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Chemistry: Buffer Solutions 
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#1
Oct610, 06:58 PM

P: 7

1. The problem statement, all variables and given/known data
What is the pH of a carbonate buffer solution prepared by mixing 1.500 mol Na2CO3 and 1.000 mol NaHCO3 and adding water to make a 1.000 L solution? (pKa of (HCO3) is 10.32) A 200.00 mL sample of 1.000 M nitric acid was added to the buffered solution above. What is the resulting pH? 2. Relevant equations HendersonHasselbalch equation (pH = pKa + log ([A]/[HA])) pH = log [H+] 3. The attempt at a solution I used the equation for the first part, and I calculated a pH of 10.50 (pH = 10.32 + log (1.500/1.000)). Is this correct? I'm not sure how to include the second part though... I'm guessing I'm supposed to calculate the resulting pH when nitric acid is added to 1.000 L of water and then somehow use that pH for the second part... If that makes any sense. Any help is appreciated! 


#2
Oct710, 02:10 AM

Admin
P: 23,578

10.50 looks OK, although you added, not multiplied (I guess that's just a typo).
What reaction takes place in the solution when you add strong acid? How does it change amounts of CO_{3}^{2} and HCO_{3}^{}? 


#3
Oct710, 01:49 PM

P: 7

Oops, that is a typo =P Apologies!
So if I add HNO_{3}, it would dissociate, creating more H_{3}O^{+}. And the addition of hydronium ions would shift the equilibrium to increase the concentration of HCO_{3}^{}, which decreases the concentration of CO_{3}^{2}. Is that right? 


#4
Oct710, 01:58 PM

Admin
P: 23,578

Chemistry: Buffer Solutions
You are on the right track. Assume carbonate protonation went to completion.



#5
Oct710, 02:04 PM

P: 7

Does that mean there wouldn't be any carbonate ions left in the solution?



#7
Oct710, 06:11 PM

P: 7

So this is what I ended up doing: 0.200 L of 1.000 M nitric acid was added, which means that 0.200 moles of nitric acid was added. Since it's a strong acid, it fully dissociates, meaning that there are 0.200 additional moles of hydrogen ions in the solution. Since there are more protons in the solution, the equilibrium shifts to create more HCO_{3}^{}, meaning that CO_{3}^{2} are used. Using the ICE method, I figured that the concentration of HCO_{3}^{} is 1.000 M and that of CO_{3}^{2} is 1.083 M. Then I used the HH equation and got this: pH = 10.32 + log (1.083 M / 1.000 M) = 10.35. Does that look right?



#8
Oct810, 02:34 AM

Admin
P: 23,578

No. As told  use just the stoichiometry of the reaction.
Simple check  you started with total concentration of carbonate and bicarbonate being 2.5M, now you are down to 2.083  so there must be something wrong, carbon can't just disappear from the solution (unless it bubbles out as CO_{2}, but you are not protonating bicarbonate, so this is not the case). 


#9
Oct810, 09:45 AM

P: 7

I'm not sure I'm understanding correctly  would the correct concentrations be 1.2 M and 1.3 M?



#10
Oct810, 11:13 AM

Admin
P: 23,578

Yes.
Assuming you assigned correct concentration to correct form. 


#11
Oct810, 11:47 AM

P: 7

Oh, I see! Thank you =)



#12
Oct810, 09:29 PM

P: 7

Just one more question: so the volume doesn't change even after adding 0.200 L of nitric acid? Wouldn't it be different?



#13
Oct910, 04:37 AM

Admin
P: 23,578

Changes  but it doesn't matter. You see, in the HendersonHasselbalch equation under log there is a ratio of two concentrations. If you add water, both concentrations change in the same way, so their ratio stays constant. In fact, it can be easily shown that volume cancels out, and it is just ratio of number of moles of acid and conjugate base that counts.
Beware: there are some traps here. Please read the page about HH equation that I linked to to find out what is the problem. 


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