Question

How many grams of glyoxylic acid and sodium glyoxylate are needed to prepare 2.25 L of a 2.50 M buffer at pH 4.15? The pKa of glyoxylic acid is 3.34. Note: Use the monohydrate forms, HCOCO2H·H2O and HCOCO2Na·H2O.

Note: Use the monohydrate forms HCOCO2H*H20 and HCOCO2Na*H2O

Answer 1

Henderson- Hasselbach equation for buffer is

pH = pKa + log (salt/acid)

4.15 = 3.34 + log (salt/acid)

log (salt/acid) = 0.81

salt/acid = 10^0.81

salt/acid = 6.45

so the ratio of salt to acid should be 6.45 molar ratio;

Total moles is 2.5 mole/L x 2.25 L = 5.625 moles

The ratio of the two is 1 : 6.45

so total moles is 7.45 should equat 5.625 moles

so glyoxylic acid will be 1/7.45 x 5.625 = 0.755 moles

0.755 x 92 g/mol = 69.26 g of glyoxylic acid in monohydrate form   (MW of HCOCO2H·H2O = 92g/mol)

sodium glyoxylate = 6.45/7.45 x 5.625 = 4.87 moles

4.87 moles x 114 g/mol = 555.17 g of sodium glyoxylate in monohydrate form.