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Calculate the pH of a buffer that is 0.250M in NaHCO3 and 0.300M in Na2CO3Asked by Michael
Calculate the pH of a buffer that is 0.130M in NaHCO3 and 0.295M in Na2CO3
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Answered by
Katlynne
Okay so first you need the dissociation constant for carbonic acid (HCO3) which is 5.6e-11. Then set up an ice table. Keep in mind that you are adding an acid to an aqueous solution of sodium bicarbonate (Na2CO3) which means you will have some CO3 ions already in the solution, preventing the carbonic acid fro dissociating entirely (which is why this a buffer problem).
NaHCO3 <-> H+ + CO3
I…….. .130 ………… 0 ……. .295
C……… -x …………. +x ……. +x
E……… .130-x …….. x …….. .295+x
Then set up your equilibrium constant
Ka=[products]/[reactants]= [.295+x][x]/[.130-x]
We can make the assumption that x will be so small that it will not really change anything so our equation becomes,
(5.6*10^-11)= [.295][x]/[.130] Then solve for x and you should get 2.47e-12, which is the concentration of H+ which you can then use to determine the pH.
pH=-log([H+])= -log(2.47e-12)= 10.61
Hope this helped!
NaHCO3 <-> H+ + CO3
I…….. .130 ………… 0 ……. .295
C……… -x …………. +x ……. +x
E……… .130-x …….. x …….. .295+x
Then set up your equilibrium constant
Ka=[products]/[reactants]= [.295+x][x]/[.130-x]
We can make the assumption that x will be so small that it will not really change anything so our equation becomes,
(5.6*10^-11)= [.295][x]/[.130] Then solve for x and you should get 2.47e-12, which is the concentration of H+ which you can then use to determine the pH.
pH=-log([H+])= -log(2.47e-12)= 10.61
Hope this helped!
Answered by
DrBob222
Also you can use the Henderson-Hasselbalch equation which is a little easier to handle.
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