Acid-base equilibria
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Tried solving but got a suspiciously small number, i.e. 0.00203cm3
The carbonate ion has a Kb of 1.8 x 10^4 moldm3. Find the volume of 0.10moldm3 of sodium hydrogen carbonate that is needed to be added to 25cm3 of 0.12cm3 of Na2CO3 to form a buffer solution of 9.5 pH.
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Originally posted by donkhead333:
Tried solving but got a suspiciously small number, i.e. 0.00203cm3
The carbonate ion has a Kb of 1.8 x 10^4 moldm3. Find the volume of 0.10moldm3 of sodium hydrogen carbonate that is needed to be added to 25cm3 of 0.12cm3 of Na2CO3 to form a buffer solution of 9.5 pH.
You've an error in the question, specifically the Kb value of the carbonate ion. It should be approx 2x10^-4, not x10^4.
Just last month, I set a very similar question for my students, which I included in my archive of H2 Chem qns here :
http://www.sgforums.com/forums/2297/topics/392932?page=3
To locate the qn quickly, do a Control-F search for "9.5" which is coincidentally the same desired pH of the solution, for both my question and your question.You can also check your answer by rephrasing the question :
"The carbonate ion has a Kb of 1.8 x 10^-4 moldm3. Find the pH of the buffer solution generated when 0.00203cm3 volume of 0.10moldm3 of sodium hydrogen carbonate is added to 25cm3 of 0.12mol/dm3 of Na2CO3."
I agree with you that your "suspiciously small number" answer is most probably wrong. Try redoing it, using my question & solution as a guide.
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Given 150 cm3 of 2.0 mol/dm3 of ammonia solution (pKa of NH4+ = 9.25) at r.t.p., what volume of 0.5 mol/dm3 of hydrochloric acid is needed to prepare a buffer solution with pH = 9.5?
Solution (pun intended) :
Let x be the volume of HCl added to achieve a pH of 9.5
NH3 + H+ ---> NH4+
I (mols) 0.3 | 0.5x | 0
C (mols) -0.5x | -0.5x | +0.5x
F (mols) 0.3 - 0.5x | 0 | 0.5x[NH3] = (0.3 - 0.5x) / (0.15 + x)
[NH4+] = (0.5x) / (0.15 + x)
At pH 9.5, pOH = 4.5 and [OH-] = 10^-4.5Hydrolysis of NH3 : NH3 + H2O <---> NH4+ + OH-
Kb = ( [NH4+] [OH-] ) / [NH3]
1.7783 x 10^-5 = [(0.5x) / (0.15 + x)] [10^-4.5] / [(0.3 - 0.5x) / (0.15 + x)]
x = 0.21596 dm3 = 216 cm3 (to 3 sf)--------------------------------------------------------------------------------------------------------------
Hi Ultima, I looked through the example - but I had a problem with understanding how 0.5x for NH4+ is obtained. Is is because all the H+ ions from HCl will react?
And with regard to my question,
"The carbonate ion has a Kb of 1.8 x 10^-4 moldm3. Find the volume of 0.10moldm3 of sodium hydrogen carbonate that needs to be added to 25cm3 of 0.12mol/dm3 of Na2CO3 to form a buffer solution of pH 9.5"
Does CO32- act as a weak base? So how should I craft my equation?
Thanks
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Originally posted by donkhead333:
Given 150 cm3 of 2.0 mol/dm3 of ammonia solution (pKa of NH4+ = 9.25) at r.t.p., what volume of 0.5 mol/dm3 of hydrochloric acid is needed to prepare a buffer solution with pH = 9.5?
Solution (pun intended) :
Let x be the volume of HCl added to achieve a pH of 9.5
NH3 + H+ ---> NH4+
I (mols) 0.3 | 0.5x | 0
C (mols) -0.5x | -0.5x | +0.5x
F (mols) 0.3 - 0.5x | 0 | 0.5x[NH3] = (0.3 - 0.5x) / (0.15 + x)
[NH4+] = (0.5x) / (0.15 + x)
At pH 9.5, pOH = 4.5 and [OH-] = 10^-4.5Hydrolysis of NH3 : NH3 + H2O <---> NH4+ + OH-
Kb = ( [NH4+] [OH-] ) / [NH3]
1.7783 x 10^-5 = [(0.5x) / (0.15 + x)] [10^-4.5] / [(0.3 - 0.5x) / (0.15 + x)]
x = 0.21596 dm3 = 216 cm3 (to 3 sf)--------------------------------------------------------------------------------------------------------------
Hi Ultima, I looked through the example - but I had a problem with understanding how 0.5x for NH4+ is obtained. Is is because all the H+ ions from HCl will react?
And with regard to my question,
"The carbonate ion has a Kb of 1.8 x 10^-4 moldm3. Find the volume of 0.10moldm3 of sodium hydrogen carbonate that needs to be added to 25cm3 of 0.12mol/dm3 of Na2CO3 to form a buffer solution of pH 9.5"
Does CO32- act as a weak base? So how should I craft my equation?
Thanks
Yes and yes.
Since this is a buffer, you can use either Kb of CO3 2- or Ka or HCO3 -, in the final part of your working to solve for x (where x be the volume of HCO3 - added to achieve a pH of 9.5).