Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: vort3x on February 21, 2009, 12:29:20 PM
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Hi,
I have a bit of a silly question but just wanted to check. I have to write an equation for when this substance acts as an acid with water (therefore donating a proton), but does it matter whether I decrease H2 or H7 in the forward reaction? I'm pretty sure it doesn't since both are counted towards the overall hydrogen count.
H2C6H7O5(aq) + H2O(l) ::equil:: HC6H7O5-(aq) + H3O+(aq)
Thanks!
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Nope if you notice why the acid is not written as C6H9O5 instead because what they are telling you it is an diprotic acid, so what you did is correct - you should not decrease H7 instead.
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Another question. Now I'm calculating the pH of strong acids (-log of molarity). Which is fine, but in this question it asks,
pH of: 5.00ml of 1.00M HCl is diluted to 0.500L.
So using the dilution equation (M1V1 = M2V2), I got a final molarity of 0.01M and therefore the pH is 2. Can anyone confirm? (just want to make sure I'm approaching these right).
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OK
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Ok, thanks.
Next question:
Calculate pH and [OH-] of solution:
10.0ml of 0.015M Ba(OH)2 with 40.0ml of 7.5x10-3M NaOH.
I just did a bunch of similar questions where I was given just molarity, but not sure how to approach it with that solution, to get a "final" molarity to calculate pH from?
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What is the total moles of OH- and what is the volume of the final solution?
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The total moles of OH- are 2, and the final volume is 50.0 ml = 0.05L. Still not sure.
Do I re-find molarity for each but just using the new volume? Ex:
Moles of Ba(OH)2 / 0.05L
Moles of NaOH / 0.05L
Then add the two molarities for my final molarity?
... wait I guess I couldn't do that since I don't have a mass, hmmm.
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Sum moles of OH- and divide by the final volume.
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So 3 mol of OH / 0.05L = 60 M is the molarity I need to proceed with calculating pH and OH-?
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On organic compounds when ever you write hydrogens on the inside you are telling your reader that those hydrogens are part of the hydrocarbon or other organic groups and not acids. These hydrogens are most inert and even the ones who do reach are often weak.
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Ok... I don't see how that helps me? My question is:
So 3 mol of OH / 0.05L = 60 M is the molarity I need to proceed with calculating pH and OH-? I don't think that's right, since the -log of 60M to find pOH gives me a negative number.
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That was a response to your first post. I didn't see your subsequent question.
For the problem
Calculate pH and [OH-] of solution:
10.0ml of 0.015M Ba(OH)2 with 40.0ml of 7.5x10-3M NaOH.
Your mol calculations seem a little off. 3 is way too much when you are dealing with mmol quantities of solution. Go back and try that again. Remember
mols = (molarity)*(liters of solution)
And in the case of Ba(OH)2 there are two moles of OH for every one of Ba(OH)2 dissolved. Do that for each of solutions and then add them up. I think you will get a more accurate answer.
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So I found moles by multiplying the total molarity and total volume given, together. I then divided that by 0.05L... I got 9 x 10-7 M.
Is that right? Find moles by multiplying total molarity x volume given, and divide that by 0.05L (total volume).
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You are getting closer, but still a few mistakes. First do not forget that it is a 0.015 M solution of Ba(OH)2 which means that it is really a 0.030M solution of OH. Second I think you are messing up on unit conversions because even with the error stated above I only get 9*10^-3.
You are making progress, keep trying. Try to type out all of your calculations so I can get a better sense of what you are doing wrong.
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So, 0.030 M BaOH + 7.5x10-3M NaOH = 0.0375 M
0.0375 M / 0.05 L = 0.75 M?
I'm using the initial molarities though of not just the hydroxides but also Ba and Na... so that can't be right for a final molarity of that solution if I'm just looking for OH?
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Just look at the units. If you divide mol/L by L doesn't that make it mol/L2
The problems asks for the OH concentration and the subsequent pH. Since it does not ask for activity coefficients the sodium and the barium concentrations are irrelevant. You only care about how much hydroxide is in solution. So for Ba(OH)2 there are 0.030 mols per liter of OH and for NaOH there is 7.5x10-3 moles per liter. You added 10mL of Ba(OH)2 so that means there are 0.030M*(10/1000)L moles of hydroxide from that source and (7.5x10-3)M*(40/1000)L moles of hydroxide from the NaOH source. So when you have the total number of OH moles you then divide by the new volume to find out the concentration. That gives you the first answer and the second is rather easy to find out as you can obtain the pOH without much hassle and then you receive the pH by subtracting the pOH from 14.
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Okay, so this is what I did. How does it look? I found the final molarity, pH, and OH- (by finding H+ and re-arranging Kw).
0.03 mol/L * 0.01 L = 0.0003 mol
7.5 x 10-3 mol/L * 0.04 L = 0.0003 mol
+ = 0.0006 mol
0.0006mol / 0.05 L = 0.012 mol/L
pOH = -log(0.012mol/L) = 1.92
pH = 14.00 - 1.92 = 12.08
[H+] = 10-12.08 = 8.32 * 10-13
[OH-] = 1.00 * 10-14 / 8.32 * 10-13 = 1.2 * 10-2
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Yep that looks very close to the answer I got so I would say you have arrived at a reasonable answer. Numerically it makes sense that a dilute solution of hydroxide would have a pH of around 10 to 12.
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What answer did you get, do we have the same numbers? Just want to make sure I'm rounding everything properly too.
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What answer did you get, do we have the same numbers? Just want to make sure I'm rounding everything properly too.
They look identical. I assume you are taking a first year college or high school Chemistry course so there should be no corrections you have to apply.