Question

Trimethylamine, (CH3)2N is a weak base (K6 = 6.3 x 10-5). What volume of this gas, measured at STP, must be dissolved in 2.5 L of solution to give that solution a pOH of 2.50?

PLEASE HELP ITS DUE IN 20 mins :(((((

Answer 1

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

`\tt [OH^-]=10^(-2.5)=0.0032=3.2* 10^(-3)`

For weak base :

`\tt [OH^-]=√(Kb.M)\\\\(3.2* 10^(-3))^2=6.3* 10^(-5)* M\\\\M=0.163`

mol = M x V

mol = 0.163 x 2.5 L

mol = 0.408

Use ideal gas law at STP(1 atm, 273 K) :

`\tt V=(nRT)/(P)=(0.408* 0.082* 273)/(1)=9.133~L`