What must be the molarity of an aqueous solution of trimethylamine, (CH3)3N, if it has a pH = 11.20?

(CH3)3N(aq)+H2O(l) double arrow (CH3)3NH+(aq)+OH-(aq)

Kb value = 6.3x10^-5 mol/L

I worked this problem for Allan below.

http://www.jiskha.com/display.cgi?id=1317528123

Find the inverse of the matrices, if they exist. Use the algorithm introduced in this section.

1 2 -1
-4 -7 3
-2 -6 4

To determine the molarity of an aqueous solution of trimethylamine (CH3)3N, you can utilize the pH and the Kb value of the compound.

First, let's establish that (CH3)3N is a weak base since it reacts with water to form hydroxide ions (OH-) and its conjugate acid (CH3)3NH+.

The Kb value represents the base dissociation constant, which can be used to find the concentration of hydroxide ions produced when the weak base is dissolved in water.

The Kb expression for the reaction (CH3)3N(aq) + H2O(l) ⇌ (CH3)3NH+(aq) + OH-(aq) is:

Kb = [OH-] * [CH3)3NH+] / [(CH3)3N]

Given that the Kb value is 6.3x10^-5 mol/L, we can use this equation to calculate [OH-].

To find [OH-], we need to know [CH3)3NH+] since the concentration of (CH3)3NH+ is the same as the concentration of OH- due to the stoichiometry of the equation.

The pH of the solution is given as 11.20. The pH is the negative logarithm (base 10) of the hydrogen ion concentration [H+]. Therefore, we can calculate [H+] as follows:

pH = -log [H+]

11.20 = -log [H+]

[H+] = 10^(-pH) = 10^(-11.20)

Now, since [H+] = [CH3)3NH+], we have:

[CH3)3NH+] = 10^(-11.20)

Next, using the Kb expression and the known value for [CH3)3NH+], we can solve for [OH-]:

6.3x10^-5 = [OH-] * 10^(-11.20) / [(CH3)3N]

At this point, we need to solve for [OH-]. To do so, we require the concentration of (CH3)3N, which is the molarity we are trying to find.

Let's assign "x" as the molarity of (CH3)3N. Consequently, [OH-] = x.

6.3x10^-5 = x * 10^(-11.20) / x

By canceling out the x on both sides of the equation, we get:

6.3x10^-5 = 10^(-11.20)

To solve for x, take the logarithm of both sides (base 10):

log(6.3x10^-5) = -11.20

x = 10^(-11.20 + log(6.3x10^-5))

Calculating the right-hand side of the equation will give you the final answer for the molarity of (CH3)3N.