Go to a short intro to what Kb is
Go to Solving Kb Problems: Part One
Go to Solving Kb Problems: Part Three
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Important note: all constants refered to: Kc, Kw, Ka, and Kb 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? Kb = 1.62 x 10¯6
These are the important equations:
Mor + H2O <==> MorH+ + OH¯
Kb = ( [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¯] = xand
[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:
x2 = 4.86 x 10¯7
Taking the square root (of both sides!!), we get:
x = 6.97 x 10¯4 M
Remember:
1) The Kb value is unitless, but x is a molarity.
2) Square root both sides. I have had students square root the x2, 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.
Go to a short intro to what Kb is
Go to Solving Kb Problems: Part One
Go to Solving Kb Problems: Part Three
Return to a listing of many types of acid base problems and their solutions