The volume of the gas = 9.14 L
Further explanation
Given
Kb = 6.3 x 10⁻⁵
2.5 L solution
pOH= 2.5
Required
Volume of gas
Solution
pOH = 2.5
[tex]\tt [OH^-]=10^{-2.5}=3.2\times 10^{-3}[/tex]
For weak base :
[tex]\tt [OH^-]=\sqrt{Kb.M}[/tex]
(
3.2 x 10⁻³)² = 6.3 x 10⁻⁵ x M
M = 0.163
mol = M x V
mol = 0.163 x 2.5 L
mol = 0.408
At STP(1 atm, 273 K) 1 mol = 22.4 L
so the volume = 0.
408 x 22.4 L = 9.14 L
