Asked by Norman
Hi, I'm working out this question for homework, but i'm a little stuck.
How many g of sodium hydrogen carbonate (NaHCO3) and 4.836 g of sodium carbonate must be dissolved in 750 mL of H20 in order to make a buffer with a pH of 10.27??
the ka of sodium hydrogen carbonate is 5.6 x 10 ^11
here is what i did:
4.836 g NaCO3(1 mol/105.99 NaCO3) = 0.04563 mols NaCO3
0.04536 mols/.750 L = 0.0608 M Na2CO3
Am i on the right track? What do i do next? Thanks, I'm completely at a loss for what to do next.
How many g of sodium hydrogen carbonate (NaHCO3) and 4.836 g of sodium carbonate must be dissolved in 750 mL of H20 in order to make a buffer with a pH of 10.27??
the ka of sodium hydrogen carbonate is 5.6 x 10 ^11
here is what i did:
4.836 g NaCO3(1 mol/105.99 NaCO3) = 0.04563 mols NaCO3
0.04536 mols/.750 L = 0.0608 M Na2CO3
Am i on the right track? What do i do next? Thanks, I'm completely at a loss for what to do next.
Answers
Answered by
DrBob222
Use the Henderson-Hasselbalch equation.
pH = pKa + log[(base)/(acid)]
You have done the Na2CO3 correctly. That goes in for base. Plug in the other numbers and calculate M for the acid (NaHCO3). Knowing M and volume and molar mass you can calculate grams. By the way, you have Ka = 5.6 x 10^11. I think that's 5.6 x 10^-11 but check me out on that.
pH = pKa + log[(base)/(acid)]
You have done the Na2CO3 correctly. That goes in for base. Plug in the other numbers and calculate M for the acid (NaHCO3). Knowing M and volume and molar mass you can calculate grams. By the way, you have Ka = 5.6 x 10^11. I think that's 5.6 x 10^-11 but check me out on that.
Answered by
Norman
thank you, i understand now! that was so much clearer than my instructor's explanation.
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