The volume of the gas = 9.133 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}=0.0032=3.2\times 10^{-3}[/tex]
For weak base :
[tex]\tt [OH^-]=\sqrt{Kb.M}\\\\(3.2\times 10^{-3})^2=6.3\times 10^{-5}\times M\\\\M=0.163[/tex]
mol = M x V
mol = 0.163 x 2.5 L
mol = 0.408
Use ideal gas law at STP(1 atm, 273 K) :
[tex]\tt V=\dfrac{nRT}{P}=\dfrac{0.408\times 0.082\times 273}{1}=9.133~L[/tex]
