Chemical Forums
Chemistry Forums for Students => Analytical Chemistry Forum => Topic started 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?
-
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?
-
No, you need to use Henderson-Hasselbalch equation (http://www.chembuddy.com/?left=pH-calculation&right=pH-buffers-henderson-hasselbalch).
-
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(-)]
?
-
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).
-
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????
-
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.
-
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.
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.