### K_{b} : Solving Weak Base Problems: Part Two

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Important note: all constants refered to: K_{c}, K_{w}, K_{a}, and K_{b} are temperature-dependant. All discussions are assumed to be at 25 °C, i.e. standard temperature.

Here's a second typical weak base problem:

What is the pH of a 0.300 M solution of morphine? K_{b} = 1.62 x 10¯^{6}

These are the important equations:

Mor + H_{2}O ⇌ MorH^{+} + OH¯

K_{b} = ( [MorH^{+}] [OH¯] ) / [Mor] = 1.62 x 10¯^{6}

Mor refers to the morphine molecule and MorH^{+} refers to the molecule after accepting a proton. It is completely unimportant what its formula is.

As before, we want the [OH¯]. So we have:

[OH¯] = x
and

[MorH^{+}] = x

This is because of the one-to-on molar ratio between [OH¯] and [MorH^{+}] that is created as Mor molecules react with the water.

Remember, the [Mor] started at 0.300 M and went down as Mor molecules reacted. In fact, due to the one-to-one ratio, it went down by 'x' amount and wound up at an ending value of 0.300 - x.

Next is our 'dropping the subtract x' trick:

0.300 - x approximately equals 0.300

We will check the validity of the trick after completing the calculation. If the approximation exceeds 5%, then we have to use the quadratic.

We now have our equation:

1.62 x 10¯^{6} = {(x) (x)} / 0.300

which is very easy to solve.
We move the 0.300 to the other side to get:

x^{2} = 4.86 x 10¯^{7}

Taking the square root (of both sides!!), we get:

x = 6.97 x 10¯^{4} M

Remember:

1) The K_{b} value is unitless, but x is a molarity.

2) Square root both sides. I have had students square root the x^{2}, but not the other side. Weird, but true.

We take the pOH to get 3.157. Converting to pH (remember pH + pOH = 14), we get a pH = 10.843

Checking for 5% we find we have 0.23%, which is acceptable.

Return to the Acid Base menu

Go to a short intro to what K_{b} is

Go to Solving K_{b} Problems: Part One

Go to Solving K_{b} Problems: Part Three

Return to a listing of many types of acid base problems and their solutions