# Chemical Forums

## Chemistry Forums for Students => Analytical Chemistry Forum => Topic started by: Marry on April 23, 2011, 07:09:39 PM

Title: Buffer Solutions
Post by: Marry 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?
Title: Re: Buffer Solutions
Post by: Marry 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?
Title: Re: Buffer Solutions
Post by: Borek on April 24, 2011, 05:02:44 PM
No, you need to use Henderson-Hasselbalch equation (http://www.chembuddy.com/?left=pH-calculation&right=pH-buffers-henderson-hasselbalch).
Title: Re: Buffer Solutions
Post by: Marry 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(-)]

?

Title: Re: Buffer Solutions
Post by: Borek on April 25, 2011, 08:00:11 AM
10,00 =  10.31 + ( log [CO3(-)] / log [HCO3(-)]

Now it should be a matter of simple concentration calculations. [HCO3-] is given (not directly, but you shouldn't haver any problems calculating it).
Title: Re: Buffer Solutions
Post by: Marry on April 26, 2011, 10:04:19 PM
10,00 =  10.31 + ( log [CO3(-)] / log [HCO3(-)]

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????

Title: Re: Buffer Solutions
Post by: AWK 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.
Title: Re: Buffer Solutions
Post by: Borek 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.