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Offline Peppersrule

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Interesting pH question
« on: October 06, 2013, 05:11:35 PM »
Captain Kirk, of the Starship Enterprise, has been told by his superiors that only a chemist can be trusted with the combination to the safe containing the dilithium crystals that power the ship. The combination is the pH of solution A described below, followed by the pH of solution C. (Example: If the pH of sol A is 3.47 and that of sol C is 8.15, then the combination to the safe is 3-47-8-15.) the chemist must determine using only information below (all solutions are at 25 degree Celsius).

Solution A is 50.0Ml of a 0.100 M solution of the weak monoprotic acid HX.

Solution B is a 0.0500 M solution of the salt NaX. It has a pH of 10.02.

Solution C is made by adding 15.0 mL of 0.250 M KOH to solution A.

What is the combination to the safe?

If in preparing solution C, 15.0 mL of water was added instead of the 15.0 mL of KOH, what effect would this have on the combination to the safe?


Can I see your calculations for this anyone? I want to compare. Thanks a mil.

Offline Borek

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Re: Interesting pH question
« Reply #1 on: October 06, 2013, 05:27:16 PM »
You have to show your attempts at solving the question to receive help. This is a forum policy.

So, give your combination, tell us how you got it, and we will see if it opens anything.
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Offline Peppersrule

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Re: Interesting pH question
« Reply #2 on: October 06, 2013, 06:26:02 PM »
What I have so far

solution B

NaX --> Na+ + X-
X- + H2O --> HX + OH-
pH = 10.02
so pOH = 14 - 10.02 = 3.98
[OH-] = 10^-3.98 = 0.000105 M
K = ( 0.000105)^2 / 0.0500 - 0.000105 = 2.21 x 10^-7
Ka = 10^-14 / 2.21 x 10^-7 =4.52 x 10^-8

Solution A
HX --> H+ + X-
4.52 x 10^-8 = x^2 / 0.100 -x
[H+] = x = 0.0000673 M, pH = 4.17.

Solution C
n HX = 0.0500 L x 0.100 = 0.00500 moles
n KOH = 0.0150 L x 0.250 M = 0.00375 moles
HX + OH- --> X- + H2O
n HX = 0.00500 - 0.00375 = 0.00125 moles
moles X- = 0.00375
volume = 0.065 L (total)
[HX] = 0.00125 / 0.065 = 0.0192 M
[X-] = 0.00375 / 0.065 = 0.0577 M
pKa = - log 4.52 x 10^-8 = 7.34,     pH = 7.34 + log 0.0577 / 0.0192 = 7.82

Is this logical?

Offline Borek

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Re: Interesting pH question
« Reply #3 on: October 07, 2013, 03:08:30 AM »
Logic looks OK to me. Don't round down intermediate results, do the calculations using full available precision (or at least use some guard digits). In this case it probably doesn't matter, but in other cases you can be nastily surprised.
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