A Combination Lock Has Been Made That Will Require a Chemist to Crack

Acid/Base problem

A combination lock has been made that will require a chemist to crack.

The combination is the pH of solution A followed by the pH of solution C

Example: If pH of A is 3.47 and C is 8.15, then the combination is 3-47-8-15.

Solution A is 50 ml of a 0.001 M solution of a weak monoprotic acid (HX)

Solution B is a 0.05 M solution of salt, NaX. It has a pH of 10.02

Solution C is made by adding 15 ml of 0.250 KOH (strong base) to solution A

What is the combination to the safe? ______

Solution : as near as I can figure

What is the pH of solution A?

Need to determine the Ka of HX Ka = [H+] [X-] / [HX] Remember that pKa + pKb =14

If we assume complete dissociation of the salt and we know the resulting solution is basic , then

Na+ + X- ßà HX + OH- The Kb of this reaction is [OH-] [HX]/ X-

We assume that the OH- = the HX concentration, then the problem looks like

B2/ X- = [OH-] 2 / [1M] =

The pOH = 3.98 so [OH] = 1.05 x 10-4

Kb = [1.05 x 10-4] 2 / [1M] = 1.1 x 10-8

Thus pKb is 7.96 pKa must be 6.04 Ka = 9.12 x 10-7

So the pH of 0.001 M HX is Ka = [H+] [X-] / [HX]

9.12 x 10-7 = X2/[0.001]

X2 = 9.12 x 10-10 X= 3.02 x 10-5 pH = 4.52

Now to calculate solution C

0.015 liters x 0.250 moles/liter KOH = 0.00375 moles KOH in a volume of 0.065 l L = 0.058 M OH-

This is more than enough OH- to eat up all the weak acid in A (the OH will convert all available H+ to water, as the free H+ gets eaten up, the equilibrium will dissociate more from the complexed weak acid. If the amount of OH exceeds the concentration of complexed acid (HX) then the equilibrium will be fully titrated

0.05 liters x 0.001 M = 0.00005 mole HX /0.065 L = [.000769] M new HX concentration (initial)

The concentration of OH will be .058 OH - -.00769 HX = .00503 M leftover free OH-

pOH = 2.298 thus pH = 11.7

The combination to the lock is 4-52-11-7