Calculate the pH of a 0.063 M (C2H5)2NH solution (Kb = 1.3 10-3).
Here is a link where I worked it for someone. I think the concn is not the same but the procedure is the same.
http://www.jiskha.com/display.cgi?id=1322093470
To calculate the pH of a solution, we need to determine the concentration of hydroxide ions (OH-) in the solution. In this case, we are given the concentration of the compound (C2H5)2NH, which is a weak base. To find the concentration of OH-, we will use the equation for the base dissociation constant (Kb):
Kb = [OH-]^2 / [C2H5)2NH]
The + signs have been omitted for simplicity.
Given that Kb = 1.3 x 10^-3 and the concentration of (C2H5)2NH = 0.063 M, we can rearrange the equation to solve for [OH-].
[OH-]^2 = Kb x [C2H5)2NH]
[OH-]^2 = (1.3 x 10^-3) x (0.063)
[OH-]^2 = 8.19 x 10^-5
Taking the square root of both sides:
[OH-] = √(8.19 x 10^-5)
= 0.00904 M
Now that we have the concentration of OH-, we can calculate the pOH using the formula:
pOH = -log[OH-]
= -log(0.00904)
= 2.04
The pH of a solution can be calculated using the following equation:
pH = 14 - pOH
= 14 - 2.04
= 11.96
Therefore, the pH of a 0.063 M (C2H5)2NH solution is approximately 11.96.