Question

How many grams of glyoxylic acid and sodium glyoxylate are needed to prepare 2.50 L of a 1.60 M buffer at pH 3.85? The pKa of glyoxylic acid is 3.34. Note: Use the monohydrate forms, HCOCo2H H20 and HCOCO2Na H20 Number 113.78 g HCOCo,H H2O Number Incorrect. You may not have included the mass of the water g HCOCO,Na H2 molecule in your calculation of the molar mass of sodium glyoxylate.

Answer 1

Answer:

86.71 g of glyoxylic acid monohydrate, and 348.56 g of sodium glyoxylate monohydrate.

Explanation:

In order to solve this problem we need to use the Henderson–Hasselbalch equation:

pH= pka + $log\frac{[A^{-} ]}{[HA]}$

In this case [A⁻] is the concentration of sodium glyoxylate, and [HA] is the concentration of glyoxylic acid.

Using the Henderson–Hasselbalch (H-H) equation, the given pH and pka we can determine a relationship between [A⁻] and [HA]:

3.85 = 3.34 +  $log\frac[{A^{-}] }{[HA]}$

0.51 = $log\frac{[A^{-} ]}{[HA]}$

$10^{0.51} =\frac{[A^{-}] }{[HA]} \\3.24=\frac{[A^{-} ]}{[HA]}$

Also from the problem, we know that

[A⁻] + [HA] = 1.60 M

We rearrange that equation to

[A⁻] = 1.60 M - [HA]

And replace the value of  [A⁻] in the H-H equation and solve for [HA]:

$3.24=\frac{1.60M-[HA]}{[HA]}\\3.24*[HA]=1.60-[HA]\\4.24*[HA]=1.60\\0.377 M = [HA]$

We substract 0.377 M to 1.60 M in order to calculate [A⁻].

1.60 M - 0.377 M= 1.223 M = [A⁻]

Lastly we calculate the mass of each reagent, using the concentration, volume and molecular weights:

• The formula of sodium glyoxylate monohydrate is C₂HO₃Na·H₂O, thus its molecular weight is 114 g/mol
• The formula of glyoxylic acid monohydrate is C₂H₂O₃·H₂O, thus its molecular weight is 92 g/mol

• Mass of glyoxylic acid:

2.50 L * 0.377 M * 92 g/mol = 86.71 g glyoxylic acid monohydrate.

• Mass of sodium glyoxylate:

2.50 L * 1.223 M * 114 g/mol = 348.56 g sodium glyoxylate monohydrate.

Why does the problem ask that we use the monohydrate forms? Because that's the available reagent in the laboratory.