ OH ^{-}\right]\right) $
$=10^{-12} $