To calculate the equilibrium constant (K) for the reaction (CH3)2NH + H2O ā (CH3)2NH2+ + OH-, we need to use the given pOH value and initial concentration of (CH3)2NH.
Let's start by converting the pOH value to pH:
pOH + pH = 14
2.58 + pH = 14
pH = 14 - 2.58
pH = 11.42
Now, we can use the pH value to find the concentration of OH-:
pOH = -log[OH-]
10^-pOH = [OH-]
10^(-2.58) = [OH-]
[OH-] = 2.51 Ć 10^-3 M
Since (CH3)2NH is a weak base, it will dissociate partially in water. Let's assume x is the concentration of (CH3)2NH that ionizes and forms (CH3)2NH2+ and OH-.
The initial concentration of (CH3)2NH is 1.59 Ć 10^-2 M. Therefore, the concentration of (CH3)2NH2+ and OH- at equilibrium will be (1.59 Ć 10^-2 - x) M.
Using the equation:
K = ([CH3)2NH2+][OH-]) / ([(CH3)2NH])
We substitute the equilibrium concentrations into the equation:
K = ([CH3)2NH2+][OH-]) / ([(CH3)2NH]) = (1.59 Ć 10^-2 - x)(2.51 Ć 10^-3) / x
Since x is small compared to 1.59 Ć 10^-2 and assuming it dissociates significantly, we can approximate the concentration of (CH3)2NH to be approximately equal to (1.59 Ć 10^-2 - x).
Now, rearrange the equation to solve for x:
K = (1.59 Ć 10^-2 - x)(2.51 Ć 10^-3) / x
Simplifying further:
K = (4.0009 Ć 10^-5 - 2.51 Ć 10^-3x) / x
At this point, you can solve the equation either by using numerical methods or by approximating the value of x to simplify the expression. Once you obtain the value of x, substitute it back into the expression to find the equilibrium constant (K). Keep in mind that this is an approximate calculation and not an exact one.
Note: If you have access to software with symbolic algebra capabilities, you can enter the equation and solve it symbolically to obtain the exact value.