Trimethylamine, (CH3)3N is a weak base (Kb = 6.3 × 10–5). What volume of this gas, measured at STP, must be dissolved in 2.5 Lof solution to give that solution a pOH of 2.50?

8.9L is the volume of the gas that must be dissolved.

Explanation:

For a weak base, we can find [(CH₃)₃N] using the equation:

Kb = [OH⁻] [[(CH₃)₃NH⁺] / [(CH₃)₃N]

As [OH⁻] = [[(CH₃)₃NH⁺] and [OH⁻] = 10^-pOH = 3.16x10⁻³M:

6.3x10⁻⁵ = [3.16x10⁻³M][3.16x10⁻³M] / [(CH₃)₃N]

[(CH₃)₃N] = 0.1587M

As the volume is 2.5L, moles are:

2.5L * (0.1587mol / L) = 0.3968moles

Using:

PV = nRT

We can solve for volume of the gas as follows:

P = 1atm at STP; n = 0.3968moles; R = 0.082atmL/molK; T = 273.15K at STP

V = 0.3968mol*0.082atmL/molK*273.15K/1atm

V = 8.9L is the volume of the gas that must be dissolved.