Chemistry
posted by Dan
Only a chemist can be trusted with the combination to the safe containing a ton of money. The combination is the pH of solution A, followed by the pH of solution C. (for example: is the pH of solution A is 3.47 and the pH of solution C is 8.15 the combination to the safe is 347815) The chemist must determine the combination using only the information below (all solutions are at 25 degrees C)
Solution A is 50.00 mL of a 0.100 M solution of the weak monoprotic acid HX.
Solution B is a 0.0500 M solution of a salt NaX. It has a pH of 10.02.
Solution C is made by adding 15.00 mL of .250 M KOH to solution A.
For solution B. 0.05 M NaX has pH = 10.02.
X^ + HOH ==> HX + OH^
Kb = (HX)(OH^)/(X^) = Kw/Ka
pH = log(H^+) = 10.02
(H^+) = 9.55E11
(OH^) = 1.047E4
(HX) = 1.047E4
(OH^) = 1.047E4
(X^) = 0.05  1.047E4 = 0.0499 (or 0.05)
[(1.047E4)^2/0.05]= Kw/Ka
Kw = 1E=14
Ka = ??
I understand all of this but for some reason I am not getting the same Ka value you got of Ka = 4.56E8. I am using (1.047x10^4)^2 / .05 = (1.0x10^14) / Ka
Sorry I'm a little slow. I had to eat supper and do my 2 mile walk. I worked it again and came up with 4.56E8 again. If you don't find your error, post your work and I'll nail it for you.
No, problem
(1.047x10^4)^2 = 1.096478196x10^8
1.096478196x10^8 / .05 = 2.192956392x10^7
2.192956392x10^7 / 1.0x10^14 = 21929563.92
HELP!!
No, problem
(1.047x10^4)^2 = 1.096478196x10^8
This step is ok.
1.096478196x10^8 / .05 = 2.192956392x10^7 This step is ok but the next step is not. You have reversed the numbers.
2.19 x 10^7 = Kw/Ka
2.19 x 10^7 = 1x10^14/Ka
Ka = 1x10^14/2.19x10^7
Ka = 4.56E8
2.192956392x10^7 / 1.0x10^14 = 21929563.92
HELP!!
I have one final question. Where did the numbers for solution C come from? I am a bit confused. Is there another way to do it?
Step 3. Go to solution C.
You may use the HH equation here.
pH = pKa + log (base)/(acid)
pH = 7.34 = (3.75/V)/(1.25/V).
should be pH = 7.34 + (3.75/V/1.25/V) = 7.82. I typed = instead of +.
pH = 7.82.
HX + KOH ==> KX + HOH
mols HX = 0.05 L x 0.1 M = 0.005 mols
(I used 5 millimols I believe)
mols KOH = 0.015 L x 0.25 M = 0.00375 mols.
(I think I used 3.75 millimols).
So 0.00375 mols KX is formed, all of the KOH is used, and the HX remaining is 0.005  0.00375 = 0.00125 (and I used 1.25 millimols for this).
The volume is 15 mL + 50 mL = 65 mL or 0.065 L.
Technically, (HX) = 0.00125 mols/0.065 L but I saved myself some work and just wrote 0.00125/V because I know the V will cancel (actually I wrote 1.25 because I used millimols).
Then (X^) = 0.00375 mols/0.065 L or 0.00375 mols/V (and I used 3.75 millimols here).
If Ka = 4.56E8, then pKa = 7.34
The HH equation is
pH = pKa + log (base)/(acid)
pH = 7.34 + log (0.00375/V)/(0.00125/V)
pH = 7.34 + log 3.00. (You see why I didn't stick the 65 mL in because the V appears both places and will cancel. I also used millimols of 3.75/1.25 because that's easier than all those zeros). And 3.75/1.25 is the same as 0.00375/0.00125 anyway.
pH = 7.34 + log 3.00
pH = 7.34 + 0.477 = 7.818 which I rounded to 7.82.
Yes, there is another way to do it and not use the HH equation. Just use the Ka expression.
Ka = (H^+)(X^)/(HX)
H+ = solve for this.
X = 0.00125 mols/V
HX = 0.00375 mols/V
(H^+) = Ka*(HX)/(X^)
(H^+) = 4.56E8*(0.00125/0.00375)
(H^+) = 1.52E8
pH = log 1.52E8
pH = 7.82
THANK YOU SO MUCH! YOU ARE MY LIFE SAVER!
Let me say that I made a mistake. The mistake I made was not asking you for Ka when you said you had it and asked what to do next. By the time we finished that session I was confused, too, about which number I had used where. If I had asked for Ka, we would have taken care of that problem RIGHT THEN, instead of waiting until we were near the end. By the time I found Ka was not right we had talked about hydrolysis, Ka, Kb, Kw/Ka, monoprotic weak acids and the HH equation, and it just got to be too much. Although I didn't give you any false directions, I had so many numbers running around in my head that I was tearing my hair out. Anyhway, I hope I helped you. Thanks for using Jiskha, and come again.

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pH=1/2pKa1/2(log (C)
first solution pH should be 4.17
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