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 HendersonHasselbalch equation (http://www.chembuddy.com/?left=pHcalculation&right=pHbuffershendersonhasselbalch).

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. [HCO_{3}^{}] 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. [HCO_{3}^{}] 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
0,58x
R 34,8g????

Log [CO3(2)]  Log [HCO3()] = 0,31
Something is missing on the right side
Note  mass of NaHCO_{3} is given and you have to calculate mass of Na_{2}CO_{3}
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.