Estimate the equilibrium constant for the weak base (CH3)2NH, if a 1.59×10-2 M aqueous solution of (CH3)2NH has a pOH 2.58 (make an exact calculation assuming that initial concentration is not equal to the equilibrium concentration).

(CH3)2NH + H2O = (CH3)2NH2+ + OH-

I know how to calculate it if initial concentraton = equilibrium concentration...but how are you suppose to calculate it if they aren't?

I've not seen a problem like this and I'm not sure exactly what it means; however, I think I would try

Keq = [(CH3)2NH2^+)][OH^-]/[(CH3)2NH)]
and plug in (0.00263)^2 for the numerator. For the denominator I would substitute 0.0159 and NOT subtract the OH^- from it (which it would have at equilibrium). However, I don't see that as an exact calculation and I don't understand talk of an exact calculation when the problem starts by saying estimate.

Does the .00263 come from antilog (-2.58)?

If so, I plugged it into my calculator and got .00275.
That was the method I initially used (.00275^2)/.0159 and I got 4.35 x 10^-4
but that answer was wrong.
The answer is suppose to be 5.21 x 10^-4.
Thank you for trying though. =]

OK. Yes, that's how I obtained 0.00263 but I just checked and that's the antilog of -2.58 so I suggest you check your calculator again. If you get a different number, perhaps you should look into a new calculator. ;-). The 5.21 number you get by

(0.00263)^2/(0.0159-0.00263) = ?? which is the exact way of doing things; but that is what it is at equilibrium. I'm still confused by the wording of the problem.

Ohh. I just tried a similar problem and used the equilibrium way and it was right...

Thank you so much for your help!

To calculate the equilibrium constant (K) for the reaction (CH3)2NH + H2O ⇌ (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 × 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 × 10^-2 M. Therefore, the concentration of (CH3)2NH2+ and OH- at equilibrium will be (1.59 × 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 × 10^-2 - x)(2.51 × 10^-3) / x

Since x is small compared to 1.59 × 10^-2 and assuming it dissociates significantly, we can approximate the concentration of (CH3)2NH to be approximately equal to (1.59 × 10^-2 - x).

Now, rearrange the equation to solve for x:

K = (1.59 × 10^-2 - x)(2.51 × 10^-3) / x

Simplifying further:

K = (4.0009 × 10^-5 - 2.51 × 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.