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Topic: Buffer Solutions  (Read 6623 times)

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Offline Marry

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Buffer Solutions
« on: April 23, 2011, 07:09:39 PM »
How many grams of Na2CO3 to be mixed with 5.00 g of NaHCO3 to
produce 100 mL buffer with pH = 10.00?

Offline Marry

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Re: Buffer Solutions
« Reply #1 on: April 24, 2011, 03:32:04 PM »
H2CO3 + H2O -----> H3O(+) + HCO3(-) Keq1º = 4,45x10^-7

HCO3(-) + H2O -----> H3O(+) + CO3(2-) Keq2º = 4,84x10^-11

[H3O+]² = Ka (Ca - [H3O+]) / Cb + [H3O+]

This is the equation that I use?

As they step 5 grams to mol / L?

Offline Borek

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Re: Buffer Solutions
« Reply #2 on: April 24, 2011, 05:02:44 PM »
No, you need to use Henderson-Hasselbalch equation.
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Offline Marry

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Re: Buffer Solutions
« Reply #3 on: April 24, 2011, 07:40:43 PM »
H2CO3 + H2O -----> H3O(+) + HCO3(-) Keq1º = 4,45x10^-7

HCO3(-) + H2O -----> H3O(+) + CO3(2-) Keq2º = 4,84x10^-11


10,00 =  10.31 + ( log [CO3(-)] / log [HCO3(-)]

?


Offline Borek

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Re: Buffer Solutions
« Reply #4 on: April 25, 2011, 08:00:11 AM »
10,00 =  10.31 + ( log [CO3(-)] / log [HCO3(-)]

Not bad, although you added a log into the equation.

Now it should be a matter of simple concentration calculations. [HCO3-] is given (not directly, but you shouldn't haver any problems calculating it).
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Offline Marry

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Re: Buffer Solutions
« Reply #5 on: April 26, 2011, 10:04:19 PM »
10,00 =  10.31 + ( log [CO3(-)] / log [HCO3(-)]

Not bad, although you added a log into the equation.

Now it should be a matter of simple concentration calculations. [HCO3-] is given (not directly, but you shouldn't haver any problems calculating it).


H2CO3 + H2O -----> H3O(+) + HCO3(-) Keq1º = 4,45x10^-7

HCO3(-) + H2O -----> H3O(+) + CO3(2-) Keq2º = 4,84x10^-11


10,00 =  10.31 + ( log [CO3(-)] / [HCO3(-)]

Log [CO3(2-)] - Log [HCO3(-)] = 0,31
------------------------------------

HCO3 = 1+ 12+48 = 61g

1 mol-----> 61g
x---------> 5g

(12,2 mol)-???
--------------------------------
Log [CO3(2-)] - Log [12,2] = - 0,31

Log [CO3(2-)]= - 0,31 +1,08

Log [CO3(2-)] = 0,77


10^0,77= 5,88mol/L


5,88:10 = 0,58 mol


CO3---> 60g/L


1 mol --------------60
0,58----------------x

R- 34,8g????







« Last Edit: April 26, 2011, 10:16:27 PM by Marry »

Offline AWK

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Re: Buffer Solutions
« Reply #6 on: April 27, 2011, 02:23:17 AM »
Quote
Log [CO3(2-)] - Log [HCO3(-)] = 0,31
Something is missing on the right side
Note - mass of NaHCO3 is given and you have to calculate mass of Na2CO3

Now convert ratio of concentrations into ratio of masses and take into account that one of them is 5.00 g.
AWK

Offline Borek

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Re: Buffer Solutions
« Reply #7 on: April 27, 2011, 02:44:24 AM »
HCO3 = 1+ 12+48 = 61g

1 mol-----> 61g
x---------> 5g

(12,2 mol)-???

There is something wrong with the math here. Looks like you are using a correct approach, but it gets twisted on the way. Besides, number of moles is NOT YET a concentration.

Quote
Log [CO3(2-)] - Log [12,2] = - 0,31

This is different from what you wrote above - sign on the right is different. Could be it was just a typo.
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